Login
Premium
Home Question Bank Math Algebra
Question
upstudy study bank question image url

Câu 9. Cho biểu thức : \[ \mathrm{P}=\left(\frac{x^{2}}{x^{3}-4 x}+\frac{6}{6-3 x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^{2}}{x+2}\right) \] a) Rút gọn p . b) Tính giá trị của biểu thức p khi \( / \mathrm{x} /=\frac{3}{4} \) c) Với giá trị nào của x thì \( \mathrm{p}=7 \)

Ask by Wyatt Gordon. in Vietnam
Jan 21,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

**Câu 9.** a) Biểu thức \( \mathrm{P} \) rút gọn là \( \frac{-1}{x - 2} \). b) Khi \( x = \frac{3}{4} \), \( \mathrm{P} = \frac{4}{5} \). c) Khi \( x = \frac{13}{7} \), \( \mathrm{P} = 7 \).

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Reviewed and approved by the UpStudy tutoring team

Mind Expander

Để giải bài toán này, chúng ta sẽ thực hiện từng phần một. ### a) Rút gọn P Đầu tiên, ta có thể rút gọn từng phần trong biểu thức \( P \): 1. Phần tử trên tử: \[ \frac{x^{2}}{x^{3}-4x} = \frac{x^{2}}{x(x^2-4)} = \frac{x^{2}}{x(x-2)(x+2)} = \frac{x}{(x-2)(x+2)} \] \[ \frac{6}{6-3x} = \frac{6}{3(2-x)} = -\frac{2}{2-x} = -\frac{2}{-(x-2)} = \frac{2}{x-2} \] \[ \frac{1}{x+2} \] Khi cộng các phân số này: \[ P_{tử} = \frac{x}{(x-2)(x+2)} + \frac{2}{x-2} + \frac{1}{x+2} \] Ta đưa chúng về cùng mẫu số: \[ P_{tử} = \frac{x + 2(x+2) + (x-2)}{(x-2)(x+2)} = \frac{4x + 2}{(x-2)(x+2)} \] 2. Phần tử dưới: \[ x - 2 + \frac{10-x^2}{x+2} = \frac{(x-2)(x+2) + (10-x^2)}{x+2} = \frac{x^2 - 4 + 10 - x^2}{x+2} = \frac{6}{x+2} \] Đưa vào biểu thức \( P \): \[ P = \frac{\frac{4x + 2}{(x-2)(x+2)}}{\frac{6}{x+2}} = \frac{4x + 2}{6(x-2)} = \frac{2(2x + 1)}{6(x-2)} = \frac{2x + 1}{3(x-2)} \] ### b) Tính giá trị của biểu thức P khi \( |x| = \frac{3}{4} \) Lấy giá trị \( x = \frac{3}{4} \): \[ P = \frac{2\left(\frac{3}{4}\right) + 1}{3\left(\frac{3}{4}-2\right)} = \frac{\frac{3}{2} + 1}{3\left(-\frac{5}{4}\right)} = \frac{\frac{5}{2}}{-\frac{15}{4}} = \frac{5 \cdot 4}{2 \cdot -15} = \frac{20}{-30} = -\frac{2}{3} \] ### c) Với giá trị nào của x thì \( P = 7 \) Giải phương trình: \[ \frac{2x + 1}{3(x-2)} = 7 \] Giải phương trình này: \[ 2x + 1 = 21(x - 2) \implies 2x + 1 = 21x - 42 \] \[ 1 + 42 = 21x - 2x \implies 43 = 19x \implies x = \frac{43}{19} \] Vậy giá trị của \( x \) cần tìm là \( x = \frac{43}{19} \).

Related Questions

1. Find the products. a) \( (x+2)(x+3) \) c) \( (a-2)(a+3) \) e) \( (2 y+5)(3 y-2) \) g) \( (2 a-3)(3 a-2) \) i) \( (x-3)(x-2) \) k) \( (2 x-3 y)(2 x+4 y) \) m) \( (7 x+3 y)(2 x-4 y) \)
Algebra South Africa Jan 22, 2025
7 Expand and simplify \( 5 x(3 x+4)(2 x-1) \)
Algebra United Arab Emirates Jan 22, 2025
\begin{tabular}{l} (a) Simplify \( \frac{2}{y^{0}} \) \\ (b) Simplify fully \( \left(16 a^{4}\right)^{\frac{3}{4}} \) \\ 5 \\ Factorise fully \( 18 c^{3} d^{2}-21 c^{2} \) \\ (Total for Question 4 is 3 marks) \\ \hline\end{tabular}
Algebra United Arab Emirates Jan 22, 2025
2. Find the products. \( \begin{array}{ll}\text { a) }(x+1)(x+3) & \text { b) }(a+3)(a+4) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { e) }(3 x y+2 z)(2 x y-3 z) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { k) }(2 a-b)(2 a+b) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \text {-n) }(x+3)^{2}\end{array} \)
Algebra South Africa Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { \b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { \l) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b) \\ \text { 2. Find the products. } & \text { b) }(a+3)(a+4) \\ \text { a) }(x+1)(x+3) & \text { (d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { (e) }(3 x y+2 z)(2 x y-3 z) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { k) }(2 a-b)(2 a+b) & \text { n) }(x+3)^{2} \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \end{array} \)
Algebra South Africa Jan 22, 2025

Latest Algebra Questions

2. Find the products. \( \begin{array}{ll}\text { a) }(x+1)(x+3) & \text { b) }(a+3)(a+4) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { e) }(3 x y+2 z)(2 x y-3 z) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { k) }(2 a-b)(2 a+b) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \text {-n) }(x+3)^{2}\end{array} \)
Algebra South Africa Jan 22, 2025
5. \( \begin{aligned} j-9 \pi= & 12 \pi \\ +9 \pi & +9 \pi\end{aligned} \)
Algebra United States Jan 22, 2025
7 Expand and simplify \( 5 x(3 x+4)(2 x-1) \)
Algebra United Arab Emirates Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { (b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { 1) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b)\end{array} \)
Algebra South Africa Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { \b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { \l) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b) \\ \text { 2. Find the products. } & \text { b) }(a+3)(a+4) \\ \text { a) }(x+1)(x+3) & \text { (d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { (e) }(3 x y+2 z)(2 x y-3 z) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { k) }(2 a-b)(2 a+b) & \text { n) }(x+3)^{2} \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \end{array} \)
Algebra South Africa Jan 22, 2025
Prev Next
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy