Question
upstudy study bank question image url

\( \left\{ \begin{array} { l } { 1.1 ) = \frac { 101 } { 100 } + \frac { 100 } { 101 } \quad b = \frac { 101 } { 100 } - \frac { 100 } { 101 } } \\ { a ^ { 2 } - b ^ { 2 } + 1 = ? } \\ { 2 x + y + z = 0 \times y z = - 18 \quad x ^ { 3 } + y ^ { 3 } + z ^ { 3 } = ? } \end{array} \right. \)

Ask by Weber Ramirez. in Uzbekistan
Dec 24,2024

Real Tutor Solution

Tutor-Verified Answer

Answer

**1.2) Hisoblash:** \[ a = \frac{101}{100} + \frac{100}{101} \] \[ b = \frac{101}{100} - \frac{100}{101} \] \[ a^2 - b^2 + 1 = 5 \] **2. Hisoblash:** \[ 2x + y + z = 0 \] \[ y \times z = -18 \] \[ x^3 + y^3 + z^3 = -7x^3 - 108x \]

Solution

**1.2) Hisoblash:** Berilgan: \[ a = \frac{101}{100} + \frac{100}{101} \] \[ b = \frac{101}{100} - \frac{100}{101} \] Topshiriq: \[ a^2 - b^2 + 1 = ? \] **YeChim:** \[ a^2 - b^2 = (a + b)(a - b) \] Dastlab \( a + b \) va \( a - b \) ni hisoblaymiz: \[ a + b = \left( \frac{101}{100} + \frac{100}{101} \right) + \left( \frac{101}{100} - \frac{100}{101} \right) = 2 \times \frac{101}{100} = \frac{202}{100} = \frac{101}{50} \] \[ a - b = \left( \frac{101}{100} + \frac{100}{101} \right) - \left( \frac{101}{100} - \frac{100}{101} \right) = 2 \times \frac{100}{101} = \frac{200}{101} \] Endi \( a^2 - b^2 \) ni hisoblaymiz: \[ a^2 - b^2 = \left( \frac{101}{50} \right) \times \left( \frac{200}{101} \right) = \frac{101 \times 200}{50 \times 101} = \frac{200}{50} = 4 \] Shunday qilib: \[ a^2 - b^2 + 1 = 4 + 1 = 5 \] **Javob:** \( 5 \) --- **2. Hisoblash:** Berilgan: \[ 2x + y + z = 0 \] \[ y \times z = -18 \] Topshiriq: \[ x^3 + y^3 + z^3 = ? \] **YeChim:** Birinchi tenglamani \( x \) bo'yicha ifodalaymiz: \[ y + z = -2x \] Qabul qilaylik: \[ S = x + y + z = x + (-2x) = -x \] \[ P = y \times z = -18 \] Kub yig'indisi formulasi: \[ x^3 + y^3 + z^3 = (x + y + z)^3 - 3(x + y + z)(xy + yz + zx) + 3xyz \] Ammo qo'shimcha ma'lumot kerak. Alternativ yondashuv: Agar \( y + z = -2x \) va \( yz = -18 \), demak \( y \) va \( z \) kvadrat tenglamasining ildizlari: \[ t^2 + 2x t - 18 = 0 \] Ilgari berilgan formulani ishlatamiz: \[ y^3 + z^3 = (y + z)^3 - 3yz(y + z) = (-2x)^3 - 3(-18)(-2x) = -8x^3 - 108x \] Shunday qilib: \[ x^3 + y^3 + z^3 = x^3 + (-8x^3 - 108x) = -7x^3 - 108x \] Agar qo'shimcha ma'lumot bo'lmasa, javobni shu shaklda ifodalash mumkin. Ammo agar \( x + y + z = 0 \) bo'lsa, unda: \[ x^3 + y^3 + z^3 = 3xyz \] Bu holda: \[ 3x \times (-18) = -54x \] Ammo bu faqat \( x + y + z = 0 \) uchun to'g'ri. **Javob:** Qo'shimcha ma'lumot yo'qligi sababli, umumiy yechim \( x^3 + y^3 + z^3 = -7x^3 - 108x \)

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Bonus Knowledge

To solve the first equation, we simplify \( a \) and \( b \): First, calculate \( a = \frac{101}{100} + \frac{100}{101} \). This gives us a common denominator of 10100, resulting in \( a = \frac{101^2 + 100^2}{10100} = \frac{10201 + 10000}{10100} = \frac{20201}{10100} \). Next, \( b = \frac{101}{100} - \frac{100}{101} = \frac{101^2 - 100^2}{10100} = \frac{1}{10100} \). Now, we can find \( a^2 - b^2 + 1 = (a-b)(a+b) + 1 \). For the second part, from the equations \( 2x + y + z = 0 \) and \( yz = -18 \), you can express \( z = -2x - y \). Substituting \( z \) into \( yz = -18 \) leads to a polynomial, enabling you to find \( x^3 + y^3 + z^3 \) using the identity for the sum of cubes. Keep an eye out for potential negative roots or complex numbers as they can often yield unexpected solutions!

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