5. Sivetai \( \eta \) пapáotaoŋ \( : ~ A=\frac{x^{2}-4 x y+4 y^{2}}{15} \cdot \frac{9}{x^{2}-4 y^{2}} \) а) Na апобгіंदॄтв о̀тı: \( A=\frac{3 x-6 y}{5 x+10 y} \). i) \( \frac{x}{y} \) ii) \( \frac{3 x+2 y}{x-y} \)

Phần III. Tư luện ( 7,0 điềm). Bài 1 ( 1,5 điém). 1) Chùng minh đäng thúc: \( (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-5 \sqrt{\frac{1}{5}}=2 \) 2) Cho biều thưrc: \( A=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{4 \sqrt{x}}{4-x} \), vơi \( x \geq 0 ; x \neq 4 \). a) Rutt gpon biêu thưc \( A \). b) Tim tất cà cấc giá trị cùa \( x \), biết \( A<0 \).

complete the square \[ 2 x^{2}-10 x-20=8 \]

Chùng minh đàng thức: \( (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-5 \sqrt{\frac{1}{5}}=2 \)

13) If \( f(x)=\frac{x-1}{x-4} \) and \( f^{-1}(x) \) is the inverse of \( f \), what is the value of \( f^{-1}(4) ? \) \( \begin{array}{llll}\text { (A) } 5 & \text { (B) } 6 & \text { (C) } 8 & \text { (D) } 10 \\ \text { a } & \\ \text { b } & \end{array} \)

\( \begin{array}{ll}\text { 2. } & 2 \cdot \mathrm{C}_{0}+3 \cdot \mathrm{C}_{1}+4 \cdot \mathrm{C}_{2}+\ldots+(\mathrm{n}+2) \cdot \mathrm{C}_{\mathrm{n}}=(\mathrm{n}+4) \cdot 2^{\mathrm{n}-1} \\ \text { 3. } & 2 \cdot \mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+8 \cdot \mathrm{C}_{2}+\ldots(\mathrm{n}+1) \text { terms }=(3 \mathrm{n}+4) \cdot 2^{\mathrm{n}-1} \\ \text { 4. } & 3 \cdot \mathrm{C}_{1}+7 \cdot \mathrm{C}_{2}+11 \cdot \mathrm{C}_{3}+\ldots \mathrm{n} \text { terms }=1+(2 \mathrm{n}-1) 2^{\mathrm{n}} \\ \text { 5. } & 7 \mathrm{C}_{1}-12 \cdot \mathrm{C}_{2}+\ldots(-1)^{\mathrm{n}-1} \cdot(5 \mathrm{n}+2) \mathrm{C}_{\mathrm{n}}=2\end{array} \)