làm mấy câu cũng được, làm hết càng tốt 1. 1.3(x+1)^2-(x-1)^2 2. x^2- 2x- 4y^2-4y 3. x^2+ 7x-8 4. x^3-2x^2+x-xy^2 5. (x+4)^2+ (3-x).(3+x)=7 6. Rút gọn

Đáp án:

$\begin{array}{l}
1)3.{\left( {x + 1} \right)^2} – {\left( {x – 1} \right)^2}\\
 = {\left( {\sqrt 3 x + \sqrt 3 } \right)^2} – {\left( {x – 1} \right)^2}\\
 = \left( {\sqrt 3 x + \sqrt 3  – x + 1} \right)\left( {\sqrt 3 x + \sqrt 3  + x – 1} \right)\\
 = \left[ {\left( {\sqrt 3  – 1} \right).x + \left( {\sqrt 3  + 1} \right)} \right].\left[ {\left( {\sqrt 3  + 1} \right).x + \left( {\sqrt 3  – 1} \right)} \right]\\
2){x^2} – 2x – 4{y^2} – 4y\\
 = {x^2} – 4{y^2} – 2x – 4y\\
 = \left( {x – 2y} \right)\left( {x + 2y} \right) – 2\left( {x + 2y} \right)\\
 = \left( {x + 2y} \right)\left( {x – 2y – 2} \right)\\
3){x^2} + 7x – 8\\
 = {x^2} + 8x – x – 8\\
 = \left( {x + 8} \right)\left( {x – 1} \right)\\
4){x^3} – 2{x^2} + x – x{y^2}\\
 = x\left( {{x^2} – 2x + 1 – {y^2}} \right)\\
 = x.\left( {x – 1 – y} \right)\left( {x – 1 + y} \right)\\
5){\left( {x + 4} \right)^2} + \left( {3 – x} \right)\left( {3 + x} \right) = 7\\
 \Rightarrow {x^2} + 8x + 16 + 9 – {x^2} = 7\\
 \Rightarrow 8x =  – 18\\
 \Rightarrow x = \dfrac{{ – 9}}{4}\\
Vay\,x = \dfrac{{ – 9}}{4}\\
6)\dfrac{{2a{x^2} – 4ax + 2a}}{{5b – 5b{x^2}}}\\
 = \dfrac{{2a\left( {{x^2} – 2x + 1} \right)}}{{ – 5b\left( {{x^2} – 1} \right)}}\\
 = \dfrac{{2a{{\left( {x – 1} \right)}^2}}}{{ – 5b\left( {x – 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2a\left( {x – 1} \right)}}{{ – 5b\left( {x + 1} \right)}}\\
 = \dfrac{{ – 2ax – 2a}}{{ – 5bx – 5b}}
\end{array}$