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![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a)
\((2a-5b)^2+(2a+5b)^2\)
\(=4a^2-2.2a.5b+25b^2+4a^2+2.2a.5b+25b^2\)
\(=8a^2+50b^2=2(4a^2+25b^2)\)
b)
\((a-2b-3c)^2-(a-2b+3c)^2\)
\(=[(a-2b-3c)-(a-2b+3c)][(a-2b-3c)+(a-2b+3c)]\)
\(=-6c(2a-4b)=12c(2b-a)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a : \(\left(2a-3b\right)^2-\left(2a+3b\right)^2\)
\(=\left(2a-3b+2a+3b\right)\left(2a-3b-2a-3b\right)\)
\(=4a.-6b=-24ab\)
Câu b : \(\left(a-2b-3c\right)^2-\left(a-2b+3c\right)^2\)
\(=\left(a-2b-3c+a-2b+3c\right)\left(a-2b-3c-a+2b-3c\right)\)
\(=\left(2a-4b\right).\left(-6c\right)\)
\(=2\left(a-2b-3c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(a-2b-3c-a+2b-3c\right)\left(a-2b-3c+a-2b+3c\right)\)
\(=-6c\cdot\left(2a-4b\right)\)
\(=-12c\left(a-2b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x2 + 2)2 - (2 + x)(x - 2)(x2 + 4) + 10
= (x2 + 2)2 - (4 - x2)(x2 + 4) + 10
= x4 + 4x2 + 4 - (16 - x4) + 10
= x4 + 4x2 + 4 - 16 + x4 + 10
= 2x4 + 4x2 - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Đ}K\text{X}\text{Đ}:x\ne\pm2\)
Ta có: \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\div\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2x+2-4}{\left(x+2\right)^2}\right):\left(\frac{2-x-2}{\left(x+2\right)\left(x-2\right)}\right)=\frac{2x-2}{\left(x+2\right)^2}\cdot\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{2\left(x-1\right)\left(x-2\right)}{-x\left(x+2\right)}\)
Lời giải:
Cách 1:
\((a-2b-3c)^2-(a-2b+3c)^2\)
\(=[(a-2b-3c)-(a-2b+3c)][(a-2b-3c)+(a-2b-3c)]\)
\(=-6c(2a-4b)=12c(2b-a)\)
Cách 2:
\((a-2b-3c)^2-(a-2b+3c)^2\)
\(=[(a-2b)-3c]^2-[(a-2b)+3c]^2\)
\(=[(a-2b)^2-6c(a-2b)+9c^2]-[(a-2b)^2+6c(a-2b)+9c^2]\)
\(=-12c(a-2b)=12c(2b-a)\)
Akai Haruma chị ơi làm sao để ra -6c(2a - 4b) vậy ạ???