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)

\((a-2b-3c)^2-(a-2b+3c)^2\)

\(=[(a-2b-3c)-(a-2b+3c)][(a-2b-3c)+(a-2b+3c)]\)

\(=-6c(2a-4b)=12c(2b-a)\)

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 : \(\left(a-2b-3c\right)^2-\left(a-2b+3c\right)^2\)

\(=\left(a-2b-3c+a-2b+3c\right)\left(a-2b-3c-a+2b-3c\right)\)

\(=\left(2a-4b\right).\left(-6c\right)\)

\(=2\left(a-2b-3c\right)\)

\(=\left(a-2b-3c-a+2b-3c\right)\left(a-2b-3c+a-2b+3c\right)\)

\(=-6c\cdot\left(2a-4b\right)\)

\(=-12c\left(a-2b\right)\)

(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

\(\text{Đ}K\text{X}\text{Đ}:x\ne\pm2\)

Ta có: \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\div\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(=\left(\frac{2x+2-4}{\left(x+2\right)^2}\right):\left(\frac{2-x-2}{\left(x+2\right)\left(x-2\right)}\right)=\frac{2x-2}{\left(x+2\right)^2}\cdot\frac{\left(x+2\right)\left(x-2\right)}{-x}\)

\(=\frac{2\left(x-1\right)\left(x-2\right)}{-x\left(x+2\right)}\)

\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)

\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)

\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)

\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)

101/12 bạn nha

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