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Ta có:
\(\left(a+b+c\right)^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2\)
\(=a^2+2ab+b^2+2ac+2bc+c^2\)
\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)\) \(\Rightarrowđpcm\)
\(1.\)
\(4x^2-12x+9\)
\(=\left(2x\right)^2-12x+3^2=\left(2x-3\right)^2\)
\(2.\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(\left(7x-5\right)\left(x-y\right)\)
\(3.\)
\(x^3-9x\)
\(=x\left(x^2-9\right)\)
\(=x\left(x-3\right)\left(x+3\right)\)
\(4.\)
\(5x\left(x-y\right)-15\left(x-y\right)\)
\(=\left(5x-15\right)\left(x-y\right)\)
\(=5\left(x-3\right)\left(x-y\right)\)
\(5.\)
\(2x^2+x\)
\(=2x\left(x+1\right)\)
\(6.\)
\(x^3+27\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
\(7.\)
\(2x^2-4xy+2y^2-32\)
\(=2\left(x^2-2xy+y^2-16\right)\)
\(=2\left[\left(x^2-2xy+y^2\right)-16\right]\)
\(=2\left[\left(x-y\right)^2-4^2\right]\)
\(=2\left(x-y+4\right)\left(x-y-4\right)\)
\(8.\)
\(x^3-4x-3x^2+12\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(9.\)
\(2x+2y+x^2-y^2\)
\(=2\left(x+y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+2\right)\)
\(10.\)
\(x^2y-2xy+y\)
\(=y\left(x^2-2x+1\right)\)
\(=y\left(x-1\right)^2\)
\(11.\)
\(y^2+2y\)
\(=y\left(y+2\right)\)
\(12.\)
\(y^2-x^2-6y-6x\)
\(=\left(y-x\right)\left(y+x\right)-6\left(y+x\right)\)
\(=\left(y+x\right)\left(y-x-6\right)\)
\(13.\)
\(x^3-3x\)
\(=x\left(x^2-3\right)\)
\(=x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
\(14.\)
\(2x-xy+2z-yz\)
\(=x\left(2-y\right)+z\left(2-y\right)\)
\(=\left(2-y\right)\left(x+z\right)\)
Xong
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)
Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó 
Có một số câu thì mình không làm được. Mong bạn thông cảm!!!
b)
Gửi câu trả lời c
a)x^2+y^2=5
<=> (x+y)^2- 2xy = 5
<=> 9-2xy=5
suy ra xy=2
mà x+y=3
Do đó x=1, y=2 hoặc x=2, y=1
Vậy x^3+y^3=1^3+2^3=2^3+1^3=9