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Ta có:
\(\left(a+b+c\right)^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2\)
\(=a^2+2ab+b^2+2ac+2bc+c^2\)
\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)\) \(\Rightarrowđpcm\)
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\(1.\)
\(4x^2-12x+9\)
\(=\left(2x\right)^2-12x+3^2=\left(2x-3\right)^2\)
\(2.\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(\left(7x-5\right)\left(x-y\right)\)
\(3.\)
\(x^3-9x\)
\(=x\left(x^2-9\right)\)
\(=x\left(x-3\right)\left(x+3\right)\)
\(4.\)
\(5x\left(x-y\right)-15\left(x-y\right)\)
\(=\left(5x-15\right)\left(x-y\right)\)
\(=5\left(x-3\right)\left(x-y\right)\)
\(5.\)
\(2x^2+x\)
\(=2x\left(x+1\right)\)
\(6.\)
\(x^3+27\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
\(7.\)
\(2x^2-4xy+2y^2-32\)
\(=2\left(x^2-2xy+y^2-16\right)\)
\(=2\left[\left(x^2-2xy+y^2\right)-16\right]\)
\(=2\left[\left(x-y\right)^2-4^2\right]\)
\(=2\left(x-y+4\right)\left(x-y-4\right)\)
\(8.\)
\(x^3-4x-3x^2+12\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(9.\)
\(2x+2y+x^2-y^2\)
\(=2\left(x+y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+2\right)\)
\(10.\)
\(x^2y-2xy+y\)
\(=y\left(x^2-2x+1\right)\)
\(=y\left(x-1\right)^2\)
\(11.\)
\(y^2+2y\)
\(=y\left(y+2\right)\)
\(12.\)
\(y^2-x^2-6y-6x\)
\(=\left(y-x\right)\left(y+x\right)-6\left(y+x\right)\)
\(=\left(y+x\right)\left(y-x-6\right)\)
\(13.\)
\(x^3-3x\)
\(=x\left(x^2-3\right)\)
\(=x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
\(14.\)
\(2x-xy+2z-yz\)
\(=x\left(2-y\right)+z\left(2-y\right)\)
\(=\left(2-y\right)\left(x+z\right)\)
Xong
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Câu 1:
\(\frac{1}{2}\)x2 . ( 2x + y) (2x - y)
= \(\frac{1}{2}\)x2 . (4x2 - y2)
= 2x4 - y2
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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!!!
(x + y)2 - (x - y)2
= [(x + y) - (x - y)].[(x + y) + (x - y)]
= 2x.2y
= 2xy
= 4xy nhé