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2x - 2y - x² + 2xy - y²
= (2x - 2y) - (x² - 2xy + y²)
= 2(x - y) - (x - y)²
= (x - y)(2 - x + y)
\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x-y\right)\left(x^2+8y^2\right)\)
\(=x^3-8y^3-\left(x^3-x^2y+8xy^2-8y^3\right)\)
\(=x^3-8y^3-x^3+x^2y-8xy^2+8y^3\)
\(=x^2y-8xy^2\)
\(=81-\left(2a-5b\right)^2=\left(9-2a+5b\right)\left(9+2a-5b\right)\)
\(a,4x^2+28x+49=\left(2x\right)^2+2.2x.7+7^2=\left(2x+7\right)^2\\ b,16y^2-8y+1=\left(4y\right)^2-2.4y.1+1^2=\left(4y-1\right)^2=\left(1-4y\right)^2\\ 4a^2+20ab+25b^2=\left(2a\right)^2+2.2a.5b+\left(5b\right)^2=\left(2a+5b\right)^2\\ d,9x^2-6xy+y^2=\left(3x\right)^2-2.3x.y+y^2=\left(3x-y\right)^2=\left(y-3x\right)^2\)
\(x^2-2xy+2y^2+2y+5=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)+4=\left(x-y\right)^2+\left(y+1\right)^2+4\)
Do \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\) ;\(\forall x;y\)
\(\Rightarrow\left(x-y\right)^2+\left(y+1\right)^2+4>0\) ; \(\forall x;y\)
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
\(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
hk tốt
^^
a) \(7x^2-63y^2\)
\(=7\cdot\left(x^2-9y^2\right)\)
\(=7\left[x^2-\left(3y\right)^2\right]\)
\(=7\left(x-3y\right)\left(x+3y\right)\)
b) \(36-4a^2+20ab-25b^2\)
\(=36-\left(4a^2-200ab+25b^2\right)\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
c) \(2x-2y-x^2+2xy-y^2\)
\(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
a: =7(x^2-9y^2)
=7(x-3y)(x+3y)
b: =36-(4a^2-20ab+25b^2)
=36-(2a-5b)^2
=(6-2a+5b)(6+2a-5b)
c: =2(x-y)-(x-y)^2
=(x-y)(2-x+y)