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Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
2x2 + 4x + 2 – 2y2 (có nhân tử chung là 2)
= 2.(x2 + 2x + 1 – y2) (Xuất hiện x2 + 2x + 1 là hằng đẳng thức)
= 2[(x2 + 2x + 1) – y2]
= 2[(x + 1)2 – y2] (Xuất hiện hằng đẳng thức (3))
= 2(x + 1 – y)(x + 1 + y)
\(x^2+3x-10\)
\(=x^2+5x-2x-10\)
\(=\left(x^2+5x\right)-\left(2x+10\right)\)
\(=x\left(x+5\right)-2\left(x+5\right)\)
\(=\left(x-2\right)\left(x+5\right)\)
Thích hđt thì chiều :))
x2 + 3x - 10
= ( x2 + 3x + 9/4 ) - 49/4
= ( x + 3/2 )2 - ( 7/2 )2
= ( x + 3/2 - 7/2 )( x + 3/2 + 7/2 )
= ( x - 2 )( x + 5 )
\(=-2\left(x^2-2xy+y^2-4\right)\)
\(=-2\left[\left(x-y\right)^2-4\right]\)
\(=-2\left(x-y-2\right)\left(x-y+2\right)\)
\(3x^2-7xy+2y^2\)
\(=3x^2-6xy-xy+2y^2\)
\(=\left(3x^2-6xy\right)-\left(xy-2y^2\right)\)
\(=3x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(3x-y\right)\)
4: \(\left(2x+3\right)^3-1\)
\(=\left(2x+3-1\right)\left(4x^2+12x+9+2x+3+1\right)\)
\(=\left(2x+2\right)\left(4x^2+14x+13\right)\)
\(=2\left(x+1\right)\left(4x^2+14x+13\right)\)
5: \(4x^2+20xy+25y^2=\left(2x+5y\right)^2\)
6: \(x^4-64xy^3\)
\(=x\left(x^3-64y^3\right)\)
\(=x\left(x-4y\right)\left(x^2+4xy+16y^2\right)\)
a) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
b) \(10ab+0,25a^2+100b^2\)
\(=\left(0,5a\right)^2+2\cdot0,5a\cdot10b+\left(10b\right)^2\)
\(=\left(0,5a+10b\right)^2\)
c) \(9x^2-xy+\frac{1}{36}y^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot\frac{1}{6}y+\left(\frac{1}{6}y\right)^2\)
\(=\left(3x-\frac{1}{6}y\right)^2\)
\(27x^3-a^3b^3\)
\(=\left(3x\right)^3-\left(ab\right)^3\)
\(=\left(3x-ab\right)\left[\left(3x\right)^2+3x\cdot ab+\left(ab\right)^2\right]\)
\(=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)