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\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
Chúc bạn học tốt và nhớ click cho mình với nhá!
= (5x-25) + (5x - x2)
= 5(x-5) + x(5-x)
= 5(x-5) - x(x-5)
= (5 - x)(x - 5)
\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)
\(b,x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
\(c,4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)
\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right).\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)
a) \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x+8y\right)\left(\frac{1}{5}x-8y\right)\)
b) \(x^3+\frac{1}{27}=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
c) \(-x^3+9x^2-27x+27\)
\(=27-x^3+9x^2-27x\)
\(=\left(3-x\right)\left(9+3x+x^2\right)+9x\left(x-3\right)\)
\(=\left(3-x\right)\left(9+3x+x^2\right)-9x\left(3-x\right)\)
\(=\left(3-x\right)\left(9+3x+x^2-9x\right)\)
\(=\left(3-x\right)\left(9-6x+x^2\right)=\left(3-x\right)\left(9-3x-3x+x^2\right)\)
\(=\left(3-x\right)\left[3\left(3-x\right)-x\left(3-x\right)\right]=\left(3-x\right)\left(3-x\right)\left(3-x\right)=\left(3-x\right)^3\)
(Nhớ k cho mình với nha!, Mình chắc chắn là mình làm đứng luôn đó! Chúc may mắn nhá!)
a/ Ta có: \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
b/ \(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
c/ Đề sai
= (3x + 1 - x - 1)(3x + 1 + x + 1)
= 2x(4x + 2)
Em áp dụng hđt số 3 trong sgk nhé.
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)
a) \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)
\(=\left[3\left(a+b\right)+2\left(a-2b\right)\right]\left[3\left(a+b\right)-2\left(a-2b\right)\right]\)
\(=\left(3a+3b+2a-4b\right)\left(3a+3b-2a+4b\right)\)
\(=\left(5a-b\right)\left(a+7b\right)\)
b) \(\left(2a-b\right)^2-4\left(a-b\right)^2\)
\(=\left[\left(2a-b\right)-2\left(a-b\right)\right]\left[\left(2a-b\right)+2\left(a-b\right)\right]\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
c) \(125-\left(x+2\right)^3\)
\(=\left(5-x-2\right)\left[25+5\left(x+2\right)+\left(x+2\right)^2\right]\)
\(=\left(3-x\right)\left(25+5x+10+x^2+4x+4\right)\)
\(=\left(3-x\right)\left(x^2+9x+39\right)\)
d) \(\left(x+3\right)^3-8=\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+4\right]\)
\(=\left(x+1\right)\left(x^2+8x+19\right)\)
e) \(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2=\left(x^6-y^2\right)\left(x^6+y^2\right)\) 9 khai triển tiếp hđt 6,7)
a, \(\left(4x+5\right)^2=\left(4x+5\right)\left(4x+5\right)=\left[\left(4x+5\right)4x\right]+\left[\left(4x+5\right)5\right]=4x^2+20x+25\)
b, \(\left(5x-2\right)^2=\left(5x-2\right)\left(5x-2\right)=\left[\left(5x-2\right)5x-\left(5x-2\right)2\right]=5x^2-10x+25\)
b, \(8^2-12x^2=\left(8^2-12x^2\right)\left(8^2+12x^2\right)\)
đúng ko :)
@No name: Bị sai rồi nhé, a,b,c sai hết :>
a) ( 4x + 5 )2
= ( 4x )2 + 2.4x.5 + 52
= 16x2 + 40x + 25
b) ( 5x - 2 )2
= ( 5x )2 - 2.5x.2 + 22
= 25x2 - 20x + 4
c) 82 - 12x2
= 64 - 12x2
= ( V8 - V12x )( V8 + V12x )
a/ x2 + 6x + 9 = (x + 3)2 = (x + 3)(x + 3)
b/ 10x - 25 - x2 = -x2 + 10x - 25 = -(x2 -10x + 25) = -(x - 5)2 = -(x - 5)(x - 5)
c/ \(8x^3+\frac{1}{8}=\left(2x\right)^3+\left(\frac{1}{2}\right)^3=\left(2x+\frac{1}{2}\right)\left(4x^2-x+\frac{1}{4}\right)\)