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a) \(x\left(x-1\right)+\left(1-x\right)^2\)
\(=x\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)\left(x+x-1\right)\)
\(=\left(x-1\right)\left(2x-1\right)\)
b) \(\left(x+1\right)^2-3\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)-3\right]\)
\(=\left(x+1\right)\left(x+1-3\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
c) \(2x\left(x-2\right)-\left(x-2\right)^2\)
\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
a) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)
b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)
Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!
Trả lời:
A=(x-1)(x+2)(x-3)(x+4)-144
A= (x2-5x-14)(x2-5x-24)-144 (1)
đặt m=x2-5x-14
=> A= m.(m-10)-144
A=m2-10m-144
A= (m-18)(m+8)
thay m vào, ta có:
A= (x2-5x-32)(x2-5x-6)
A=(x2-5x-32)(x+1)(x-6)
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(t=x^2+5x+4\)
(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)
Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)
a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
a: \(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
b: \(x^2-1-xy+y\)
\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y+1\right)\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)
\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)
\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)
a) \(x^2-xy+x-y\)
\(=\left(x^2+x\right)-\left(xy+y\right)\)
\(=x\left(x+1\right)-y\left(x+1\right)\)
\(=\left(x+1\right)\left(x-y\right)\)
b) \(x^2+2xy-4x-8y\)
\(=x\left(x+2y\right)-4\left(x+2y\right)\)
\(\left(x-4\right)\left(x+2y\right)\)
c) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
a: Ta có: \(x^5-x^3+x^2-1\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x-1\right)\cdot\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
b: Ta có: \(5x^3-45x\)
\(=5x\left(x^2-9\right)\)
\(=5x\left(x-3\right)\left(x+3\right)\)
c: Ta có: \(16x^4y^2+2xy^5\)
\(=2xy^2\left(8x^3+y^3\right)\)
\(=2xy^2\cdot\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d: Ta có: \(a^3-8+6a^2-12a\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+8a+4\right)\)
e: Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
`b)x^3+y^3+z^3-3xyz`
`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`
`=(x+y)^3+z^3-3xy(x+y+z)`
`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`
`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`
`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`
\(a,=ab\left(a+3\right)\\ b,=\left(x-1\right)^2\\ c,=x\left[\left(x-3\right)^2-y^2\right]=x\left(x-y-3\right)\left(x+y-3\right)\)
\(a.3x^2-3y^2-2\left(x-y\right)^2\\ =3\left(x^2-y^2\right)-2\left(x-y\right)^2\\ =3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left[3\left(x+y\right)-2.\left(x-y\right)\right]=\left(x-y\right)\left(x+5y\right)\\ b.x^2-y^2-2x-2y\\ =\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-y-2\right)\\ c.\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\\ =\left(x-1\right)\left(2x+1\right)\left[1+3\left(x+2\right)\right]\\ =\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\\ d.\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\\ =\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\\ =\left(x-5\right)\left[\left(x-5\right)+\left(x+5\right)+\left(2x+1\right)\right]\\ =\left(x-5\right)\left(4x+1\right)\)
\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2-5x+6\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2-5x+6\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)\(A=\left(x-1\right)\left(x^2-5x+6+x-2\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2-4x+4\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x-2\right)^2-\left(x-1\right)\)
\(A=\left(x-1\right)\left[\left(x-2\right)^2-1\right]\)
\(A=\left(x-3\right)\left(x-1\right)^2\)
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https://olm.vn/hoi-dap/detail/9212510579.html
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