phân tích thành nhân tử
a)(x^2+2x)^2+2(x^2+2x)+1
b)16x^2-9(x+y)^2
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\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
a: \(x^8+x^4+1\)
\(=x^8+2x^4+1-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1+x^2\right)\left(x^4+1-x^2\right)\)
\(=\left(x^4+2x^2+1-x^2\right)\left(x^4-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\cdot\left[\left(x^2+1\right)^2-x^2\right]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
b: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2\)
\(=\left(x^2+1\right)\left(x^2+x+1\right)+2x\left(x^2+1+x\right)\)
\(=\left(x^2+x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+x+1\right)\left(x+1\right)^2\)
a) \(B=\left(x^2+2x+1\right)+\left(y^2-2.2.y+2^2\right)=\left(x+1\right)^2+\left(y-2\right)^2\)
thay x=99 và y=102 vào B ta có:
\(B=\left(99+1\right)^2+\left(102-2\right)^2=100^2-100^2=0\)
b)
b) \(2x^2+16x+32-2y^2=2\left(x^2+8x+16-y^2\right)=2\left(\left(x+4\right)^2-y^2\right)=2\left(x+4-y\right)\left(x+4+y\right)\)
\(a,=\left(4x^2\right)^2\left(x-y\right)-\left(x-y\right)\)
\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(4x^2-1\right)\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)\left(x-y\right)\)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a)\(=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
b)\(=\left(x+9\right)^2-\left(6x\right)^2=\left(x+9-6x\right)\left(x+9+6x\right)=\left(-5x+9\right)\left(7x+9\right)\)
c)\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\\ =\left(x-y+z-t\right)\left(x-y-z+t\right)\)
`@` `\text {Ans}`
`\downarrow`
`x^2 + 4x + 3`
`= x^2 + 3x + x + 3`
`= (x^2 + 3x) + (x + 3)`
`= x(x + 3) + (x + 3)`
`= (x+1)(x+3)`
____
`2x^2 + 3x - 5`
`= 2x^2 + 5x - 2x - 5`
`= (2x^2 - 2x) + (5x - 5)`
`= 2x(x - 1) + 5(x - 1)`
`= (2x + 5)(x - 1)`
____
`16x - 5x^2 - 3`
`= 15x + x - 5x^2 - 3`
`= (15x - 5x^2) + (x - 3)`
`= 5x(3 - x) + (x - 3)`
`= -5x(x - 3) + (x - 3)`
`= (1 - 5x)(x - 3)`
\(-5x^2+15x+x-3\) thì phải bằng \(-5x\left(x-3\right)+\left(x-3\right)\) chứ ạ
a)\(\left(x^2+2x\right)^2+2\left(x^2+2x\right)+1\)
\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
b) \(16x^2-9\left(x+y\right)^2\)
\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)
\(=\left(4x+3x+3y\right)\left(4x-3x-3y\right)\)
\(=\left(7x+3y\right)\left(x-3y\right)\)