phân tích thành nhân tử
4x^3 - 4x^2 + x
x^2 - y^2 + 4 - 4x
x^2 -7x + 12
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\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(---\)
\(x^2-16y^2+4x+4\)
\(=\left(x^2+4x+4\right)-16y^2\)
\(=\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left(x+2-4y\right)\left(x+2+4y\right)\)
\(=\left(x-4y+2\right)\left(x+4y+2\right)\)
\(---\)
\(3x^2+6xy+3y^2-12\)
\(=3\left(x^2+2xy+y^2-4\right)\)
\(=3\left[\left(x+y\right)^2-2^2\right]\)
\(=3\left(x+y-2\right)\left(x+y+2\right)\)
\(---\)
\(4x^3+4x^2+x\)
\(=x\left(4x^2+4x+1\right)\)
\(=x\left(2x+1\right)^2\)
a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
Ta có : 5x(x - 2y) + 2(2y - x)2
= 5x(x - 2y) + 2(x - 2y)2 (vì (2y - x)2 = (x - 2y)2 )
= (x - 2y)[5x + 2(x - 2y)]
= (x - 2y)(5x + 2x - 4y)
= (x - 2y)(7x - 4y)
b) 7x(y - 4)2 - (4 - y)3
= 7x(y - 4)2 - (4 - y)2(4 - y)
= 7x(y - 4)2 - (y - 4)2(4 - y)
= (y - 4)2(7x - 4 + y)
c) (4x - 8)(x2 + 6) - (4x - 8)(x + 7) + 9(8 - 4x)
= (4x - 8)(x2 + 6) - (4x - 8)(x + 7) - 9(4x - 8)
= (4x - 8)(x2 + 6 - x - 7 - 9)
= 2(x - 4)(x2 - x - 10)
mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk
\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
câu thứ 2 bạn ktra lại đề
\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
\(x^3-x^2-4=\left(x-2\right)\left(x^2+x+2\right)\)
\(x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
a) \(x^5-7x^4-x^3+43x^2-36\)
\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-6\right)\)
c) \(x^4+2x^3-15x^2-18x+64\)
\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(b,x^3-3x^2-4x+12\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(c,3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
\(\text{d) 2x}^4- 7x^3 - 2x^2 + 13x + 6\)
\(\text{= (2x^4 + 2x^3) - (9x^3 + 9x^2) + (7x^2 + 7x) + (6x + 6)}\)
\(\text{= 2x^3(x + 1) - 9x^2(x + 1) + 7x(x + 1) + 6(x + 1)}\)
\(\text{= (x + 1)(2x^3 - 9x^2 + 7x + 6)}\)
\(\text{= (x + 1)(2x + 1)(x - 3)(x - 2)}\)
\(a,=x\left(x+5\right)\\ b,=\left(y-1\right)^2-4x^2=\left(y-1-2x\right)\left(y-1+2x\right)\\ c,\text{Không phân tích đc}\)
a) \(x^{2}+5x=x(x+5)\)
b)\(y^{2}-4x^{2}-2y+1 = (y-1)^{2}-4x^{2}=(y-1-2x)(y-1+2x)\)
c)\(x^{2}-8x+16=(x-4)^{2}\)
4x3 - 4x2 + x
= x.(4x2 - 4x + 1)
= x.[(2x)2 - 2.2.x + 12]
= x.(2x+1)2
x2 - y2 + 4 - 4x
= (x2 - 4x) - (y2 - 4)
= x.(x-4) - (y+4).(y-4)
x2 - 7x + 12
= x2 - 3x - 4x + 12
= x.(x-3) - 4.(x-3)
= (x-3).(x-4)
a) 4x3 - 4x2 + x
= x(4x2 - 4x + 1)
= x[(2x)2 - 2.2x.1 + 12)
= x(2x + 1)2
b) x2 - y2 + 4 - 4x
= (x2 - 4x + 4) - y2
= (x2 - 2.x.2 + 22) - y2
= (x - 2)2 - y2
= (x - 2 - y)(x - 2 + y)
c) x2 - 7x + 12
= x2 - 4x - 3x + 12
= (x2 - 4x) - (3x - 12)
= x(x - 4) - 3(x - 4)
= (x - 4)(x - 3).