Phân tích các đa thức sau thành nhân tử:
a) \(4{x^2} - 1\)
b) \({\left( {x + 2} \right)^2} - 9\)
c) \({\left( {a + b} \right)^2} - {\left( {a - 2b} \right)^2}\)
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`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
a: \(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2+3\left(x-y\right)-4\)
\(=\left(x-y+4\right)\left(x-y-1\right)\)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)
\(A=y^2+2y+1-25\)
\(A=\left(y+1\right)^2-5^2\)
\(A=\left(y+1-5\right)\left(y+1+5\right)\)
\(A=\left(y-4\right)\left(y+6\right)\)
\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)
\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)
\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)
Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)
b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)
c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)
\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)
\(=3b\left(2a-b\right)\)
`a, 4x^2-1 = (2x+1)(2x-1)`
`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`
`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`