Phân tích thành nhân tử:
a) \(B=\left(4x+1\right)\left(12x-1\right)\left(12x-1\right)\left(x+1\right)-4\)
b) \(C=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
PHân tích các đa thức sau thành nhân tử
a) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
b) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
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 \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
Phân tích các đa thức sau thành nhân tử:
a) \({x^3} + 4x\)
b) \(6ab - 9a{b^2}\)
c) \(2a\left( {x - 1} \right) + 3b\left( {1 - x} \right)\)
d) \({\left( {x - y} \right)^2} - x\left( {y - x} \right)\)
`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)`
Phân tích thành nhân tử:
a) \(B=\left(4x+1\right)\left(12x-1\right)\left(12x-1\right)\left(x+1\right)-4\)
b) \(C=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
c) \(D=\left(x^2+2x\right)^2+9x^2+18x+20\)
d) \(E=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
e) \(F=\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
vào câu hỏi tương tự mà lm tương tự như thế nha
Phân tích đa thức thành nhân tử:
a) \(A=\left(4x+1\right)\left(12x-1\right)\left(12x-1\right)\left(x+1\right)-4\)
b) \(B=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
c) \(C=\left(x^2+2x\right)^2+9x^2+18x+20\)
d) \(D=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
e) \(E=\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
Phân tích đâ thức thành nhân tử
a)\(x^2-8y^2+6x+9\)
b)\(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
a) \(x^2-8y^2+6x+9\)
\(=\left(x^2+6x+9\right)-8y^2\)
\(=\left(x+3\right)^2-\left(\sqrt{8}\cdot y\right)^2\)
\(=\left(x+3+\sqrt{8}y\right)\left(x+3-\sqrt{8}y\right)\)
Phân tích đa thức thành nhân tử = phương pháp đổi biến:
a) \(\left(x^2+x\right)-2\left(x^2+x\right)-15\)
b) \(x^2+2xy+y^2-x-y-12\)
c) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
e) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
Phân tích nhân tử:
\(\left(4x-1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)
Phân tích các đa thức sau thành nhân tử:
\(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(P\left(x\right)=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right].\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+8x+3x+2\right).\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right).\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x=t\), ta có:
\(\left(t+2\right)\left(t-1\right)-4\)
\(=t^2-t+2t-2-4=t^2+t-6\)
\(=t^2-2t+3t-6\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
Thay \(t=12x^2+11x\), ta được:
\(P\left(x\right)=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
Đs...
Phân tích đa thức sau thành nhân tử:
a) \(x^2-2xy+3x-3y+y^2-4\)
b) \(2\left(x^2-6x+1\right)^2+5\left(x^2-6x+1\right)\left(x^2+1\right)+2\left(x^2+1\right)^2\)
a: \(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2+3\left(x-y\right)-4\)
\(=\left(x-y+4\right)\left(x-y-1\right)\)