Đặt \(x^2+4x+8=t\)
Khi đó PT có dạng:
\(t^2+3xt+2x^2=t^2-tx-2xt^2+2x^2=t\left(t-x\right)-2x\left(t-x\right)=\left(t-x\right)\left(t-2x\right)\)
\(=\left(x^2+4x+8-x\right)\left(x^2+4x+8-2x\right)=\left(x^2+3x+8\right)\left(x^2+2x+8\right)\)
Tìm x
a)\(3x\left(2x+1\right)=5\left(2x+1\right)\)
b)\(\left(3x-8\right)^2=\left(2x-7\right)^2\)
c)\(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
d)\(\left(9x^2-16\right)^2-4\left(3x+4\right)^2\)
e)\(\left(2x-1\right)\left(4x^2+2x+1\right)=x\left(x-8\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)
Phân tích thành nhân tử:
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
TÌM X
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)
a)\(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)
b)\(\left(x-y\right)^2+4\left(x-y\right)-12\)
c)\(x^2-2xy+y^2+3x-3y-10\)
d)\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
Phân tích đa thức thành nhân tử:
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
Phân tích đa thức thành nhân tử
a) \(\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+x^4\)
b) \(\left(x^2+4x+2\right)^2-3x\left(x^2+4x+2\right)+2x^2\)
c) \(4x^4-8x^3+3x^2-8x+4\)
d)\(2x^4-15x^3+35x^3-30x+8\)
\(1.\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2=0\)
\(2.\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)
\(3.4\left(x^2+x+1\right)^2+5x\left(x^2+x+1\right)+x^2=0\)
Giair phương trình hộ mik nhé đúng mik tick cho
\(4x\left(x^2-5\right)-\left(2x-3\right)\left(2x^2+x-8\right)=4x^2+20\)
