\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-1-2x-5=0\\x^2+3x-1+2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(x-2\right)=0\\\left(x+4\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
1,phân tích đa thức thành nhân tử
a,\(^{x^2-2x+2y-xy}\)
b,\(\left(x^2+1\right)^2-4x^2\)
2,tìm x,biết:
\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
giúp mk vs mai nộp rùi hu hu
A)\(^{ }\left(^{ }x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
B)\(^{ }\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
C)\(^{ }x^2-6x+11=0\)
D)(\(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
GIẢI PHƯƠNG TRÌNH SAU
A) \(\frac{X^2+2X+1}{X^2+2X+2}+\frac{X^2+2X+2}{X^2+2X+3}=\frac{7}{6}\)
B) \(\frac{\left(X^2-3X-4\right)^4}{\left(X-3\right)^5\left(X+2\right)^3}+\frac{\left(X^2+4X+3\right)^6}{\left(X-3\right)^3\left(X+2\right)^5}=0\)
rút gọn
a) \(\frac{1}{x-y}-\frac{3xy}{x^2-y^2}+\frac{x-y}{x^2+x+y^2}\)
b) \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+4x+4}+\frac{1}{x^2+5x+6}\)
c) \(\frac{4.\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\frac{x^2-25}{9x^2.\left(2x+5\right)^2}-\frac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right).\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\) thực hiện phép tính
Tìm x biết : \(\left(2x+5\right)^2+\left(4x+10\right)\left(3-x\right)+x^2-6x+9=0\)
giải jùm vs
Rút gọn Pthức:
A= \(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2+x+1}\right)\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
giải phương trình sau :
a) \(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
b) \(3x^4-13x^3+16x^2-13x+3=0\)
c) \(\left(x+3\right)^4+\left(x+5\right)^4=16\)
Tìm x biết
a) \(4x^2-12x=-9\)
b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
