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\(\left(x-5\right)\left(x-1\right)=2x\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-5-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy............
\(5\left(x+3\right)\left(x-2\right)-3\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow5\left(x^2+x-6\right)-3\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow2x^2-16x-60=0\)
\(\Leftrightarrow x^2-8x-30=0\)
làm tiếp nhé!!!!!
a)(2x+1)(3x-2)=(5x-8)(2x+1)
⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0
⇔(2x+1)(3x-2-5x+8)=0
⇔(2x+1)(-2x+6)=0
⇔2x+1=0 hoặc -2x+6=0
1.2x+1=0⇔2x=-1⇔x=-1/2
2.-2x+6=0⇔-2x=-6⇔x=3
phương trình có 2 nghiệm x=-1/2 và x=3
TA CÓ:
\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)
\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)
\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)
\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
a, \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow x^4+2x^3+x^2+4x^2+4x+12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+x^2+2x+6x+12\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)=0\)
có : \(x^2+x+6>0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
b, \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]-297=0\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+7x-21\right)-297=0\)
đặt \(x^2+4x-13=t\)
\(\Leftrightarrow\left(t+8\right)\left(t-8\right)-297=0\)
\(\Leftrightarrow t^2-64-297=0\)
\(\Leftrightarrow t^2=361\)
\(\Leftrightarrow t=\pm19\)
có t rồi tìm x thôi
Ta xét các trường hợp :
TH1 : \(x< 1\Rightarrow x-1,x-2< 0\Rightarrow\left|x-1\right|=1-x;\left|x-2\right|=2-x\)
\(\Rightarrow1-x+2-x=5\Leftrightarrow2x=-2\Leftrightarrow x=-1\left(t.m\right)\)
TH2 : \(1\le x< 2\Rightarrow x-2< 0\le x-1\Rightarrow\left|x-1\right|=x-1;\left|x-2\right|=2-x\\ \Rightarrow x-1+2-x=5\Leftrightarrow1=5\left(VL\right)\)
TH3: \(x\ge2\Rightarrow x-1,x-2\ge0\Leftrightarrow\left|x-1\right|=x-1;\left|x-2\right|=x-2\)
\(\Rightarrow x-1+x-2=5\Leftrightarrow2x=8\Leftrightarrow x=4\left(t.m\right)\)