Question

c,|x² -3x|=5x

c,|x² +5x|=6x

c,|x² +2x|=-x

Answer 1

c: |x^2-3x|=5x

=>(x^2-3x)^2=(5x)^2 và x>=0

=>(x^2-3x-5x)(x^2-3x+5x)=0 và x>=0

=>x^2(x-8)(x+2)=0

=>x=0 và x=8

c: |x^2+5x|=6x

=>(x^2+5x)^2=(6x)^2 và x>=0

=>(x^2+5x-6x)(x^2+5x+6x)=0 và x>=0

=>x^2(x-1)(x+11)=0 và x>=0

=>x=0 hoặc x=1

c: |x^2+2x|=-x

=>(x^2+2x)^2=(-x)^2 và x<=0

=>(x^2+2x+x)(x^2+2x-x)=0 và x<=0

=>(x^2+x)(x^2+3x)=0 và x<=0

=>\(x\in\left\{0;-1;-3\right\}\)

Answer 2

\(\left|x^2-3x\right|=5x\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x=5x\\x^2-3x=-5x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-8x=0\\x^2+2x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-8\right)=0\\x\left(x+2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-2\end{matrix}\right.\)

 

\(\left|x^2+5x\right|=6x\\ \Leftrightarrow\left[{}\begin{matrix}x^2+5x=6x\\x^2+5x=-6x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-x=0\\x^2+11x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\x\left(x+11\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-11\end{matrix}\right.\)

 

\(\left|x^2+2x\right|=-x\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2x=-x\\x^2+2x=x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3x=0\\x^2+x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x+3\right)=0\\x\left(x+1\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-3\\x=-1\end{matrix}\right.\)

 

Answer 3

\(c1:\left|x^2-3x\right|=5x\)

Th1:x=0 \(=>\left|x^2-3x\right|=5x\)

\(=>\left|0^2-3.0\right|=5.0\)

\(=>0=0\)

\(=>x=0\) thỏa mãn

Th2:x>0 \(=>\left[{}\begin{matrix}x^2-3x=5x\\x^2-3x=-5x\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x^2=8x\\x^2=-2x\end{matrix}\right.\)             

\(=>\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

\(=>x\in\left\{-2;8\right\}\)

\(c2,\left|x^2+5x\right|=6x\)

\(=>\left[{}\begin{matrix}x^2+5x=6x\\x^2+5x=-6x\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x^2=x\\x^2=-11x\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x^2-x=0\\x=-11\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x\left(x-1\right)=0\\x=-11\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=0\\x-1=0\\x=-11\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=0\\x=1\\x=-11\end{matrix}\right.\)

\(=>x\in\left\{0;1;-11\right\}\)

 

\(c3,\left|x^2+2x\right|=-x\)

Loại vì GTTĐ luôn luôn ≥ 0