tim x biet /3x-2/-x>1
HA
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tim x biet
|x+1|+|3x+2|=3x
|x-2|+4|x+1|=x-3
tim x biet x^3-3x^2+3x-1=0
x^3 - 3x^2 + 3x - 1 = 0
( x- 1)^3 = 0
=> x -1 = 0
=> x = 1
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tim x biet /x^2+/3x-1//=x^2+7
|x2+|3x-1||=x2+7
Vì |x2+|3x-1|| \(\ge\) 0;
x2+7 \(\ge\) 0
nên VT=VP
<=>x2+|3x-1|=x2+7
<=>|3x-1|=x2+7-x2=7
<=>\(\int^{3x-1=7\Rightarrow3x=8\Rightarrow x=\frac{8}{3}}_{3x-1=-7\Rightarrow3x=-6\Rightarrow x=-2}\)
Vậy x \(\in\) (-2;8/3}
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(tim x biet (x+1)(2-x)-(3x+5)(x+2)=-4x^2+1
tim x biet |2-3x|=x-1
|2-3x|=x-1
2-3x=x-1 hoặc 2-3x=x-1
TH1:2-3x=x-1
3x-x=2-(-1)
2x=2+1
2x=3
x=3/2
TH2:2-3x=-(x-1)
2-3x=-x+1
3x+x=2-1
4x=1
x=1/4
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Ta có : |2 - 3x| = x - 1
\(\Leftrightarrow\orbr{\begin{cases}2-3x=x-1\\2-3x=-x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x-x=-1-2\\-3x+x=1-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-4x=-3\\-2x=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\end{cases}}\)
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tim x biet x^2+3x+1=0
tim x biet (x^2+2x+1)(x+2)-x^2(x-3)-7x(x-1)=3x-9
tim x biet (x-2)^3 -(x-3)(x^2+3x+9)+6(x+1)^2
help: tim x biet: | x+1| +|2x-3|=|3x-2|