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a: =>|2x+3|=2+2x-5=2x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(2x-3-2x-3\right)\left(2x-3+2x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: Trường hợp 1: x<-3
Pt sẽ là -x-3+1-2x=10
=>-3x-2=10
=>-3x=8
hay x=-8/3(loại)
Trường hợp 2: -3<=x<1/2
Pt sẽ la x+3+1-2x=10
=>4-x=10
hay x=-6(loại)
Trường hợp 3: x>=1/2
Pt sẽ là x+3+2x-1=10
=>3x+2=10
hay x=8/3(nhận)
II3x-3I+2x+1I=3x+2021^0
II3x-3I+2x+1I=3x+1
\(\)ĐK:3x+1\(\ge\)0
3x\(\ge\)-1
x\(\ge\frac{-1}{3}\)
\(\Rightarrow\)I3x-3I+2x+1=3x+1
I3x-3I=x
\(\Rightarrow\)3x-3=\(\pm\)x
TH1:3x-3=x TH2:3x-3=-x
2x=3 4x=3
x=\(\frac{3}{2}\) x=\(\frac{3}{4}\)
Vậy x=\(\frac{3}{2}\); x=\(\frac{3}{4}\)
x+|x|=2x
\(\Rightarrow\)|x|=2x-x=x
\(\Rightarrow\)x\(\ge\)0 thỏa mãn đề bài
a,\(2x^2-8x=0\)
\(2x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)
\(=2x^2-8x-3x-2x^2\)
=\(-11x\)
c,\(-11x=0\)
\(\Rightarrow x=0\)
\(A\left(x\right)=2x^2-8x\)
\(\Rightarrow2x^2-8x=0\)
\(\Rightarrow x\left(2x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)
\(B\left(x\right)=-3x+2x^2\)
\(B\left(x\right)=2x^2-3x\)
\(2x^2-3x=0\)
\(\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
\(\frac{2x-3}{2}=\frac{32}{2x-3}\)
\(\Rightarrow\left(2x-3\right).\left(2x-3\right)=2.32\)
\(\left(2x-3\right)^2=64=8^2=\left(-8\right)^2\)
=> 2x-3 = 8 => 2x = 11 => x = 11/2
2x -3 = -8 => 2x = -5 => x = -5/2
KL:...
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
a) \(\frac{3}{4}-\frac{2}{5}.x=x\)
\(\Rightarrow\frac{-2}{5}.x-x=\frac{-3}{4}\)
\(x.\left(\frac{-2}{5}-1\right)=\frac{-3}{4}\)
\(x.\frac{-7}{5}=\frac{-3}{4}\)
\(x=\frac{-3}{4}:\left(\frac{-7}{5}\right)\)
\(x=\frac{15}{28}\)
b) (2x-1).(3x-1/5).(4-2x) = 0
=> 2x - 1 = 0 => 2x = 1 => x = 1/2
3x-1/5 = 0 => 3x = 1/5 => x = 1/15
4-2x = 0 => 2x = 4 => x = 2
KL: x = 1/2 hoặc x = 1/15 hoặc x = 2
=>|5x-3|=2x+14
Vì 5x-3>=0 với mọi x
=>2x+14>=0
=>2x>=-14
=>x>=-7
Th1:5x-3=2x+14
=>5x-2x=3+14
=>3x=17
=>x=17/3 (thỏa mãn điều kiện x>=-7)
Th2:3-5x=2x+14
=>3-14=2x+5x
=>-11=7x
=>x=-11/7 (thỏa mãn điều kiện x>=-7)