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Bài 1:
a: \(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}=x\)
=>\(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)
=>\(x-\dfrac{1}{2}>=0\)
=>\(x>=\dfrac{1}{2}\)
b: \(\left|1-3x\right|+1=3x\)
=>\(\left|1-3x\right|=3x-1\)
=>\(1-3x< =0\)
=>3x-1>=0
=>3x>=1
=>\(x>=\dfrac{1}{3}\)
Bài 2:
a: \(C=\left|5-x\right|+x=\left|x-5\right|+x\)
TH1: x>=5
\(C=x-5+x=2x-5\)
TH2: x<5
C=5-x+x=5
b: D=|2x-1|-x
TH1: x>=1/2
\(D=2x-1-x=x-1\)
TH2: \(x< \dfrac{1}{2}\)
D=1-2x-x=1-3x
Bài 2:
a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=6\)
hay x=3
b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)
\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)
\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)
a) \(\left(x+1\right)\left(x-1\right)\)
\(=x^2-1^2\)
\(=x^2-1\)
b) \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\)
\(=\left(x^2\right)^2-1^2\)
\(=x^4-1\)
c) \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)-x^8\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)-x^8\)
\(=\left(x^4-1\right)\left(x^4+1\right)-x^8\)
\(=\left(x^4\right)^2-1-x^8\)
\(=x^8-1-x^8\)
\(=-1\)
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
a) \(|x|-x\)
\(\Rightarrow\orbr{\begin{cases}x< 0\rightarrow\left|x\right|-x=2\left|x\right|\\x>0\rightarrow\left|x\right|-x=0\end{cases}}\)
\(\Rightarrow x=0\rightarrow x=0\)
\(A=\left|x\right|+x\)
Trường hợp 1: \(x\ge0\Rightarrow\left|x\right|=x\Rightarrow A=x+x=2x\)
Trường hợp 2: \(x< 0\Rightarrow\left|x\right|=-x\Rightarrow A=-x+x=0\)
\(B=\left|x\right|-x\)
Trường hợp 1: \(x\ge0\Rightarrow\left|x\right|=x\Rightarrow A=x-x=0\)
Trường hợp 2: \(x< 0\Rightarrow x=\left|-x\right|\Rightarrow A=-x-x=-2x\)
\(C=\left|x-1\right|+x\)
Nếu: \(\orbr{\begin{cases}\left|x-1\right|=x-1\Rightarrow x-1\ge0hayx\ge1\\\left|x-1\right|=-\left(x-1\right)=-x+1\Rightarrow x-1< 0hayx< 1\end{cases}}\)
Trường hợp 1: \(x\ge1\Rightarrow\left|x-1\right|+x=x-1+x=2x-1\)
Trường hợp 2: \(x< 1\Rightarrow\left|x-1\right|+x=-x+1+x=1\)