\(\dfrac{\left(2x+1\right)}{2}-\dfrac{1}{3}=\dfrac{\left(x+1\right)}{6}\left(MC=12\right)\\ \Leftrightarrow\dfrac{\left(2x+1\right)2}{2.6}-\dfrac{1.4}{3.4}=\dfrac{\left(x+1\right)2}{6.2}\\ \Leftrightarrow\dfrac{4x+2}{12}-\dfrac{4}{12}=\dfrac{2x+2}{12}\\ \Leftrightarrow4x+2-4=2x+2\\ \Leftrightarrow4x-2x=2+2-4\\ \Leftrightarrow2x=0\\ \Leftrightarrow x=0\)

\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

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

a) Ta có: \(\sqrt{2x-1}\)

Biểu thức này có nghĩa là: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)

b) Ta có: \(\sqrt{4-x}\)

Biểu thức này có nghĩa là: \(4-x\ge0\Leftrightarrow x\le4\)

c) Ta có: \(\sqrt{\dfrac{3x}{2}}\)

Biểu thức này có nghĩa là: \(\dfrac{3x}{2}\ge0\Leftrightarrow x\ge0\)

d) Ta có: \(\sqrt{2x^2}\)

Biểu thức có nghĩa là: \(2x^2\ge0\Leftrightarrow x^2\ge0\) với mọi x

a, \(9x+3x\left(2x^2+x-3\right)=9x+6x^3+3x^2-9x\)

b, \(\left(3x-1\right)^2-9x\left(x+1\right)=9x^2-6x+1-9x^2-9x=1-15x\)

c, \(\left(x-1\right)^2-x\left(x+1\right)=x^2-2x+1-x^2-x=1-3x\)

1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)

2: \(C=A:B\)

\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)

\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)

=>C>=-1

Bài 2:

Diện tích khu vườn là:

\(\left(14+x\right)\left(18-x\right)\)

\(=252-14x+18x-x^2\)

\(=-x^2+4x+252\)

\(=-\left(x^2-4x+4-256\right)\)

\(=-\left(x-2\right)^2+256\le256\forall x\)

Dấu '=' xảy ra khi x=2

Chu vi hình chữ nhật là:

\(C=2\left[14+x+18-x\right]=2\cdot32=64\left(cm\right)\)

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)