\(a,\Rightarrow\dfrac{1}{2}x=-2\Rightarrow x=-4\\ b,3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{24}{8}=3\\ \Rightarrow\left\{{}\begin{matrix}x=15\\y=9\end{matrix}\right.\)

\(PT\Leftrightarrow\sqrt{x^2+24}-\sqrt{x^2+8}=3x-1\)

Mà \(\sqrt{x^2+24}>\sqrt{x^2+8}\) nên \(3x-1>0\Leftrightarrow x>\frac{1}{3}\)

Ta có : \(PT\Leftrightarrow\left(\sqrt{x^2+24}-5\right)-\left(\sqrt{x^2+8}-3\right)-3x+3=0\)

\(\Leftrightarrow\frac{x^2+24-25}{\sqrt{x^2+24}+5}-\frac{x^2+8-9}{\sqrt{x^2+8}+3}-3\left(x-1\right)=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+24}+5}-\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+8}+3}-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{\left(x+1\right)}{\sqrt{x^2+24}+5}-\frac{\left(x+1\right)}{\sqrt{x^2+8}+3}-3\right]=0\)

Suy luận dựa \(ĐK\) ta được \(x=1\)

1) x-12=(-8)+(-17)
    x-12=(-25)
    x      = (-25)+12
    x      = (-13)

\(a,\Rightarrow12x-91=101\\ \Rightarrow12x=192\\ \Rightarrow x=16\\ b,\Rightarrow x:23+45=133\\ \Rightarrow x:23=88\\ \Rightarrow x=\dfrac{88}{23}\\ c,\Rightarrow\left(6x-39\right):7=3\\ \Rightarrow6x-39=21\\ \Rightarrow6x=60\\ \Rightarrow x=10\\ d,\Rightarrow3x-24=\dfrac{148}{73}\\ \Rightarrow3x=\dfrac{1900}{73}\\ \Rightarrow x=\dfrac{1900}{219}\\ e,\Rightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\\ f,\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ d,\left(9-x\right)^3=64=4^3\\ \Rightarrow9-x=4\\ \Rightarrow x=5\\ h,\Rightarrow x=27\\ i,\Rightarrow6x=312\cdot12=624\cdot6\\ \Rightarrow x=624\\ j,\Rightarrow\left(19x+104\right):14=25-42=-17\\ \Rightarrow19x+104=-238\\ \Rightarrow19x=-342\\ \Rightarrow x=-18\)

a:x=3/5-7/8=24/40-35/40=-11/40

b: =>1/3:(2x-1)=-1/6

=>2x-1=-2

=>2x=-1

=>x=-1/2

c: =>x-3/4=17/2+7/4=34/4+7/4=41/4

=>x=11

d: =>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-7:2/5=-35/2

a) \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=-\frac{6}{11}\)

b) \(\Leftrightarrow\left(\frac{43}{8}+x-\frac{173}{24}\right).-\frac{3}{50}=1\Leftrightarrow\frac{-11}{6}+x=-\frac{50}{3}\Leftrightarrow x=-\frac{37}{2}\)

1.

 \(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

2.

\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

3.

\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)

Vì (x-1)²≥0

23>0

\(\Rightarrow\left(x-1\right)^2+23>0\)

Vậy x vô nghiệm

4.

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

5.

\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)