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1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)
b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)
\(-3x=\frac{47}{10}\)
\(x=\frac{-47}{30}\)
c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)
\(-7x=\frac{-1}{6}\)
\(x=\frac{1}{42}\)
d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(3x=\frac{1}{6}\)
\(x=\frac{1}{18}\)
Học tốt nhé bn!
a) Ta có: \(\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)
TH1: \(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
TH2: \(2x+\frac{3}{5}=\frac{-3}{5}\Rightarrow2x=\frac{-6}{5}\Rightarrow x=\frac{-3}{5}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
Mà \(\frac{-1}{27}=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\Leftrightarrow3x=\frac{1}{6}\Rightarrow x=\frac{1}{18}\)
c) \(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{2}{3}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{2}{3}x+\frac{5}{6}=0\)
\(\Rightarrow-\frac{37}{6}x=\frac{-1}{6}\Rightarrow x=\frac{1}{37}\)
d) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3-x-\frac{1}{5}=0\)
\(\Rightarrow-3x=\frac{47}{10}\Rightarrow x=\frac{-47}{30}\)
a. x:(1/2+2/3)=6/5
=>x:7/6=6/5
=>x=6/5*7/6=>x=7/5
b.(x-1/2)-5(x-2/3)=3/2x
=>x-1/2-5x+10/3=3/2x
=>-4x+17/6=3/2x
=>17/6x=3/2x--4x
=>17/6=x(3/2+4)=>17/6=11/2x=>x=17/33
c.-5(x+1/5)-1/2(x-2/3)=x
=>-5x-1-1/2x+1/3=x=>-11/2x-2/3=x
=>-2/3=x+11/2x=>-2/3=x(1+11/2)=>-2/3=13/2x
=>x=-4/39
a) x : (1/2 + 2/3) = 6/5
=> x : 7/6 = 6/5
=> x = 6/5 x 7/6
x = 7/5
b) , c) ko bít hihi