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a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10
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`(1 1/3 - 25% - 5/12) + 2x=1,6 : 3/5`
`(4/3 - 1/4 - 5/12) +2x=8/5 : 3/5`
`2/3 + 2x = 8/3`
`2x=2`
`x=1`
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\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
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a) \(2x-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{1}{6}\)
\(2x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(2x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(2x=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}\div2\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{1}{3}+\dfrac{2}{3}\cdot\left(x-2\right)=3\)
\(\dfrac{2}{3}\cdot\left(x-2\right)=3-\dfrac{1}{3}\)
\(\dfrac{2}{3}\cdot\left(x-2\right)=\dfrac{8}{3}\)
\(x-2=\dfrac{8}{3}\div\dfrac{2}{3}\)
\(x-2=\dfrac{24}{6}\)
\(x-2=4\)
\(x=4+2\)
\(x=6\)
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\(a,\dfrac{x}{8}=\dfrac{7}{-2}\\ \Rightarrow x=-28\\ b,\dfrac{1-2x}{6}=\dfrac{-1}{2}\\ \Leftrightarrow2-4x=-6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\\ c,\dfrac{x+2}{3}=\dfrac{x+3}{4}\\ \Leftrightarrow4x+8=3x+9\\ \Leftrightarrow x=1\\ d,\dfrac{10}{2-x}=2\\ \Leftrightarrow4-2x=10\\ \Leftrightarrow2x=-6\\ \Leftrightarrow x=-3\)
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a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
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\(1+5^2+5^4+...+5^{2x}\left(1\right)=\dfrac{25^6-1}{24}\)
Đặt \(\left(1\right)=A\)
\(\Rightarrow A=1+5^2+...+5^{2x}\)
\(\Rightarrow5^2A=5^2+5^4+...+5^{2x+2}\)
\(\Rightarrow25A=5^2+5^4+...+5^{2x+2}\)
\(\Rightarrow25A-A=5^2+5^4+...+5^{2x+2}-1-5^2-...-5^{2x}\)
\(\Rightarrow24A=5^{2x+2}-1\)
\(\Rightarrow A=\dfrac{5^{2x+2}-1}{24}\)
Mà: \(A=\dfrac{25^6-1}{24}\)
\(\Rightarrow\dfrac{5^{2x+2}-1}{24}=\dfrac{\left(5^2\right)^6-1}{24}\)
\(\Rightarrow5^{2x+2}-1=5^{12}-1\)
\(\Rightarrow5^{2x+2}=5^{12}\)
\(\Rightarrow2x+2=12\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=\dfrac{10}{2}\)
\(\Rightarrow x=5\)
\(\dfrac{25}{30}=\dfrac{2x+3}{6}\)
\(\Rightarrow2x+3=\dfrac{25\cdot6}{30}=5\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
Ta có: \(\dfrac{25}{30}=\dfrac{2x+3}{6}\)
\(\Leftrightarrow\dfrac{2x+3}{6}=\dfrac{5}{6}\)
\(\Leftrightarrow2x+3=5\)
\(\Leftrightarrow2x=2\)
hay x=1
Vậy: x=1