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1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
a) \(\frac{-2}{3}x+\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{1}{10}-\frac{1}{5}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{-2}{3}\)
\(\Leftrightarrow x=\frac{3}{20}\)
1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x-5-5x+10-4x-4=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)
\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)
\(\Leftrightarrow-3x=-9\)
hay x=3
3) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
4) Ta có: \(x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) Ta có: \(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
6) Ta có: \(\left(5x-1\right)^3=125\)
\(\Leftrightarrow5x-1=5\)
\(\Leftrightarrow5x=6\)
hay \(x=\dfrac{6}{5}\)
7) Ta có: \(3^{x+1}=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
a; - \(\dfrac{1}{3}\).(15\(x-9\)) + \(\dfrac{2}{7}\).(- \(x-34\)) = 1 - \(\dfrac{3}{4}\).(-16\(x+4\))
- 5\(x\) + 3 - \(\dfrac{2}{7}\)\(x\) - \(\dfrac{68}{7}\) = 1 + 12\(x\) - 3
12\(x\) + 5\(x\) + \(\dfrac{2}{7}x\) = 3 - \(\dfrac{68}{7}\) - 1 + 3
17\(x\) + \(\dfrac{2}{7}x\) = (3 - 1 + 3) - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\)\(x\) = 5 - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\) \(x\) = - \(\dfrac{33}{7}\)
\(x\) = - \(\dfrac{33}{7}\): \(\dfrac{121}{7}\)
\(x\) = - \(\dfrac{3}{11}\)
Vậy \(x\) = - \(\dfrac{3}{11}\)
\(2^x:4=32\\ \Rightarrow2^x=128\\ \Rightarrow2^x=2^7\\ \Rightarrow x=7\)
\(3^{x-2}:3=243\\ \Rightarrow3^{x-2}=729\\ \Rightarrow3^{x-2}=3^6\\ \Rightarrow x-2=6\\ \Rightarrow x=8\)
\(256:4^{x+1}=4^2\\ \Rightarrow4^{x+1}=4^2\\ \Rightarrow x+1=2\\ \Rightarrow x=1\)
\(4^{2x-1}:4=4^4\\ \Rightarrow4^{2x-1}=4^5\\ \Rightarrow2x-1=5\\ \Rightarrow x=3\)
\(5^{x-1}:5=5^3\\ \Rightarrow5^{x-1}=5^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
\(3^{2x+1}:3=3^4\\ \Rightarrow3^{2x+1}=3^5\\ \Rightarrow2x+1=5\\ \Rightarrow x=3\)
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
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\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
