1)Tìm x biết:
A)\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
B)\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
2)So sánh:
A)\(5^{222}và\)\(2^{555}\)
B)\(3^{48}và4^{36}\)
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1/a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{4}{5}\)
Vậy .........
2/ a/
Ta có :
\(5^{222}=\left(5^2\right)^{111}=25^{111}\)
\(2^{555}=\left(2^5\right)^{111}=32^{111}\)
Vì \(25^{111}< 32^{111}\Leftrightarrow5^{222}< 2^{555}\)
b/ Ta có :
\(3^{48}=\left(3^4\right)^{12}=81^{12}\)
\(4^{36}=\left(4^3\right)^{12}=64^{12}\)
Vì \(81^{12}>64^{12}\Leftrightarrow3^{48}>4^{36}\)
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
1/ a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...........
b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\left(\dfrac{1}{3}\right)^5\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{24}{30}\)
Vậy ....