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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
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
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
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: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)
=> \(2x+7=4\)
=> 2x= -3
=> x=-3/2 . Vậy x=-3/2
e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)
=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)
= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
=> \(7^x=5^{2x}\)
Đến đoạn này là mik nghĩ không ra nhé
Cô làm tiếp giúp Linh Đan:
\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
a: =>-3/2+x-7=5-1/3x+4/15
=>4/3x=413/30
hay x=413/40
b: \(\Leftrightarrow5-\dfrac{3}{2}x=-\dfrac{22}{3}\cdot\dfrac{-11}{8}=\dfrac{121}{12}\)
=>3/2x=-61/12
hay x=-61/18
c: (3x+2)2+|3x+2y|=0
=>3x+2=0 và 3x=-2y
=>x=-2/3 và -2y=-2
=>(x,y)=(-2/3;1)
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)
\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)