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d: =>-x-5/6=7/12-4/12=3/12=1/4
=>-x=1/4+5/6=13/12
hay x=-13/12
e: =>x+3=-5
hay x=-8
f: =>4,5-2x=-1/2
=>2x=5
hay x=5/2
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
a:=>0,75x-x+1,25x=0,2
=>x=0,2
b: =>1/3-x=-3/6+4/6=1/6
=>x=1/3-1/6=1/6
c: =>(x-1)/45=-6/30=-1/5
=>x-1=-9
=>x=-8
d: =>(2/5x-1)=-5/7
=>2/5x=2/7
=>x=5/7
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: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{32}=\dfrac{2}{x+1}\left(đk:x\ne1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=64\)
\(\Leftrightarrow\left(x+1\right)^2-64=0\)
\(\Leftrightarrow x^2+2x+1-64=0\)
\(\Leftrightarrow x^2+6x-63=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+16}{2}\\x=\dfrac{-2-16}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\left(đk:x\ne-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)
Vậy \(x_1=-9;x_2=7\)
b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\Leftrightarrow\dfrac{x+1}{5}=\dfrac{7}{x-1}\left(đk:x\ne1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=35\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)-35=0\)
\(\Leftrightarrow x^2-1-35=0\)
\(\Leftrightarrow x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\left(đk:x\ne1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Vậy \(x_1=-6;x_2=6\)
c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)
\(\Leftrightarrow\left|4,5-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|4,5-2x\right|\cdot\dfrac{4}{11}=\dfrac{11}{14}\)
\(\Leftrightarrow\dfrac{4}{11}\cdot\left|4,5-2x\right|=\dfrac{11}{14}\)
\(\Leftrightarrow\left|4,5-2x\right|=\dfrac{121}{56}\)
\(\Leftrightarrow\left[{}\begin{matrix}4,5-2x=\dfrac{121}{56}\\4,5-2x=-\dfrac{121}{56}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{131}{112};x_2=\dfrac{373}{112}\)
a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)=32.2\)
\(\Rightarrow\left(x+1\right)^2=64\)
\(\Rightarrow\left(x+1\right)^2=8^2\)
\(\Rightarrow x+1=8\)
\(\Rightarrow x=8-1\)
\(\Rightarrow x=7\left(TM\right)\)
Vậy \(x=7\) là giá trị cần tìm
b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\Rightarrow\left(x+1\right)\left(x-1\right)=7.5\)
\(\Rightarrow\left[{}\begin{matrix}x+1=7\\x-1=5\end{matrix}\right.\) \(\Rightarrow x=6\left(TM\right)\)
Vậy \(x=6\) là giá trị cần tìm
c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)
\(\left|\dfrac{45}{10}-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)
\(\left|\dfrac{9}{2}-2x\right|=\dfrac{11}{14}.\dfrac{11}{4}\)
\(\left|\dfrac{9}{2}-2x\right|=\dfrac{121}{56}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{9}{2}-2x=\dfrac{121}{56}\\\dfrac{9}{2}-2x=\dfrac{-121}{56}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{131}{56}\\2x=\dfrac{373}{56}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{131}{112};\dfrac{373}{112}\right\}\) là giá trị cần tìm