Tìm x:
a)x+3/4=5/3
b)x-2/3=7/2
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6 tháng 11 2021
\(a,\Rightarrow x^2+4x+25-x^2=3\\ \Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\\ b,\Rightarrow\left(2x-3-4x-3\right)\left(2x-3+4x+3\right)=0\\ \Rightarrow6x\left(-2x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
18 tháng 1
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
G
11 tháng 5 2023
\(a,x-\dfrac{5}{7}=\dfrac{3}{5}\times\dfrac{1}{2}\)
\(x-\dfrac{5}{7}=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}+\dfrac{5}{7}\)
\(x=\dfrac{21}{70}+\dfrac{50}{70}\)
\(x=\dfrac{71}{70}\)
\(Vayx=\dfrac{71}{70}\)
\(b,x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{3}{4}\)
\(x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{6}{8}\)
\(x\times\dfrac{2}{3}=\dfrac{11}{8}\)
\(x=\dfrac{11}{8}:\dfrac{2}{3}\)
\(x=\dfrac{11}{8}\times\dfrac{3}{2}\)
\(x=\dfrac{33}{16}\)
\(Vayx=\dfrac{33}{16}\)
8 tháng 7 2023
a: x=2/3-4/5=10/15-12/15=-2/15
b: 1/2-x=7/12
=>x=1/2-7/12=-1/12
c: =>7/2:x=-7/2
=>x=-1
d: =>1/6x=3/8-5/2=3/8-20/8=-17/8
=>x=-17/8*6=-102/8=-51/4
e: =>1,5x=-1,5
=>x=-1
NM
2
28 tháng 8 2021
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
15 tháng 9 2021
\(a,\sqrt{x}=3\Leftrightarrow x=9\\ b,\sqrt{x}=6\Leftrightarrow x=36\\ c,\sqrt{x}=8\Leftrightarrow x=64\\ d,\sqrt{x}=12\Leftrightarrow x=144\\ e,2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\\ f,3\sqrt{x}=21\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\\ g,\sqrt{x}=8\Leftrightarrow x=64\\ h,2+\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)
HP
15 tháng 9 2021
a. \(\sqrt{x}=3\)
<=> \(\left(\sqrt{\sqrt{x}}\right)^2-\left(\sqrt{3}\right)^2=0\)
<=> \(\left(\sqrt{\sqrt{x}}-\sqrt{3}\right)\left(\sqrt{\sqrt{x}}+\sqrt{3}\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt{\sqrt{x}}-\sqrt{3}=0\\\sqrt{\sqrt{x}}+\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\left(Vnghiêm\right)\end{matrix}\right.\)
Vậy nghiệm của PT là S = \(\left\{9\right\}\)
a) x + \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)
x = \(\dfrac{5}{3}\) - \(\dfrac{3}{4}\)
x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)
x = \(\dfrac{11}{12}\)
b) x - \(\dfrac{2}{3}\) = \(\dfrac{7}{2}\)
x = \(\dfrac{7}{2}\) + \(\dfrac{2}{3}\)
x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\)
x = \(\dfrac{25}{6}\)
nhanh lên nhé