tim x
5/6x−3/4=−1/4+2/3
−1va1/2−2/3x=5/6−(−2/5)
(4/5:x+1,5):2/3=−1,5
4/3x−2/3=1/4−x
giup minh nhe minh dang can gap
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\(\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{5}{6}x=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{7}{5}\)
b) \(-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)\)
\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{41}{15}\)
\(\Rightarrow x=-\dfrac{41}{10}\)
c) \(\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5\)
\(\Leftrightarrow\dfrac{8+15x}{10x}.\dfrac{3}{2}=\dfrac{-3}{2}\)
\(\Leftrightarrow\dfrac{24+45x}{20x}=\dfrac{-3}{2}\)
\(\Leftrightarrow-60x=48+90x\)
\(\Rightarrow x=-0,32\)
d) \(\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x\)
\(\Leftrightarrow\dfrac{4x-2}{3}=\dfrac{1-4x}{4}\)
\(\Rightarrow16x-8=3-12x\)
\(\Rightarrow x=\dfrac{11}{28}\)
a: \(\dfrac{2.75}{x}=\dfrac{0.4}{1.5}=\dfrac{4}{15}\)
\(\Leftrightarrow x=\dfrac{11}{4}\cdot\dfrac{15}{4}=\dfrac{165}{16}\)
b: \(3\dfrac{1}{2}:\left(2x-3\right)=\dfrac{-3}{4}:0.2\)
\(\Leftrightarrow\dfrac{7}{2}:\left(2x-3\right)=\dfrac{-3}{4}:\dfrac{1}{5}=\dfrac{-15}{4}\)
\(\Leftrightarrow2x-3=\dfrac{7}{2}:\dfrac{-15}{4}=\dfrac{-7}{2}\cdot\dfrac{4}{15}=\dfrac{-28}{30}=\dfrac{-14}{15}\)
=>2x=-14/15+3=45/45-14/15=31/45
=>x=31/90
c: \(\dfrac{3x+2}{27}=\dfrac{3}{3x+2}\)
\(\Leftrightarrow\left(3x+2\right)^2=81\)
=>3x+2=9 hoặc 3x+2=-9
=>3x=7 hoặc 3x=-11
=>x=7/3 hoặc x=-11/3
d: \(\dfrac{5-x}{4}=\dfrac{2x+3}{2}\)
=>10-2x=8x+12
=>-10x=2
hay x=-1/5
a.\(\left(3x-2\right)^2=16\)
Ta có: \(\left(3x-2\right)^2=16\)
\(\Rightarrow\left(3x-2\right)^2=\left(4\right)^2\)
\(\Rightarrow3x-2=4\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)
\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)
\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)
\(\Rightarrow x=\dfrac{7}{16}\)
TA CÓ:
\(\frac{3}{x-5}=\frac{4}{x+2}=\frac{3-\left(-4\right)}{x-5-x-2}=\frac{7}{-7}=-1\)
=>\(\frac{3}{x-5}=-1\)=>3=-x+5 =>x=2
=>\(\frac{4}{x+2}=-1\)=>4=-x-2=>x=-6
Vì ko thể có 2 giá trị x trong 1 trường hợp nên ko tồn tại x thỏa mãn đề bài
a) \(\left(x+1\right)\left(x+3\right)-x\left(x+2\right)=7\)
\(\Leftrightarrow x^2+4x+3-x^2-2x=0\)
\(\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)
\(2x\left(3x+5\right)-x\left(6x-1\right)=33\)
\(\Leftrightarrow6x^2+10x-6x^2+x=33\)
\(\Leftrightarrow11x=33\Leftrightarrow x=3\)