Các bạn giúp mình với a, -2/3x=4/15 b, x+1/3=2/5-(-1/3) c, -2/3x+-3/7+1/2x=-5/6 d, 2x.(x-1/7)=0
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29 tháng 6 2019
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
29 tháng 6 2019
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
17 tháng 5 2022
a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
12 tháng 6 2018
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
a, \(\frac{-2}{3}x=\frac{4}{15}\)
x = \(\frac{4}{15}:\frac{-2}{3}\)
x = \(\frac{-2}{5}\)
b, x + \(\frac{1}{3}\) = \(\frac{2}{5}-\left(\frac{-1}{3}\right)\)
x = \(\frac{11}{15}-\frac{1}{3}\)
x = \(\frac{2}{5}\)
c, \(\frac{-2}{3}\)x + \(\frac{-3}{7}\) + \(\frac{1}{2}\)x = \(\frac{-5}{6}\)
\(\frac{-1}{6}\)x + \(\frac{-3}{7}\) = \(\frac{-5}{6}\)
\(\frac{-1}{6}\)x = \(\frac{-5}{6}+\frac{3}{7}\)
\(\frac{-1}{6}\)x = \(\frac{-17}{42}\)
x = \(\frac{-17}{42}:\frac{-1}{6}\)
x = \(\frac{17}{7}\)
d, 2x(x - \(\frac{1}{7}\)) = 0
\(\Rightarrow\) 2x = 0 hoặc x - \(\frac{1}{7}\) = 0
TH1: 2x = 0
x = 0
TH2: x - \(\frac{1}{7}\) = 0
x = \(\frac{1}{7}\)
Vậy x = 0; x = \(\frac{1}{7}\)
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