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1) = \(\frac{3}{5}\)
2) =\(\frac{6}{7}\)
3)\(\frac{9}{13}\)
4)\(\frac{4}{13}\)
a. -3/4 x 12/-5 x (-25/6)=-15/2
b. -2 x -38/21 x -7/4 x (-3/8)=-19/8
c. (11/12: 33/16) x 3/5=4/15
d. 7/23 x [(-8/6)- 45/18]=-7/6
\(\left(x-\dfrac{4}{7}\right):-\dfrac{1}{3}=1\)
\(\Rightarrow x-\dfrac{4}{7}=1\cdot-\dfrac{1}{3}\)
\(\Rightarrow x-\dfrac{4}{7}=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{1}{3}+\dfrac{4}{7}\)
\(\Rightarrow x=\dfrac{5}{12}\)
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\(\dfrac{2}{3}-\dfrac{7}{4}:x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{7}{4}:x=\dfrac{2}{3}-\dfrac{5}{6}\)
\(\Rightarrow\dfrac{7}{4}:x=-\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{7}{4}:-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{21}{2}\)
(x - 4/7) : (-1/3) = 1
x - 4/7 = -1/3
x = -1/3 +4/7
x = 5/21
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2/3 + 7/4 : x = 5/6
7/4 : x = 5/6 - 2/3
7/4 : x = 1/6
x = 7/4 : 1/6
x = 21/2
1) \(\frac{17}{6}-\left(x-\frac{7}{6}\right)=\frac{7}{4}\)
\(\Rightarrow x-\frac{7}{6}=\frac{17}{6}-\frac{7}{4}\)
\(\Rightarrow x=\frac{13}{12}+\frac{7}{6}=\frac{9}{4}\)
2) \(\frac{3}{35}-\left(\frac{3}{5}-x\right)=\frac{2}{7}\)
\(\Rightarrow\)\(\frac{3}{5}-x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)
\(\Rightarrow x=\frac{3}{5}-\left(-\frac{1}{5}\right)=\frac{4}{5}\)
3) 4) Hjhj^_^^_^
a: =>x-2=6,3
=>x=8,3
d:=>|x-3|=14
=>x-3=14 hoặc x-3=-14
=>x=17 hoặc x=-11
`17/18+(x-7)/6=(x+12)/3`
`=>17+3(x-7)=6(x+12)`
`=>17+3x-21=6x+72`
`=>3x-4=6x+72`
`=>3x=-76`
`=>x=-76/3`
Vậy `x=-76/3`
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
c: Ta có: \(\left|x+\dfrac{5}{6}\right|:\dfrac{4}{5}=\dfrac{3}{8}\)
\(\Leftrightarrow\left|x+\dfrac{5}{6}\right|=\dfrac{3}{8}\cdot\dfrac{4}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{6}=\dfrac{3}{10}\\x+\dfrac{5}{6}=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-8}{15}\\x=-\dfrac{17}{15}\end{matrix}\right.\)
ĐKXĐ: \(x\ne3\)
\(\dfrac{12}{\left|x-3\right|}=\dfrac{6}{7}\)
\(\Leftrightarrow\left|x-3\right|=12:\dfrac{6}{7}\)
\(\Leftrightarrow\left|x-3\right|=14\)
\(\Leftrightarrow x-3=14\) hoặc \(x-3=-14\)
\(\Leftrightarrow x=17\) hoặc \(x=-11\)