tìm x biết
a,3/35-(3/5+x)=2/7
b,5/6+1/6:x=3/4
c,|x-3 4/7|=-0.5
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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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\) ĐKXĐ: x \(\ne\) 0
<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)
<=> 15x + 28 = 3x
<=> 15x - 3x = -28
<=> 12x = -28
<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)
b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)
<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)
<=> -5x . 6 = 12 . 2
<=> -30x = 24
<=> x = \(-\dfrac{4}{5}\)
a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)
Vậy....
b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)
\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)
Vậy....
c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)
\(\dfrac{x}{182}=-\dfrac{5}{26}\)
\(=>x\cdot26=-5\cdot182\)
\(26x=-910\)
\(x=-910:26=-35\)
Vậy....
a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)
hay \(x=\dfrac{37}{70}\)
Vậy: \(x=\dfrac{37}{70}\)
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
a: =>x=3/7+3/5=15/35+21/35=36/35
b: =>x/35=4/5-5/7=28/35-25/35=3/35
=>x=3
c: =>x<3/4+8/4=11/4
=>\(x\in\left\{0;1;2;3\right\}\)
d: =>5/3<x<5/6+24/6=29/6
=>\(x\in\left\{2;3;4\right\}\)
e: =>x<10/12-9/12=1/12
=>x=0
f: =>2/3<x<12/6-5/6=7/6
=>x=1
a: 5-3x=6x+7
=>-3x-6x=7-5
=>-9x=2
=>\(x=-\dfrac{2}{9}\)
b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)
=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)
=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)
=>3x-2+3x+14=48
=>6x+12=48
=>6x=36
=>\(x=\dfrac{36}{6}=6\)
c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
=>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
a, \(x=-\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{-10-7}{12}=-\dfrac{17}{12}\)
b, \(\dfrac{2}{9}-x=-\dfrac{4}{3}.\dfrac{5}{6}=-\dfrac{24}{18}=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{2}{9}+\dfrac{4}{3}=\dfrac{14}{9}\)
c, \(-3=x-1\Leftrightarrow x=-2\)
d, \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{4}{5}:\dfrac{1}{5}=4\Leftrightarrow\dfrac{3}{5}x=4+\dfrac{2}{3}=\dfrac{14}{3}\Leftrightarrow x=\dfrac{14}{3}:\dfrac{3}{5}=\dfrac{70}{9}\)
\(a,\frac{3}{35}-\left[\frac{3}{5}+x\right]=\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2\cdot5}{35}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{10}{35}=\frac{-7}{35}\)
\(\Rightarrow x=\frac{-7}{35}-\frac{3}{5}=\frac{-7}{35}-\frac{3\cdot7}{35}=\frac{-7}{35}-\frac{21}{35}=\frac{-28}{35}=\frac{-4}{5}\)
\(b,\frac{5}{6}+\frac{1}{6}:x=\frac{3}{4}\)
\(\Rightarrow\frac{1}{6}:x=\frac{3}{4}-\frac{5}{6}\)
\(\Rightarrow\frac{1}{6}:x=\frac{18}{24}-\frac{20}{24}\)
\(\Rightarrow\frac{1}{6}:x=\frac{-2}{24}\)
\(\Rightarrow x=\frac{1}{6}:\frac{-2}{24}=\frac{1}{6}:\frac{-1}{12}=\frac{12}{-6}=\frac{-12}{6}=-2\)
\(c,\left|x-3\frac{4}{7}\right|=-0,5\)
Vì \(\left|x-3\frac{4}{7}\right|\ge0\)mà \(-0,5< 0\)nên \(x\in\varnothing\)
Vậy không có giá trị x thỏa mãn
a) 3/35-(3/5+x)=2/7
<=> 3/5+x=3/35-2/7
<=> 3/5+x=3/35-10/35
<=> 3/5+x=-7/35
<=> 3/5+x=-1/5
<=> x=-1/5-3/5
<=> x=-4/5
Vậy x=-4/5