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2 tháng 4 2020
1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)
\(x+\frac{4}{5}=\pm\frac{4}{5}\)
\(TH1:x+\frac{4}{5}=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)
\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)
Vậy x ∈ {0; \(\frac{-8}{5}\)}
Hai câu cuối khó nhìn nên ko giải
TT
1 tháng 10 2017
a. \(5.\left(x-2\right)+3.\left(x-2\right)=0\)
\(\Rightarrow8.\left(x-2\right)=0\)
\(\Rightarrow x-2=0:8\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy...
b. \(\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{2}{4}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{2}{4}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{-1}{6}\)
\(\Rightarrow x=\dfrac{5}{2}:\dfrac{-1}{6}=-15\)
Vậy...
c. \(2.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow x-\dfrac{1}{7}=0:2\)
\(\Rightarrow x-\dfrac{1}{7}=0\)
\(\Rightarrow x=\dfrac{1}{7}\)
Vậy...
d. \(\dfrac{11}{20}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}:\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}\)
Vậy...
e. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)
Vậy...
g. \(\dfrac{2}{3}x+\dfrac{5}{7}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{-29}{70}\)
\(\Rightarrow x=\dfrac{-29}{70}:\dfrac{2}{3}=\dfrac{-87}{140}\)
Vậy...
S
12 tháng 7 2018
a, \(\frac{x+1}{5}=\frac{3}{7}\Rightarrow7\left(x+1\right)=15\Rightarrow7x+7=15\Rightarrow7x=8\Rightarrow x=\frac{8}{7}\)
b, \(\frac{x-2}{3}=\frac{3}{8}\Rightarrow8\left(x-2\right)=9\Rightarrow8x-16=9\Rightarrow8x=25\Rightarrow x=\frac{25}{8}\)
c, \(\frac{-x-1}{2}=\frac{-3}{5}\Rightarrow5\left(-x-1\right)=-6\Rightarrow-5x-5=-6\Rightarrow-5x=-1\Rightarrow x=\frac{1}{5}\)
d, \(\frac{4}{5-x}=\frac{1}{3}\Rightarrow5-x=12\Rightarrow x=-7\)
e, \(2x\left(x-\frac{1}{7}\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\)
\(\left(\frac{-7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}=\frac{-7}{4}\cdot\frac{8}{5}\cdot\frac{11}{16}=\frac{-7.11}{4.5.2}=\frac{-77}{40}\)
17 tháng 6 2022
a: =>2x-1=4 hoặc 2x-1=-4
=>2x=5 hoặc 2x=-3
=>x=5/2 hoặc x=-3/2
d: =>x=|2|=2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\Rightarrow x=y=1\)
NH
6 tháng 9 2019
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
6 tháng 9 2019
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
22 tháng 6 2017
1, \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=5\)
\(\Leftrightarrow4x^2+12x+9-4x^2-1=5\)
\(\Leftrightarrow12x=-3\)
\(\Leftrightarrow x=\dfrac{-1}{4}\)
Vậy \(x=\dfrac{-1}{4}\)
2, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\)
\(\Leftrightarrow x^3+27-x^3-5x=20\)
\(\Leftrightarrow5x=7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy...
5, \(x^2-9+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
Vậy...
Q
22 tháng 6 2017
1) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=5\) (1)
\(\Leftrightarrow4x^2+12x+9-\left(4x^2-1\right)=5\)
\(\Leftrightarrow4x^2+12x+9-4x^2+1=5\)
\(\Leftrightarrow12x+10=5\)
\(\Leftrightarrow12x=5-10\)
\(\Leftrightarrow12x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{5}{12}\right\}\)
2) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\) (2)
\(\Leftrightarrow x^3+27-x^3-5x=20\)
\(\Leftrightarrow27-5x=20\)
\(\Leftrightarrow-5x=20-27\)
\(\Leftrightarrow-5x=-7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{7}{5}\right\}\)
3) \(\left(x+2\right)^3-x\left(x^2+6x\right)=15\) (3)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=15\)
\(\Leftrightarrow12x+8=15\)
\(\Leftrightarrow12x=15-8\)
\(\Leftrightarrow12x=7\)
\(\Leftrightarrow x=\dfrac{7}{12}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{7}{12}\right\}\)
4) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+10\right)\left(x-1\right)=7\) (4)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x\left(x+10\right)\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2-10x\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(-9x+1\right)=7\)
\(\Leftrightarrow-9x^2+x+9x-1=7\)
\(\Leftrightarrow-9x^2+10-1=7\)
\(\Leftrightarrow-9x^2+10x-1-7=0\)
\(\Leftrightarrow-9x^2+10x-8=0\)
\(\Leftrightarrow9x^2-10x+8=0\)
\(\Leftrightarrow x\notin R\)
5) \(x^2-9+5\left(x+3\right)=0\) (5)
\(\Leftrightarrow x^2-9+5x+15=0\)
\(\Leftrightarrow x^2+5x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+1}{2}\\x=\dfrac{-5-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình (5) là \(S=\left\{-3;-2\right\}\)
8 tháng 9 2019
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5