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\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
1/ a) \(x^2-x-1⋮x-1\)
=>\(x.\left(x-1\right)-1⋮x-1\)
=>\(-1⋮x-1\)(vì x.(x-1)\(⋮\)x-1)
=>x-1\(\inƯ\left(-1\right)\)
Đến đay tự làm
b/c/d/e/ tương tự
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
a) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)
b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)
c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)
d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\\ \dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}\\ \dfrac{1}{2}x=2\\ x=2:\dfrac{1}{2}\\ x=4\)
a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)
c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)
d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)
e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)
\(\Leftrightarrow-8x+20-2=18x-45\)
\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x+1 | 1 | -1 | 2 | -2 |
| x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)
=> \(\frac{2}{3}:x=-7-\frac{1}{3}\)
=> \(\frac{2}{3}:x=-\frac{22}{3}\)
=> \(x=\frac{2}{3}:\left(-\frac{22}{3}\right)\)
=> \(x=-\frac{1}{11}\)
b) \(\frac{1}{3}x+\frac{2}{5}x=0\)
=> \(\frac{11}{15}x=0\)
=> \(x=0\)
c) \(\left(2x-3\right)\left(6-2x\right)=0\)
=> \(\left(2x-3\right)\left(3-x\right).2=0\)
=> \(\left(2x-3\right)\left(3-x\right)=0\)
=> \(\orbr{\begin{cases}2x-3=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)
\(\Rightarrow\frac{2}{3}.\frac{1}{x}=-7-\frac{1}{3}\)
\(\Rightarrow\frac{2}{3x}=\frac{-21-1}{3}\)
\(\Rightarrow\frac{2}{3x}=\frac{-22}{3}\)
\(\Rightarrow-22.3x=6\)
\(\Rightarrow3x=\frac{-6}{22}=\frac{-3}{11}\)
\(\Rightarrow x=\frac{-3}{11}:3=\frac{-3}{11}.\frac{1}{3}\)
\(\Rightarrow x=\frac{-1}{11}\)
b) \(\frac{1}{3}x+\frac{2}{5}x=0\)
\(\Rightarrow x.\left(\frac{1}{3}+\frac{2}{5}\right)=0\)
\(\Rightarrow x=0\)
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=3\\2x=6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
d) \(x:\frac{3}{4}+\frac{1}{4}=\frac{-2}{3}\)
\(\Rightarrow x.\frac{4}{3}=\frac{-2}{3}-\frac{1}{4}\)
\(\Rightarrow x.\frac{4}{3}=\frac{-11}{12}\)
\(\Rightarrow x=\frac{-11}{12}:\frac{4}{3}=\frac{-11}{12}.\frac{3}{4}=\frac{-11}{16}\)
e) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{4}\\x-\frac{2}{3}=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)
a:x=3/5-7/8=24/40-35/40=-11/40
b: =>1/3:(2x-1)=-1/6
=>2x-1=-2
=>2x=-1
=>x=-1/2
c: =>x-3/4=17/2+7/4=34/4+7/4=41/4
=>x=11
d: =>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-7:2/5=-35/2
b, \(\left(x+\frac{1}{2}\right)^2-\frac{1}{3}=\frac{1}{9}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{4}{9}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{3}\right)^3\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
c, \(\left(x-\frac{1}{2}\right)^3+\frac{1}{4}=\frac{3}{8}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{2}\\x-\frac{1}{2}=-\frac{1}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
d, \(2^{x+3}=32\)
\(\Leftrightarrow2^{x+3}=2^5\Leftrightarrow x+3=5\Rightarrow x=2\)
e, \(3^{x+2}+3^x=90\)
\(\Rightarrow\)\(3^x.\left(3^2+1\right)=90\)
\(\Rightarrow\)\(3^x.10=90\Rightarrow3^x=9\Rightarrow3^x=3^2\Rightarrow x=2\)