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Ta có: 2x - ( 1/7 - x ) = 0
=> 2x - 1/7 + x = 0
=> 3x - 1/7 = 0
=> 3x = 1/7
=> x = 1/7 : 3
=> x = 1/21
Nhưng bạn ơi trong sách giải ghi là x = 0 va x = 1/7
Tai ko có ghi cách giải nên mình ms hỏi mn
Theo đầu bài ta có:
\(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x=0\\x-\frac{1}{7}\end{cases}=0}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)
\(2x.\left(x-\frac{1}{7}\right)=0\Rightarrow2x=0\)hoặc \(x-\frac{1}{7}=0\)
\(\Rightarrow x=0\)hoặc \(x=\frac{1}{7}\)
a) \(\left|x\right|+2=9\Leftrightarrow\left|x\right|=7\Leftrightarrow x=\pm7\).
b) \(\left|x-4\right|=1\Leftrightarrow\orbr{\begin{cases}x-4=-1\\x-4=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\).
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) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{4}\\x-\frac{2}{3}=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2008}{2010}.\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
=> x + 1 = 2010
=> x = 2009
Ta có : \(\frac{2}{2\times3}+\frac{2}{3\times4}+....+\frac{2}{x\times\left(x+1\right)}=\frac{2008}{2010}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+.....+\frac{1}{x\times\left(x+1\right)}\right)=\frac{1004}{1005}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{1005}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{502}{1005}=\frac{1}{2010}\)
\(\Rightarrow x+1=2010\)
\(\Rightarrow x=2010-1=2009\)
\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)
\(\Leftrightarrow2x^2-4x-2x^2-6x-3x+3-29=0\)
\(\Leftrightarrow-13x-26=0\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=26:-13\)
\(\Leftrightarrow x=-2\)
Vậy ...
\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)
\(2x^2-4x-2x^2-6x-3x+3-29=0\)
\(2x^2-4x-2x^2-6x-3x=0+29-3\)
\(\left(2x^2-2x^2\right)+\left(-4x-6x-3x\right)=26\)
\(0+\left(-4-6-3\right)x=26\)
\(\Rightarrow-13x=26\rightarrow x=-2\)
\(a)\)\(\left(50-6.x\right).18=2^3.3^2.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=8.9.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=360\)
\(\Leftrightarrow\)\(\left(50-6.x\right)=360\div18\)
\(\Leftrightarrow\)\(50-6.x=20\)
\(\Leftrightarrow\)\(6.x=50-20\)
\(\Leftrightarrow\)\(6.x=30\)
\(\Leftrightarrow\)\(x=5\)
\(b)\)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7450\)
\(\Leftrightarrow\)\(100x+\left(1+2+3+...+100\right)=7450\)
\(\Leftrightarrow\)\(100x+5050=7450\)
\(\Leftrightarrow\)\(100x=7450-5050\)
\(\Leftrightarrow\)\(100x=2400\)
\(\Leftrightarrow\)\(x=24\)
b.
(x+1)+(x+2)+...+(x+100)=7450
=> 100x + (1+2+3+...+100)=7450
=>100x + (100+1).50=7450
=>100x=2400
=>x=24
vì (x-2/3)(x+1/4)=0 nên x-2/3=0 hoặc x+1/4=0
x-2/3=0
x=o+2/3=2/3
x+1/4=0
x=0-1/4=-1/4