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1) \(\frac{2}{3}+x=-\frac{4}{5}\)
\(x=\left(-\frac{4}{5}\right)-\frac{2}{3}\)
\(x=-1\frac{7}{15}\)
Vậy \(x=-1\frac{7}{15}\)
2) \(\frac{2}{5}-x=-\frac{1}{3}\)
\(x=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
\(x=\frac{11}{15}\)
Vậy \(x=\frac{11}{15}\)
3) \(1-\frac{x}{3}=1\frac{1}{2}\)
\(\frac{x}{3}=1-1\frac{1}{2}\)
\(\frac{x}{3}=-\frac{1}{2}\)
\(\Rightarrow x=\frac{\left(-1\right)\cdot3}{2}\)
\(x=-1\frac{1}{2}\)
4) \(1-\left(\frac{2x}{3}+2\right)=-1\)
\(\frac{2x}{3}+2=1-\left(-1\right)\)
\(\frac{2x}{3}+2=2\)
\(\frac{2x}{3}=2-2\)
\(\frac{2x}{3}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
1. \(\frac{25}{100}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{4}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\left(\frac{1}{4}+1-\frac{1}{5}\right)x=\frac{1}{5}\)
\(\Leftrightarrow\frac{21}{20}x=\frac{1}{5}\)
\(\Leftrightarrow x=\frac{1}{5}:\frac{21}{20}\)
\(\Leftrightarrow x=\frac{4}{21}\)
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\(2\frac{3}{5}x-\frac{1}{7}=1\frac{9}{35}\)
\(\frac{13}{5}x=\frac{44}{35}+\frac{1}{7}\)
\(\frac{13}{5}x=\frac{7}{5}\)
\(x=\frac{7}{5}:\frac{13}{5}\\ x=\frac{7}{13}\)
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
\(5,\left(x\cdot0,5-\frac{3}{7}\right):\frac{1}{2}=1\frac{1}{7}\)
\(\Leftrightarrow x\cdot0,5:\frac{1}{2}-\frac{3}{7}:\frac{1}{2}=1\frac{1}{7}\)
\(\Leftrightarrow x-\frac{6}{7}=\frac{8}{7}\)
\(\Leftrightarrow x=2\)
\(6,x\cdot1,75=1\frac{3}{10}+45\%\)
\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{13}{10}+\frac{9}{20}\)
\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{7}{4}\)
\(\Leftrightarrow x=1\)
\(7,\frac{5-x}{15}+\frac{5}{12}-\frac{1}{8}=\frac{3}{8}\)
\(\Leftrightarrow\frac{5-x}{15}=\frac{3}{8}-\frac{5}{12}+\frac{1}{8}\)
\(\Leftrightarrow\frac{5-x}{15}=\frac{1}{12}\)
\(\Leftrightarrow60-12x=15\)
\(\Leftrightarrow12x=45\)
\(\Leftrightarrow x=\frac{15}{4}\)
\(8,\left|x-\frac{25}{33}\right|-\frac{3}{11}=\frac{2}{3}\)
\(\Leftrightarrow\left|\frac{x-25}{33}\right|=\frac{31}{33}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{25}{33}=\frac{31}{33}\\x-\frac{25}{33}=-\frac{31}{33}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{56}{33}\\x=-\frac{2}{11}\end{cases}}\)
\(9,-\frac{9}{8}+\frac{-3}{8}\cdot x=-\frac{1}{8}\)
\(\Leftrightarrow\frac{-9}{8}+\frac{-3}{8}\cdot x+\frac{1}{8}=0\)
\(\Leftrightarrow-1-\frac{3}{8}x=0\)
\(\Leftrightarrow\frac{3}{8}x=-1\)
\(\Rightarrow x=-\frac{8}{3}\)
Mấy câu trên dễ , bạn có thể tự làm được
Chứng minh \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< 1\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1}-\frac{1}{10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{9}{10}\)
Lại có : \(\frac{9}{10}< 1\)
=> \(A< \frac{9}{10}< 1\)
=> \(A< 1\left(đpcm\right)\)
1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
=> x + 1 = 0
=> x = - 1
b) \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)
=> \(\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)
=> \(\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)
=> \(\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)
=> x + 2010 = 0
=> x = -2010
c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)
=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)
=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)
=> x = -1900
d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)
=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)
=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)
=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)
=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)
=> x = -2028
1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
+ TH1: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)
+ TH2: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)
Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)
mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-1\)
2) Ta có: \(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)
\(\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0\)
\(\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0\)
\(\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
+ TH1: \(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)
+ TH2: \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)
Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)
\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)
mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-2010\)
3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)
\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)
\(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
+ TH1: \(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)
+ TH2: \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)
Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)
\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)
mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-1900\)
4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)
\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)
\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)
\(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)
+ TH1: \(x-104=0\)\(\Leftrightarrow\)\(x=104\)
+ TH2: \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)
Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)
\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)
mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=104\)
5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)
\(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)
\(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)
\(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)
+ TH1: \(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)
+ TH2: \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)
Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)
mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)
\(\Rightarrow\)Phương trình trên vô nghiệm
Vậy \(x=-2028\)
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