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Ta có: \(2x+\frac76+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)
=>\(2x+1+\frac16+1+\frac{1}{12}+\ldots+1+\frac{1}{90}=10\)
=>\(2x+8+\left(\frac16+\frac{1}{12}+\cdots+\frac{1}{90}\right)=10\)
=>\(2x+8+\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac19-\frac{1}{10}\right)=10\)
=>\(2x+8+\left(\frac12-\frac{1}{10}\right)=10\)
=>\(2x+8+\frac{4}{10}=10\)
=>\(2x=10-8-\frac{4}{10}=2-\frac{4}{10}=2-\frac25=\frac85\)
=>\(x=\frac85:2=\frac45\)
a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
=>2007-x=0
hay x=2007
b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)
\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)
=>x+7+1/3-1/10=0
hay x=-217/30
a, <
b, <
c,<
d, =
e,>
g, <
Bn bấm mấy tính r so các số ở cuối, giả sử = 5...,... x\(10^4\) còn số còn lại là 5...,...x\(10^5\) thì số kia to hơn do x\(10^5\) > x\(10^4\) ( mẹo thôi chứ m ko bt cách làm dạng bài này)
a: \(5^{28}=\left(5^2\right)^{14}=25^{14}\)
mà 25<26
nên \(5^{28}=25^{14}<26^{14}\)
c: \(31^{11}<32^{11}=\left(2^5\right)^{11}=2^{55}\)
\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}\)
mà \(2^{56}>2^{55}\)
nên \(17^{14}>31^{11}\)
d: \(64^7=\left(4^3\right)^7=4^{3\cdot7}=4^{21}\)
e: \(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
mà 8192>3125
nên \(2^{91}>5^{35}\)
g: \(21^{12}=\left(21^3\right)^4=9261^4>54^4\)
\(\Rightarrow\left(\frac{\frac{31}{36}}{\frac{11}{9}:4-\frac{3}{4}}:31\right).x=\frac{-1}{16}\Rightarrow\left(\frac{\frac{31}{36}}{\frac{-4}{9}}:31\right)x=-\frac{1}{16}\Rightarrow\left(-\frac{31}{16}:31\right)x=-\frac{1}{16}\Rightarrow-\frac{1}{16}x=-\frac{1}{16}\Rightarrow x=1\)
1. \(\Leftrightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{51-x}{49}+1=-5+5\)
\(\Leftrightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
2. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)
a: =-45-91+17-45+91
=-90+17=-73
b: =-57-84-31+57+84-31
=-31-31=-62
a) \(-\left(45+91\right)-\left(-17+45-91\right)\)
\(=-45-91+17-45+91\)
\(=17+-\left(45+45\right)-\left(91-91\right)\)
\(=17-90-0\)
\(=-73\)
b) \(\left(-57-84\right)-\left(31-57\right)-\left(-84+31\right)\)
\(=-57-84-31+57+84-31\)
\(=\left(-57+57\right)-\left(84-84\right)+\left(-31+-31\right)\)
\(=0+0-62\)
\(=-62\)