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=26x + ( 1+ 2+....+25) = 1235
=26x + 325 = 1235
=26x = 910
suy ra x =35
bn có thể trình bày ( 1+2+3 +...+25 rõ ràng hơn
\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)
\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)
\(\Leftrightarrow x=-22\)
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
a: =>14x+2*(14*13/2)=14
=>14x+14*13=14
=>14x=-12*14
=>x=-12
b: =>3x+(0-1-2-3+1+2+...+25)=25
=>3x+4+...+25=25
=>3x+4+...+24=0
=>3x+(24-4+1)*28/2=0
=>3x+21*28/2=0
=>3x=-21*14
=>x=-7*14=-98
x+(x+1)+(x+2)+....+(x+25)=1235
= \(\left(x+x+...+x\right)+\left(1+2+3+...+25\right)=1235\)
= \(26x+325=1235\)
=> \(26x=1235+325=1560\)
= \(26x=1560\)
=> x = 1560 : 26 = 60
=> x = 60
x + (x+1) + (x+2) + ...+ (x+25 ) = 1235
x + x + 1 + x + 2 +...+ x + 25 = 1235
(x + x + x +...+ x) + (1 + 2 +...+ 25) = 1235
26x + 325 = 1235
26x = 1235 - 325 = 910
x=910:26=35