tim x
\(\frac{x-10}{2}=\frac{x+2}{5}\)
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\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{10}\)\(\Leftrightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{10}\)
\(\Leftrightarrow\frac{11}{10}x=\frac{-33}{10}\)\(\Leftrightarrow x=\frac{-33}{10}:\frac{11}{10}=\frac{-33}{10}.\frac{10}{11}=-3\)
Vậy \(x=-3\)
\(\frac{1}{2}\cdot x+\frac{3}{5}\cdot x=-\frac{33}{10}\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)\cdot x=-\frac{33}{10}\)
\(\left(\frac{5}{10}+\frac{6}{10}\right)\cdot x=-\frac{33}{10}\)
\(\frac{11}{10}\cdot x=-\frac{33}{10}\)
\(x=-\frac{33}{10}:\frac{11}{10}=-\frac{33}{10}\cdot\frac{10}{11}\)
\(x=-\frac{33}{11}=-3\)
Ta có : \(\frac{x-5}{5x-1}=\frac{4x-10}{20x+4}\)
=> \(\frac{x-5}{5x-1}=\frac{2x-5}{10x+2}\)
=> (x - 5)(10x + 2) = (2x - 5)(5x - 1)
=> 10x2 + 2x - 50x - 10 = 10x2 - 2x - 25x + 5
=> 10x2 - 48x - 10x2 + 27x = 5 + 10
=> -21x = 15
=> x = 15 : (-21) = -5/7
Thay x = -5/7 vào \(\frac{x-5}{5x-1}=\frac{y}{3}\)
=> \(\frac{-\frac{5}{7}-5}{5.\left(-\frac{5}{7}\right)-1}=\frac{y}{3}\)
=> \(\frac{-\frac{40}{7}}{-\frac{32}{7}}=\frac{y}{3}\)
=> \(\frac{5}{4}=\frac{y}{3}\)
=> 4y = 15
=> y = 15/4
Vậy ...
Ta có: \(\frac{5}{y}=\frac{3}{x}\) => \(\frac{x}{3}=\frac{y}{5}\) => \(\frac{x^2}{9}=\frac{y^2}{25}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{9}=\frac{y^2}{25}=\frac{y^2+x^2}{25+9}=\frac{125}{34}\)
=> \(\hept{\begin{cases}\frac{x^2}{9}=\frac{125}{34}\\\frac{y^2}{25}=\frac{125}{34}\end{cases}}\) => \(\hept{\begin{cases}x^2=\frac{125}{34}.9=\frac{1125}{34}\\y^2=\frac{125}{34}.25=\frac{3125}{34}\end{cases}}\) => \(\hept{\begin{cases}x=\pm\frac{15\sqrt{170}}{34}\\y=\pm\frac{25\sqrt{170}}{34}\end{cases}}\)
a, \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{3}{2}x=\frac{-5}{4}\)
\(\Leftrightarrow x=\frac{-5}{6}\)
b, \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
\(\Leftrightarrow\frac{3}{5}.\left(3x-\frac{37}{10}\right)=\frac{-57}{10}\)
\(\Leftrightarrow3x-\frac{37}{10}=\frac{-19}{2}\)
\(\Leftrightarrow3x=\frac{-29}{5}\)
\(\Leftrightarrow x=\frac{-29}{15}\)
a) \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=-\frac{3}{4}\)
\(x\left(\frac{2}{3}+\frac{5}{6}\right)+\frac{1}{2}=-\frac{3}{4}\)
\(x\cdot\frac{3}{2}=\frac{-5}{4}\)
\(x=-\frac{5}{6}\)
\(\frac{2}{5}+\frac{3}{5}\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\left(3x-\frac{37}{10}\right)=-\frac{57}{10}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=\frac{-29}{5}\)
\(x=\frac{-29}{15}\)
a) \(\left(\frac{1}{16}\right)^x=\left(\frac{1}{2}\right)^{10}\)
\(\left(\frac{1}{2}\right)^{4x}=\left(\frac{1}{2}\right)^{10}\)
\(\Rightarrow4x=10\)
x = 2,5
\(a\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\\ =>\left(x-\frac{1}{2}\right)=\frac{1}{3}\\ =>x=\frac{1}{3}+\frac{1}{2}\\ =>x=\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\\ =>\left(x+\frac{1}{2}\right)=\frac{2}{5}\\ =>x=\frac{-1}{10}\)
d) (2x+3)2016=(2x+3)2018 khi 2x+3=0 hoặc 1
Nếu 2x+3=0
=2x=-3 ( loại )
Nếu 2x+3=1
=>2x=-2
=>x=-1 ( thỏa )
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
\(\Rightarrow5\left(x-10\right)=2\left(x+2\right)\)
\(\Rightarrow5x-50=2x+4\)
\(\Rightarrow3x=54\Rightarrow x=18\)
5x - 50 = 2x + 4
=> 5x - 50 - 2x - 4 = 0
x( 5 - 2 ) - ( 50 + 4 ) = 0
3x - 54 = 0
3x = 54
x = 18