\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{10}\)\(\Leftrightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{10}\)

\(\Leftrightarrow\frac{11}{10}x=\frac{-33}{10}\)\(\Leftrightarrow x=\frac{-33}{10}:\frac{11}{10}=\frac{-33}{10}.\frac{10}{11}=-3\)

Vậy \(x=-3\)

\(\frac{1}{2}\cdot x+\frac{3}{5}\cdot x=-\frac{33}{10}\)

\(\left(\frac{1}{2}+\frac{3}{5}\right)\cdot x=-\frac{33}{10}\)

\(\left(\frac{5}{10}+\frac{6}{10}\right)\cdot x=-\frac{33}{10}\)

\(\frac{11}{10}\cdot x=-\frac{33}{10}\)

\(x=-\frac{33}{10}:\frac{11}{10}=-\frac{33}{10}\cdot\frac{10}{11}\)

\(x=-\frac{33}{11}=-3\)

\(\frac{1}{2}\).x+\(\frac{3}{5}\).x=\(\frac{-33}{10}\)

(\(\frac{1}{2}\)+\(\frac{3}{5}\)).x=\(\frac{-33}{10}\)

(\(\frac{5}{10}\)+\(\frac{6}{10}\)).x=\(\frac{-33}{10}\)

\(\frac{11}{10}\).x=\(\frac{-33}{10}\)

x=\(\frac{-33}{10}\):\(\frac{11}{10}\)

x=-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, \(\frac{2}{3}x+\frac{5}{6}x+\frac{1}{2}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{3}{2}x+\frac{5}{4}=0\Leftrightarrow\frac{3}{2}x=-\frac{5}{4}\Leftrightarrow x=\frac{-5}{6}\)

b, \(\frac{2}{5}+\frac{3}{5}\left(3x-3,7\right)=-\frac{53}{10}\)

\(\Leftrightarrow\frac{3}{5}\left(3x-\frac{37}{10}\right)=-\frac{57}{10}\Leftrightarrow3x-\frac{37}{10}=-\frac{10}{9}\Leftrightarrow x=-\frac{29}{15}\)

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

làm chi tiết cho mk nhé

ai làm chi tiết mk k cho nhìu