x = -1
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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 \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

\(\Rightarrow5\left(x-10\right)=2\left(x+2\right)\)

\(\Rightarrow5x-50=2x+4\)

\(\Rightarrow3x=54\Rightarrow x=18\)

5x - 50 = 2x + 4

=> 5x - 50 - 2x - 4 = 0

x( 5 - 2 ) - ( 50 + 4 ) = 0

3x - 54 = 0

3x = 54

x = 18

\(A=\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}+\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-x^4y^4-2x^2y^2-1\)

Áp dụng Côsi

\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)\ge\frac{1}{2}.2\sqrt{\frac{x^{10}}{y^2}.\frac{y^{10}}{x^2}}=x^4y^4\)

\(\frac{1}{4}\left(x^{16}+y^{16}+1+1+1+1+1+1\right)\ge\frac{1}{4}.8\sqrt[8]{x^{16}y^{16}}=2x^2y^2\)

\(\Rightarrow A+\frac{6}{4}\ge x^4y^4+2x^2y^2-x^4y^4-2x^2y^2-1=-1\)

\(\Rightarrow A\ge-1-\frac{6}{4}=-\frac{5}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(x^2=y^2=1\)

Vậy GTNN của A là -2,5 khi x² = y² = 1

\(\dfrac{3}{x}=2:\dfrac{9}{10}\Rightarrow\dfrac{3}{x}=\dfrac{20}{9}\Rightarrow x=3:\dfrac{20}{9}=\dfrac{27}{20}\)

\(\dfrac{-2}{\dfrac{3}{x}}=\dfrac{9}{10}\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{9}{10}\Leftrightarrow x=\dfrac{-20}{27}\)

Bạn ơi ! nếu có sai sót gì,mong bạn thông cảm.

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

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

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

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