x + 1 : 0,75 = 1,4 : 0,25

<=> \(x+\dfrac{4}{3}=5,6\)

<=> \(x=\dfrac{64}{15}\)

nhầm rồi

1cong1=

Vì x, y > 0

Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)( k > 0 ) 

x² - y² = 4

<=> ( 5k )² - ( 4k )² = 4

<=> 25k² - 16k² = 4

<=> 9k² = 4

<=> k² = 4/9

<=> k = 2/3 ( vì k > 0 )

=> \(\hept{\begin{cases}x=5\cdot\frac{2}{3}=\frac{10}{3}\\y=4\cdot\frac{2}{3}=\frac{8}{3}\end{cases}}\)

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Ta có x/5=y/5=k
=>x=5k;y=4k
Theo tính chất dãy tỉ số = nhau ta có
x/5=y/4=x^2-y^2/5^2-4^2=4/9
x/5=4/9=>x=4/9.5=20/9
y/4=4/9=>y=4/9.4=16/9
=>x=20/9;y=16/9

Trả lời đúng , mình sẽ cho 5 vote

\(\Rightarrow x-\left(-x+x+3\right)-\left(x+3-x+2\right)=0\\ \Rightarrow x-3-5=0\\ \Rightarrow x=8\)

Sai rồi bạn ơi, bằng -2 mới đúng,bn thử lại coi

\(x.2-5.x=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

\(2x-5x=0\)

\(=>2x=5x\)

\(=>x=0\)

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x+\frac{1}{6}=0\)  

\(\Rightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)

k cho minh

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

\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x-\frac{1}{6}=0\)

\(\Leftrightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=0\)

Tính ra nhé !

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

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=x+\frac{1}{6}\)

\(\Rightarrow4x+\frac{77}{60}=x+\frac{1}{6}\)

\(\Rightarrow3x=\frac{1}{6}-\frac{77}{60}\)

\(\Rightarrow3x=-\frac{67}{60}\)

\(\Rightarrow x=-\frac{67}{60}\div3=\frac{-67}{60.3}=-\frac{67}{180}\)

Vậy x = .........

\(\frac{x+2}{x-5}< 0\) <=> x+2 và x-5 trái dấu

Mà x+2 > x-5

Nên x+2 > 0 và x-5 < 0

=>x > -2 và x < 5

Vậy -2 <x <5

Ta có: \(x+2\sqrt{2}.x^2+2x^3=0\)

\(\Leftrightarrow x\left(1+2\sqrt{2}.x+2x^2\right)=0\)

\(\Leftrightarrow x\left[1^2+2.x\sqrt{2}.1+\left(x\sqrt{2}\right)^2\right]=0\)

\(\Leftrightarrow x\left(1+x\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+x\sqrt{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{\sqrt{2}}\end{cases}}\)

Vậy\(x\in\left\{0;\frac{-1}{\sqrt{2}}\right\}\)

\(x+2\sqrt{2}x^2+2x^3=0\)

\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)

\(x\left(2\sqrt{2}x+1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{2}x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2x\sqrt{2}}\end{cases}}\)

Thôi nhầm chỗ HĐT rồi , bạn làm giống bạn Lê Tài Bảo Châu  nhé ! NHầm