\(\left\{{}\begin{matrix}u^2+v^2=13\\uv=16\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}u^2+v^2=13\\u^2v^2=256\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}u^2=13-v^2\\\left(13-v^2\right)v^2=256\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}u^2=13-v^2\\13v^2-v^4-256=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}u^2=13-v^2\\v^4-13v^2+256=0\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow\) v⁴ - 2.\(\dfrac{13}{2}\)v² + \(\dfrac{169}{4}\) + \(\dfrac{855}{4}\) = 0

\(\Leftrightarrow\) (v² - \(\dfrac{13}{2}\))² + \(\dfrac{855}{4}\) = 0 (Vô nghiệm)

\(\Rightarrow\) Pt vô nghiệm

\(\Rightarrow\) Hpt vô nghiệm

Chúc bn học tốt!

x4=16

=>x=16:4

=>x=4

x*4=16

x=16:4

x=4

x4=16

=> x=4

:0 chắc v

#m.thu

\(|8-2x|=x+6\)

\(\Leftrightarrow\orbr{\begin{cases}2x-8=x+6\left(x< 0\right)\\8-2x=x+6\left(x\ge0\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=14\left(loai\right)\\-3x=-2\Leftrightarrow x=\frac{2}{3}\left(nhan\right)\end{cases}}\)

vay \(x=\frac{2}{3}\)

nếu có sai bn thông cảm nha

\(\left|8-2x\right|=x+6\)

\(\Rightarrow8-2x=x+6\)

\(\Leftrightarrow\hept{\begin{cases}2x-8=x+6\left(x\le0\right)\\8-2x=x+6\left(x\ge0\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=14\\x=\frac{2}{3}\end{cases}}\)

Mà theo đúng biểu thức, \(\left|8-2x\right|=x+6\)

\(\Rightarrow x=\frac{2}{3}\)

Vậy x = \(\frac{2}{3}\)

Ta có: \(\left(x+2\right)^4+\left(x+4\right)^4\)

\(=\left(x^2+4x+4\right)^2+\left(x^2+8x+16\right)^2\)

\(=x^4+16x^2+16+8x^3+8x^2+32x+x^4+64x^2+256+16x^3+32x^2+256x\)

\(=2x^4+24x^3+120x^2+288x+272\)

(2x-1)³=8

(2x-1)³=2³

Suy ra 2x-1=2

 2x=2+1

  2x=3

  x=3:2

   x=1,5

`2^(x-1)=16`

`=> 2^(x-1)=2^4`

`=>x-1=4`

`=> x=4+1`

`=>x=5`

2^{x-1 =} 16

2^x-1 = 2⁴

x - 1 = 4

x      = 4 + 1

x      = 5

Vì \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow2x-15=\orbr{\begin{cases}0\\1\end{cases}}\)vì chỉ có \(0^5=0^3;1^5=1^3\)

\(\Rightarrow2x=\orbr{\begin{cases}15\\16\end{cases}}\)

\(\Rightarrow x=\orbr{\begin{cases}\frac{15}{2}\\8\end{cases}}\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3.\)

\(\Rightarrow\hept{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow\hept{\begin{cases}2x=15\\2x=16\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)

\(\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)

\(\Rightarrow\left(2x-15\right)^3=0\)hoac \(\orbr{\begin{cases}2x-16=0\\2x-14=0\end{cases}}\)

\(\Rightarrow2x-15=0\)hoac \(\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\)

\(\Rightarrow2x=15\) hoac \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

\(\Rightarrow x=\frac{15}{2}\)hoac \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

      vay \(x=\frac{15}{2}\)hoac \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)