\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\left(x< 0\right)\)

\(D=x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)

\(=x-4+\sqrt{\left(x-4\right)^2}\)

\(=x-4+\left|x-4\right|\)

\(=x-4+x-4\)

\(=2x-8\)

\(\left(8x-16\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{2;4\right\}\) là nghiệm của phương trình.

\(\left(8x-16\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(2\left(x^2+8x+16\right)-x^2+4=0\)

\(2x^2+16x+32-x^2+4=0\)

\(x^2+16x+36=0\)

\(x^2+16x+64-28=0\)

\(\left(x+8\right)^2-\left(\sqrt{28}\right)^2=0\)

\(\left(x+8-\sqrt{28}\right)\left(x+8+\sqrt{28}\right)=0\)

đến đây tự làm tiếp

\(=\dfrac{\left(x-4\right)\cdot\left(x+4\right)}{x}\cdot\dfrac{x}{\left(x-4\right)^2}=\dfrac{x+4}{x-4}\)

\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\)

\(D=x-4+\sqrt{16-8x+x^2}=x-4+\left|4-x\right|=x-4+x-4=2x-8\)

\(C=\sqrt{9x^2}-2x=-3x-2x=-5x\)

\(D=x-4+\sqrt{x^2-8x+16}=x-4+x-4=2x-8\)

\(\left(x+4\right)\left(3x-1\right)+\left(x^2+8x+16\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+4\right)\left(3x-1\right)=0\\x^2+8x+16=0\end{cases}}\)

Xét PT 1 : \(\left(x+4\right)\left(3x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{1}{3}\end{cases}}\)

Xét PT 2 : \(x^2+8x+16=0\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

\(x=\frac{-3}{4}\)hoặc \(x=-4\)

giúp mình vk mình hơi dốt toán

\(\left(x-4\right)^2=\left(x-4\right)^2\)

vô nghiệm

cac ban giai giup minh voi

:(((

Ta có: \(x^4+2x^3+8x+16\)

\(=x^3\left(x+2\right)+8\left(x+2\right)\)

\(=\left(x+2\right)\left(x^3+8\right)=\left(x+2\right)^2\cdot\left(x^2-2x+4\right)\)

Ta có: \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

\(=x^2\left(x^2-2x+4\right)+4\left(x^2-2x+4\right)=\left(x^2-2x+4\right)\left(x^2+4\right)\)

Ta có: \(E=\frac{x^4+2x^3+8x+16}{x^4-2x^3+8x^2-8x+16}\)

\(=\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\cdot\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}=\frac{x^2+4x+4}{x^2+4}=1+\frac{4x}{x^2+4}\)

Đặt \(A=\frac{4x}{x^2+4}\)

=>\(A\left(x^2+4\right)=4x\)

=>\(x^2\cdot A-4x+4A=0\) (1)

\(\Delta=\left(-4\right)^2-4\cdot A\cdot4A=16-16A^2\)

Để (1) có nghiệm thì Δ>=0

=>\(16-16\cdot A^2\ge0\)

=>\(A^2\le1\)

=>-1<=A<=1

=>0<=A+1<=2

=>0<=E<=2

=>GTNN của E là E=0 khi A=-1

(1) khi đó sẽ trở thành:

\(-x^2+4x-4=0\)

=>\(x^2-4x+4=0\)

=>\(\left(x-2\right)^2=0\)

=>x-2=0

=>x=2(nhận)

1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)