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Trường hợp 1: x<2/3

Pt sẽ là 2-3x+11-3x=9

=>-6x+13=9

=>-6x=-4

hay x=2/3(loại)

Trường hợp 2: 2/3<=x<11/3

Pt sẽ là 3x-2+11-3x=9

=>9=9(luôn đúng)

Trường hợp 3: x>=11/3

Pt sẽ là 3x-2+3x-11=9

=>6x-13=9

=>6x=22

hay x=11/3(nhận)

\(\text{(3x+9)*(11-x)=-2}\)

\(\text{(3x+9)*(11-x)=-2}.1=1.-2=-1.2=2.-1\)

TH1, \(\text{(3x+9)*(11-x)=-2}.1\)

\(\Rightarrow\orbr{\begin{cases}3x+9=-2\\11-x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-11}{3}\\x=10\end{cases}}}\)

Th2, \(\text{(3x+9)*(11-x)=1.-2}\)

\(\orbr{\begin{cases}3x+9=1\\11-x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-8}{3}\\x=13\end{cases}}}\)

Th3, \(\text{(3x+9)*(11-x)=-1.2}\)

\(\Rightarrow\orbr{\begin{cases}3x+9=-1\\11-x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-10}{3}\\x=9\end{cases}}}\)

TH4, \(\text{(3x+9)*(11-x)=2.-1}\)

\(\Rightarrow\orbr{\begin{cases}3x+9=2\\11-x=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-7}{3}\\x=12\end{cases}}}\)

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+) \(\left(2x-1\right)^9-\left(2x-1\right)^{11}=0\)

\(\Leftrightarrow\left(2x-1\right)^9=\left(2x-1\right)^{11}\)

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

+) \(\left(2-3x\right)^{13}=\left(3x-2\right)^{12}\)

\(\Leftrightarrow\left(2-3x\right)^{13}=\left(2-3x\right)^{12}\)

\(\Leftrightarrow\orbr{\begin{cases}2-3x=0\\2-3x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{1}{3}\end{cases}}\)

a:

ĐKXĐ: \(x^2+3x>=0\)

=>x(x+3)>=0

=>\(\left[{}\begin{matrix}x>=0\\x< =-3\end{matrix}\right.\)

 \(\sqrt{16}-\sqrt{x^2+3x}=0\)

=>\(\sqrt{x^2+3x}=\sqrt{16}\)

=>x^2+3x=16

=>x^2+3x-16=0

\(\text{Δ}=3^2-4\cdot1\cdot\left(-16\right)=9+64=73>0\)

Do đó: Phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{73}}{2}\\x_2=\dfrac{-3+\sqrt{73}}{2}\end{matrix}\right.\)

b:

ĐKXĐ: \(x\in R\)

 \(3x-1-\sqrt{4x^2-12x+9}=0\)

=>\(\sqrt{\left(2x-3\right)^2}=3x-1\)

=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(2x-3\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(x+2\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{4}{5}\left(nhận\right)\end{matrix}\right.\)

c:

ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+8>=0\\2x^2-10x+11>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\\\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=4\end{matrix}\right.\)

 \(\sqrt{2x^2-10x+11}=\sqrt{x^2-6x+8}\)

\(\Leftrightarrow2x^2-10x+11=x^2-6x+8\)

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>x=3(loại) hoặc x=1(nhận)

\(\dfrac{3x-11}{11}-\dfrac{x}{3}=\dfrac{3x-5}{7}-\dfrac{5x-3}{9}\)

\(\Rightarrow\dfrac{3\left(3x-11\right)}{33}-\dfrac{11x}{33}=\dfrac{9\left(3x-5\right)}{63}-\dfrac{7\left(5x-3\right)}{63}\)

\(\Rightarrow\dfrac{3\left(3x-11\right)-11x}{33}=\dfrac{9\left(3x-5\right)-7\left(5x-3\right)}{63}\)

\(\Rightarrow\dfrac{9x-33-11x}{33}=\dfrac{27x-45-35x+21}{63}\)

\(\Rightarrow\dfrac{-2x-33}{33}=\dfrac{-8x-24}{63}\)

\(\Rightarrow63\left(-2x-33\right)=33\left(-8x-24\right)\)

\(\Rightarrow-126x-2079=-264x-792\)

\(\Rightarrow-126x+264x=-792+2079\)

\(\Rightarrow138x=1287\)

\(\Rightarrow x=\dfrac{1287}{138}\)

\(\Rightarrow x=\dfrac{429}{46}\)