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ĐKXĐ: \(\hept{\begin{cases}x^2+5x+3\ge0\\x^2+3x+2\ge0\end{cases}}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2+5x+3}=a\\\sqrt{x^2+3x+2}=b\end{cases}\left(a;b\ge0\right)}\)

\(\Rightarrow a^2-b^2=x^2+5x+3-x^2-3x-2=2x+1\)

Pt trở thành

\(a-b=a^2-b^2\)

\(\Leftrightarrow a-b-\left(a-b\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(1-a-b\right)=0\)

\(pt\Leftrightarrow2\sqrt{\frac{x-1}{4}}-6=2\sqrt{\frac{4\left(x-1\right)}{9}}-\frac{1}{3}\)

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

\(\Leftrightarrow-\frac{1}{3}\sqrt{x-1}=\frac{17}{3}\)(vô lý )

Vậy pt vô nghiệm

\(=\sqrt{\left(2\sqrt{5}-7\right)^2}-\sqrt{\left(5-2\sqrt{5}\right)^2}\)

\(=7-2\sqrt{5}-5+2\sqrt{5}\)

\(=2\)

giúp mình với

+ \(n=2k\left(k\in N\right)\)

=> \(3^n-1=3^{2k}-1=9^k-1⋮8\)luôn đúng

+ \(n=2k+1\left(k\inℕ^∗\right)\)

=>\(3^n-1=3^{2k+1}-1=3.9^k-3+4=3\left(9^k-1\right)+4\)

MÀ \(9^k-1⋮8\); 4 không chia hết cho 8

=> VT không chia hết cho 8( loại )

Vậy n chẵn

\(\sqrt{\frac{2x-3}{x-1}}=2=>\frac{2x-3}{x-1}=4=>2x-3=4x-4=>1=2x=>x=\frac{1}{2}\)

Mình nghĩ vậy .

Hok tốt nha !

\(ĐKXĐ:\hept{\begin{cases}2x-3>0\\x-1>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{3}{2}\\x>1\end{cases}\Leftrightarrow}x>\frac{3}{2}}\)

\(\text{pt}\Leftrightarrow\frac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\left(x-1\right)\)

\(\Leftrightarrow2x-3=4x-4\)

\(\Leftrightarrow2x-4x=-4+3\)

\(\Leftrightarrow-2x=-1\)

\(\Leftrightarrow x=\frac{1}{2}\left(\text{loại}\right)\)

Vậy \(S=\varnothing\)

a) \(\sqrt{x^2-6x+9}+x=11\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}+x=11\)

\(\Rightarrow x-3+x=11\) 

\(\Rightarrow2x=14\Rightarrow x=7\) 

Vậy........

b) \(\sqrt{3x^2-4x+3}=1-2x\)

\(3x^2-4x+3=1-4x+4x^2\) 

\(3x^2-4x^2-4x+4x=-2\) 

\(-x^2=-2\) 

\(2=x^2\Rightarrow\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\) 

Vậy.........

d) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\) 

\(\Rightarrow2x-1=x-3\) 

\(\Rightarrow x=1-3\) 

\(\Rightarrow x=-2\) 

Vậy  x=-2