a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

\(\Leftrightarrow\sqrt{x+3}=3\)

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>=0\end{matrix}\right.\)

=>3<=x<=5

\(\sqrt{x-3}+\sqrt{5-x}=2\)

=>\(\sqrt{x-3}-1+\sqrt{5-x}-1=0\)

=>\(\dfrac{x-3-1}{\sqrt{x-3}+1}+\dfrac{5-x-1}{\sqrt{5-x}+1}=0\)

=>\(\left(x-4\right)\left(\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}\right)=0\)

=>x-4=0

=>x=4

 ĐK: \(x\ge\frac{3}{2}\)

 \(\sqrt{2x-3}+3=x\) 

<=> \(\sqrt{2x-3}=x-3\) (đk: \(x\ge3\)) 

=> \(2x-3=\left(x-3\right)^2\) 

<=> \(2x-3=x^2-6x+9\) 

<=> \(x^2-8x+12=0\) <=> \(\left(x-6\right)\left(x-2\right)=0\) 

=> \(\orbr{\begin{cases}x=6\left(TMĐK\right)\\x=2\left(KTMĐK\right)\end{cases}}\) 

Hai câu sau tương tự nhé bn 

\(x\sqrt{12}+\sqrt{18}=x\sqrt{8}+\sqrt{27}\)

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

<=> \(2x\sqrt{3}-2x\sqrt{2}=3\sqrt{3}-3\sqrt{2}\) 

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

<=> \(2x=3=>x=\frac{3}{2}\)

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

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

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

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

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)

Bài 1:

a) \(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)

\(=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\)

\(=\sqrt{2}\)

b) \(B=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{4-4\sqrt{5}+5}-\sqrt{5}\)

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

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

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

\(=-2\)

Bài 2:

a) \(\left\{{}\begin{matrix}2x-3y=4\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

b) ĐKXĐ: \(x\ne\pm2\)

Với \(x\ne\pm2\), ta có:

\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\)

\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)

\(\Leftrightarrow\dfrac{10-x-2}{x^2-4}=1\)

\(\Leftrightarrow\dfrac{8-x}{x^2-4}=1\)

\(\Rightarrow x^2-4=8-x\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\) (TM)

Vậy phương trình có tập nghiệm là: S ={3; -4}

\(a,3^{x-1}=27\\ \Leftrightarrow3^{x-1}=3^3\\ \Leftrightarrow x-1=3\\ \Leftrightarrow x=4\\ b,100^{2x^2-3}=0,1^{2x^2-18}\\ \Leftrightarrow10^{4x^2-6}=10^{-2x^2+18}\\ \Leftrightarrow4x^2-6=-2x^2+18\\ \Leftrightarrow6x^2=24\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)

\(c,\sqrt{3}e^{3x}=1\\ \Leftrightarrow e^{3x}=\dfrac{1}{\sqrt{3}}\\ \Leftrightarrow3x=ln\left(\dfrac{1}{\sqrt{3}}\right)\\ \Leftrightarrow x=\dfrac{1}{3}ln\left(\dfrac{1}{\sqrt{3}}\right)\)

\(d,5^x=3^{2x-1}\\ \Leftrightarrow2x-1=log_35^x\\ \Leftrightarrow2x-1-xlog_35=0\\ \Leftrightarrow x\left(2-log_35\right)=1\\ \Leftrightarrow x=\dfrac{1}{2-log_35}\)

Đk:\(x\ge0\)

Pt \(\Leftrightarrow2\sqrt{x}+5=36+3\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow-\sqrt{x}=22\) (vô nghiệm)

Vậy phương trình vô nghiệm

a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)