Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

1: ĐKXĐ: x>=8/3

\(\sqrt{3x-8}-\sqrt{x+1}=\frac{2x-11}{5}\)

=>\(\sqrt{3x-8}-1+2-\sqrt{x+1}=\frac{2x-11}{5}+1\)

=>\(\frac{3x-8-1}{\sqrt{3x-8}+1}+\frac{4-x-1}{2+\sqrt{x+1}}=\frac{2x-11+5}{5}\)

=>\(\left(x-3\right)\left(\frac{3}{\sqrt{3x-8}+1}-\frac{1}{2+\sqrt{x+1}}-\frac25\right)=0\)

=>x-3=0

=>x=3(nhận)

3: ĐKXĐ: -5/2<=x<=5/2

Đặt \(a=\sqrt{5+2x};b=\sqrt{5-2x}\)

=>\(ab=\sqrt{\left(5+2x\right)\left(5-2x\right)}=\sqrt{25-4x^2}\)

Theo đề, ta có: a+b+5=3ab

=>3ab-a-b-5=0

=>a(3b-1)-b+1/3-16/3=0

=>\(3a\left(b-\frac13\right)-\left(b-\frac13\right)=\frac{16}{3}\)

=>\(\left(b-\frac13\right)\left(3a-1\right)=\frac{16}{3}\)

=>(3a-1)(3b-1)=16

=>(3a-1;3b-1)∈{(1;16);(16;1);(2;8);(8;2);(4;4)}

=>(3a;3b)∈{(2;17);(17;2);(3;9);(9;3);(5;5)}

=>(a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1);(5/3;5/3)}

mà a<>b

nên (a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1)}

TH1: a=2/3 và b=17/3

=>\(\begin{cases}5+2x=\frac49\\ 5-2x=\frac{289}{9}\end{cases}\Rightarrow\begin{cases}2x=\frac49-5=\frac49-\frac{45}{9}=-\frac{41}{9}\\ 2x=5-\frac{289}{9}=-\frac{244}{9}\end{cases}\)

=>x∈∅

TH2: a=17/3 và b=2/3

=>\(\begin{cases}5+2x=\frac{289}{9}\\ 5-2x=\frac49\end{cases}\Rightarrow\begin{cases}2x=\frac{289}{9}-5=\frac{244}{9}\\ 2x=5-\frac49=\frac{41}{9}\end{cases}\)

=>x∈∅

TH3: a=1 và b=3

=>5+2x=1 và 5-2x=9

=>2x=-4 và 2x=5-9=-4

=>x=-2(nhận)

TH4: a=3 và b=1

=>5+2x=9 và 5-2x=1

=>2x=4 và 2x=4

=>x=2(nhận)

ĐKXĐ : \(x\ge3\)

\(\sqrt{8+\sqrt{x-3}}+\sqrt{5+\sqrt{x-3}}\ge\sqrt{8}+\sqrt{5}>2\sqrt{8.5}=4\sqrt{10}>4\sqrt{\frac{25}{16}}=5\)

pt vô nghiệm

PS : làm thử thui chưa bít đúng sai nhé 

Nhìn vế trái \(\ge\sqrt{8}+\sqrt{5}>5\) là biết pt vô ngiệm rồi

\(ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{8+\sqrt{x}}-3\right)+\left(\sqrt{5-\sqrt{x}}-2\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{8+\sqrt{x}}+3}+\dfrac{-\sqrt{x}+1}{\sqrt{5-\sqrt{x}}+2}=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)\left(\dfrac{1}{\sqrt{8+\sqrt{x}}+3}-\dfrac{1}{\sqrt{5-\sqrt{x}}+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\\dfrac{1}{\sqrt{8+\sqrt{x}}+3}-\dfrac{1}{\sqrt{5-\sqrt{x}}+2}=0\left(vô.n_0,\forall x\ge0\right)\end{matrix}\right.\)

Vậy PT có nghiệm duy nhất \(x=1\)

a.

ĐKXĐ: \(x\ge0\)

\(\sqrt{2x^2+13x+5}-5\sqrt{x}+\sqrt{2x^2-3x+5}-3\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2-12x+5}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{2x^2-12x+5}{\sqrt{2x^2-3x+5}+3\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-12x+5\right)\left(\dfrac{1}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-3x+5}+3\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-12x+5=0\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x^2\ge\dfrac{4}{3}\)

\(\sqrt{x^2-\dfrac{4}{3}}+\sqrt{4x^2-4}-x=0\)

\(\Leftrightarrow\sqrt{\dfrac{3x^2-4}{3}}+\dfrac{3x^2-4}{\sqrt{4x^2-4}+x}=0\)

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

\(\Leftrightarrow3x^2-4=0\)

\(\Leftrightarrow...\)

ĐKXĐ: 5-x>=0 và x+8>=0

=>-8<=x<=5

Ta có: \(13\sqrt{5-x}+18\sqrt{x+8}=61+x+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\sqrt{5-x}-26+18\sqrt{x+8}-54=x-19+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\left(\sqrt{5-x}-2\right)+18\left(\sqrt{x+8}-3\right)=x-1-18+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\frac{5-x-4}{\sqrt{5-x}+2}+18\cdot\frac{x+8-9}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{\left(5-x\right)\left(x+8\right)}-6\right)\)

=>\(13\cdot\frac{1-x}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{5x+40-x^2-8x}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{-x^2-3x+40}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\cdot\frac{-x^2-3x+40-36}{\sqrt{-x^2-3x+40}+6}\)

=>(x-1)\(\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-3x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-4x+x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{\left(x+4\right)\left(-x+1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1-3\cdot\frac{\left(x+4\right)\left(x-1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}-1+3\cdot\frac{x+4}{\sqrt{-x^2-3x+40}+6}\right)=0\)

=>x-1=0

=>x=1(nhận)

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)

\(\Leftrightarrow\sqrt{\left(2-\sqrt{x-1}\right)^2}+\sqrt{\left(3+\sqrt{x-1}\right)^2}=5\)

\(\Leftrightarrow|2-\sqrt{x-1}|+3+\sqrt{x-1}=5\)

\(\Leftrightarrow\orbr{\begin{cases}2-\sqrt{x-1}+\sqrt{x-1}=2\\\sqrt{x-1}-2+\sqrt{x-1}=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}1\le x\le5\\x=5\end{cases}}\)

\(\Rightarrow1\le x\le5\)

1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))

\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)

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

\(\Leftrightarrow\left(x^2-4\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)

Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)

2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)

Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)

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

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy pt có no x=2

Thắng Chó Râm tặc

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)