\(\sqrt{29-x}+\sqrt{x+3}=x^2-26x+177\left(1\right)\)

ĐK -3 =<x =<29

Với mọi a,b >=0 ta có:

\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge2ab\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Thay \(a=\sqrt{29-x};b=\sqrt{x+3}\)ta có:

\(\sqrt{29-x}+\sqrt{x+3}\le\sqrt{2\left(29-x+x+3\right)}=8\)

\(x^2-26x+177=\left(x-13\right)^2+8\ge8\)

\(\Rightarrow\sqrt{29-x}+\sqrt{x+3}\le x^2-26x+177\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{29-x}=\sqrt{x+3}\\x-13=0\end{cases}\Leftrightarrow x=13}\)

Do đó (1) <=> x=13 (tm)

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{26x+5}=a\ge0\\\sqrt{x^2+30}=b>0\end{matrix}\right.\)

\(\Rightarrow\frac{a^2}{b}+2a=3b\)

\(\Leftrightarrow a^2+2ab-3b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+3b\right)=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow\sqrt{26x+5}=\sqrt{x^2+30}\)

\(\Leftrightarrow x^2-26x+25=0\Rightarrow\left[{}\begin{matrix}x=1\\x=25\end{matrix}\right.\)

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

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

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

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

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

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

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

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

=>x-5=4

=>x=9(nhận)

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

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)

\(\Leftrightarrow25x-4x=-8-75\)

\(\Leftrightarrow21x=-83\)

hay \(x=-\dfrac{83}{21}\)

b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)

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

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

c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)

\(\Leftrightarrow\left|2x+1\right|=3x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)

d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

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

\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)

\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)

\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)

\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)

\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)

\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)

\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)

vậy: Phương trình vô nghiệm

c: \(x^2-6\sqrt{x^2+5}+x=2\sqrt{x-1}-14\)

=>\(x^2-4-6\left(\sqrt{x^2+5}-3\right)+x-2-2\sqrt{x-1}+2=0\)

=>\(\left(x-2\right)\left(x+2\right)-6\cdot\dfrac{x^2+5-9}{\sqrt{x^2+5}+3}+\left(x-2\right)-2\cdot\dfrac{x-1-1}{\sqrt{x-1}+1}=0\)

=>\(\left(x-2\right)\left(x+2\right)-\dfrac{6}{\sqrt{x^2+5}+3}\cdot\left(x-2\right)\left(x+2\right)+\left(x-2\right)-2\cdot\dfrac{x-2}{\sqrt{x-1}+1}=0\)

=>\(\left(x-2\right)\left[\left(x+2\right)-\dfrac{6}{\sqrt{x^2+5}+3}\cdot\left(x+2\right)+1-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

=>x-2=0

=>x=2

d: \(x^2-\sqrt{\left(x^2-8\right)\left(x-2\right)}+x=\sqrt{x^2-8}+\sqrt{x-2}+9\)

=>\(x^2-9-\sqrt{\left(x^2-8\right)\left(x-2\right)}+x-\sqrt{x^2-8}-\sqrt{x-2}=0\)

=>\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\sqrt{x^3-2x^2-8x+16}+x-3+1-\sqrt{x^2-8}+2-\sqrt{x-2}=0\)

=>\(\left(x-3\right)\left(x+3\right)+\left(x-3\right)-\sqrt{x^3-2x^2-8x+16}+1+\dfrac{1-x^2+8}{1+\sqrt{x^2-8}}+1-\sqrt{x-2}=0\)

=>\(\left(x-3\right)\left(x+4\right)-\dfrac{x^3-2x^2-8x+16-1}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2-8}+1}+\dfrac{1-x+2}{1+\sqrt{x-2}}=0\)

=>\(\left(x-3\right)\left(x+4\right)-\dfrac{x^3-2x^2-8x+15}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2-8}+1}-\dfrac{x-3}{1+\sqrt{x-2}}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)-\dfrac{\left(x-3\right)\left(x^2+x-5\right)}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2-8}+1}-\dfrac{x-3}{1+\sqrt{x-2}}=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x+4\right)-\dfrac{x^2+x-5}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{x+3}{\sqrt{x^2-8}+1}-\dfrac{1}{\sqrt{x-2}+1}\right]=0\)

=>x-3=0

=>x=3

\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)

\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)

\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)

`\sqrt{14-x}-\sqrt{x-4}=\sqrt{x-1}`      `ĐK: 4 <= x <= 14`

`<=>\sqrt{14-x}=\sqrt{x-1}+\sqrt{x-4}`

`<=>14-x=x-1+x-4+2\sqrt{(x-1)(x-4)}`

`<=>14-x=2x-5+2\sqrt{x^2-5x+4}`

`<=>2\sqrt{x^2-5x+4}=19-3x`    với `4 <= x <= 19/3`

`<=>4(x^2-5x+4)=361-114x+9x^2`

`<=>4x^2-20x+16=361-114x+9x^2`

`<=>5x^2-94x+345=0`

`<=>5x^2-25x-69x+345=0`

`<=>(x-5)(5x-69)=0`

`<=>x=5` hoặc `x=69/5`

       (t/m)                    (ko t/m)

Vậy `S={5}`