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

ĐKXĐ: \(x\ge1\).

Phương trình đã cho tương đương:

\(\sqrt{x+3}+\sqrt{x-1}=\dfrac{8}{\sqrt{4x^4-12x^3+9x^2+16}-\left(2x^2-3x\right)}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-1}=\dfrac{\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)}{2}\)

\(\Leftrightarrow\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)-2\sqrt{x+3}-2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(\sqrt{4x^4-12x^3+9x^2+16}-2\sqrt{x+3}\right)+\left(2x^2-3x-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\dfrac{4x^4-12x^3+9x^2-4x+4}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{4x^4-12x^3+9x^2-4x+4}{2x^2-3x+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3-4x^2+x-2\right)\left(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}\right)=0\).

Do \(x\ge1\) nên ta có \(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}>0\).

Do đó \(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\left(TMĐK\right)\\4x^3-4x^2+x-2=0\left(1\right)\end{matrix}\right.\).

Giải phương trình bậc 3 ở (1) ta được \(x=\dfrac{\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}{\sqrt[6]{279936}}+\dfrac{1}{\sqrt[6]{7776}\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}+\dfrac{1}{3}\approx1,157298106\left(TMĐK\right)\).

Vậy...

Vì trong bài làm của mình có một số dòng khá dài nên bạn có thể vào trang cá nhân của mình để đọc tốt hơn!

Thấy : \(x^2-4x+16=\left(x-2\right)^2+12>0\forall x\)

P/t \(\Leftrightarrow2\left(x^2-4x+16\right)-36+\sqrt{x^2-4x+16}=0\)

Đặt \(t=\sqrt{x^2-4x+16}>0\) ; khi đó : 

\(2t^2+t-36=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-\dfrac{9}{2}\left(L\right)\end{matrix}\right.\)

Với t = 4  hay \(\sqrt{x^2-4x+16}=4\Leftrightarrow x^2-4x=0\)

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

Vậy ... 

Câu 1 bạn trên giải rồi mik k giải nx nha

2/ \(3\left(x^2+2\right)=10\sqrt{x^3+1}\)

\(3\left(x^2-x+1\right)+3\left(x+1\right)=10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b\ge0\end{matrix}\right.\)

pt⇔ \(3a^2+3b^2-10ab=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3b=b\\a=3b\end{matrix}\right.\)

Đến đây bạn tự giải tiếp nha

3/ \(\sqrt{3-3x}-\sqrt{3+x}=2\)

\(\left(\sqrt{3-3x}-3\right)-\left(\sqrt{3+x}-1\right)=0\)

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

+) \(x=-2\left(TM\right)\)

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

Vì VT<0 => ptvn

2 ) ĐK : \(x\ge-1\)

P/t \(\Leftrightarrow9\left(x^2+2\right)^2=100\left(x^3+1\right)\)

\(\Leftrightarrow9x^4+36x^2+36=100x^3+100\)

\(\Leftrightarrow9x^4-100x^3+36x^2-64=0\)

\(\Leftrightarrow\left(x^2-10x-8\right)\left(9x^2-10x+8\right)=0\)

\(\Leftrightarrow x^2-10x-8=0\) ( 9x^2 - 10x + 8 > 0 )

\(\Leftrightarrow x=5\pm\sqrt{33}\) ( t/m ) 

Vậy ... 

a: Ta có: \(\sqrt{1-x^2}=x-1\)

\(\Leftrightarrow1-x^2=x-1\)

\(\Leftrightarrow1-x^2-x+1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(x\ge-2\right)\\x+2=2-x\left(x< -2\right)\end{matrix}\right.\Leftrightarrow2x=0\)

hay x=0(loại)

mong mọi người giải giúp em vs 

1.

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

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

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

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

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)

\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)

\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)

\(\Leftrightarrow x^2-5x+2=0\)