a.\(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

=>\(4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

=>\(17\sqrt{3x}=17\)

=>\(\sqrt{3x}=1\)

=>\(x=\dfrac{1}{3}\)

b.Ta có:\(\sqrt{x^2-6x+9}=1\)

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

=>\(\left|x-3\right|=1\)

Vậy có hai trường hợp:

TH1:\(x-3=1\)

=>\(x=4\)

TH2:\(x-3=-1\)

=>\(x=2\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

\(\Leftrightarrow2\cdot2\cdot\sqrt{3x}-3\cdot\sqrt{3x}+4\cdot4\cdot\sqrt{3x}=17\)

\(\Leftrightarrow4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

\(\Leftrightarrow17\sqrt{3x}=17\)

\(\Leftrightarrow\sqrt{3x}=1\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)

b) ĐKXĐ: \(x\in R\)

Ta có: \(\sqrt{x^2-6x+9}=1\)

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

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

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

Vậy: S={2;4}

\(5\sqrt{12x}-4\sqrt{3x}+2\sqrt{48x}=14\)

\(\Leftrightarrow5.2.\sqrt{3x}-4.\sqrt{3x}+2.4.\sqrt{3x}=14\)

\(\Leftrightarrow10\sqrt{3x}-4\sqrt{3x}+8\sqrt{3x}=14\)

\(\Leftrightarrow\sqrt{3x}\left(10-4+8\right)=14\)

\(\Leftrightarrow14.\sqrt{3x}=14\)

\(\Leftrightarrow\sqrt{3x}=1\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\frac{1}{3}\)

a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)

b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)

\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

ĐKXĐ: ...

\(VT\le\sqrt{2\left(2x-3+5-2x\right)}=2\)

\(VP=3\left(x-2\right)^2+2\ge2\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}2x-3=5-2x\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow x=2\)

\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\left(1\right)\)

ĐKXĐ: \(\frac{3}{2}\le x\le\frac{5}{2}\)

Áp dụng BĐT Bunhiacopxki ta có:

\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)

Dấu "=" xảy ra <=> \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)

Ta lại có VP=3x²-12x+14=3(x-2)²+2 >=2

Dấu "=" xảy ra khi x=2

Do đó VT=VP <=> x=2 (ttmđk)

Vậy S={2}

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

`<=>\sqrt{3x}=1`

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

`<=>\sqrt{x+1}=3`

`<=>x+1=9`

`<=>x=8` (t/m)

pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)

đặt\(\sqrt{x+6}=a;x^2+4x=b\)

\(a,\) Sửa đề: \(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)

Ta thấy \(3x^2-12x+16=3\left(x-2\right)^2+4\ge4\Leftrightarrow\sqrt{3x^2-12x+16}\ge\sqrt{4}=2\)

\(y^2-4y+13=\left(y-2\right)^2+9\ge9\Leftrightarrow\sqrt{y^2-4y+13}\ge\sqrt{9}=3\)

Cộng vế theo vế 2 BĐT trên:

\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}\ge2+3=5\)

Dấu \("="\Leftrightarrow x=y=2\)

Vậy pt có nghiệm \(\left(x;y\right)=\left(2;2\right)\)

\(b,x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ \Leftrightarrow x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\\ \Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5+6\sqrt{z-5}+9\right)=0\\ \Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y-3=4\\z-5=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)

câu d tách hđt r đánh giá . VP=(x-6)^2+2>=2 còn VP <=2 =>....
câu c tương tự 
câu b c bình phương oặc đặt ẩn :3