2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

=>\(-2\sqrt{3x}=-4\)

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

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

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

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

\(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)

Ta đánh giá vế phải \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=\sqrt{2\left(x-4\right)^2+9}+\sqrt{3\left(x-4\right)^2+16}\ge\sqrt{9}+\sqrt{16}=3+4=7\)(Do \(\left(x-4\right)^2\ge0\forall x\))

Như vậy, để \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)(hay dấu "=" xảy ra) thì \(\left(x-4\right)^2=0\)hay x = 4

Vậy nghiệm duy nhất của phương trình là 4

f, \(\sqrt{8+\sqrt{x}}+\sqrt{5-\sqrt{x}}=5\left(đk:25\ge x\ge0\right)\)

\(< =>\sqrt{8+\sqrt{x}}-\sqrt{9}+\sqrt{5-\sqrt{x}}-\sqrt{4}=0\)

\(< =>\frac{8+\sqrt{x}-9}{\sqrt{8+\sqrt{x}}+\sqrt{9}}+\frac{5-\sqrt{x}-4}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)

\(< =>\frac{\sqrt{x}-1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{\sqrt{x}-1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)

\(< =>\left(\sqrt{x}-1\right)\left(\frac{1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}\right)=0\)

\(< =>x=1\)( dùng đk đánh giá cái ngoặc to nhé vì nó vô nghiệm )

a,1,A=\(\sqrt{2x^2-8x+17}\)=\(\sqrt{2\left(x^2-4x+4\right)+9}\)=\(\sqrt{2\left(x-2\right)^2+9}\)

Có \(\left(x-2\right)^2\ge0\) vs mọi x

=> \(2\left(x-2\right)^2+9\ge9\) vs mọi x

<=> \(A=\sqrt{2\left(x-2\right)^2+9}\ge\sqrt{9}=3\)

Dấu "=" xảy ra <=> x=2

Vậy min A=3 <=> x=2

2,C=\(x-2\sqrt{x-4}+3\)( x\(\ge4\))

= \(\left(x-4\right)-2\sqrt{x-4}+1+6\)

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

Có \(\left(\sqrt{x-4}-1\right)^2\ge0\) với mọi \(x\ge4\)

=> C= \(\left(\sqrt{x-4}-1\right)^2+6\ge6\) với mọi x\(\ge4\)

Dấu "=" xảy ra <=> \(\sqrt{x-4}=1\) <=> \(x=5\) (t/m)

Vậy minC=6 <=>x=5

3,D=\(\sqrt{3x^2-12x+16}+\sqrt{x^4-8x^2+17}\)

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

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

Có \(\sqrt{3\left(x-2\right)^2+4}\ge\sqrt{0+4}=2\)

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

=> \(D=\sqrt{3\left(x-2\right)^2+4}+\sqrt{\left(x^2-4\right)^2+1}\ge2+1\)

<=> D \(\ge3\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-2=0\\x^2-4=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\x^2=4\end{matrix}\right.\) (t/m)

=> x=2

Vậy minD=3 <=>x=2

b, B=\(\sqrt{-3x^2+18x+22}=\sqrt{49-3\left(x^2-6x+9\right)}=\sqrt{49-3\left(x-3\right)^2}\)

Có \(3\left(x-3\right)^2\ge0\) vs mọi x

<=> 49\(-3\left(x-3\right)^2\le49\) vs mọi x

<=> \(\sqrt{49-3\left(x-3\right)^2}\le\sqrt{49}=7\)

<=> B\(\le7\)

Dấu "=" xảy ra <=> x=3

Vậy max B=7 <=> x=3

a/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+5x+2}=2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)\)

\(\Leftrightarrow6x^2+15x-26=0\)

b/ ĐKXĐ: ...

Đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)

\(a+\frac{1}{a}=\frac{5}{2}\Leftrightarrow a^2-\frac{5}{2}a+1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[5]{\frac{16x}{x-1}}=2\\\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}16x=32\left(x-1\right)\\16x=\frac{1}{32}\left(x-1\right)\end{matrix}\right.\)

c/ĐKXĐ: ...

\(\Leftrightarrow x^2-2x-\sqrt{6x^2-12x+7}=0\)

Đặt \(\sqrt{6x^2-12x+7}=a\ge0\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

\(\frac{a^2-7}{6}-a=0\Leftrightarrow a^2-6a-7=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=7\end{matrix}\right.\) \(\Rightarrow\sqrt{6x^2-12x+7}=7\)

\(\Leftrightarrow6x^2-12x-42=0\)

d/ \(\Leftrightarrow x^2+x+4-\sqrt{x^2+x+4}-2=0\)

Đặt \(\sqrt{x^2+x+4}=a>0\)

\(a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+x+4}=2\Rightarrow x^2+x=0\)

e/ \(\Leftrightarrow x^2+2x+\sqrt{3x^2+6x+4}-2=0\)

Đặt \(\sqrt{3x^2+6x+4}=a>0\Rightarrow x^2+2x=\frac{a^2-4}{3}\)

\(\frac{a^2-4}{3}+a-2=0\)

\(\Leftrightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2+6x+4}=2\Rightarrow3x^2+6x=0\)

ĐKXĐ:...

a/ \(\sqrt{2x^2+5x+2}=1+2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)+1+4\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow3\left(2x^2+6x-6\right)+4\sqrt{2x^2+5x-6}-7=0\)

Đặt \(\sqrt{2x^2+5x-6}=a\ge0\)

\(3a^2+4a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+5x-6}=1\)

\(\Leftrightarrow2x^2+5x-7=0\)