6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

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a,Ta có:\(4x^2-4x-10=\sqrt{8x^2-6x-5}\)

\(\Leftrightarrow16x^4+16x^2+100-80x^2-32x^3+80x=8x^2-6x-5\)

\(\Leftrightarrow16x^4-32x^3-64x^2+80x+100-8x^2+6x+5=0\)

\(\Leftrightarrow16x^4-32x^3-72x^2+86x+110=0\)

\(\Leftrightarrow2\left(x+1\right)\left(2x-5\right)\left(4x^2-2x-11\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{2}\\4x^2-2x-11=0\Rightarrow\left[{}\begin{matrix}\dfrac{1+3\sqrt{5}}{4}\\\dfrac{1-3\sqrt{5}}{4}\end{matrix}\right.\end{matrix}\right.\)

Vậy ....

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

⇔\(\text{3x-1=2}^2\)

⇔\(3x=5\)

⇔\(x=\dfrac{5}{3}\)

b)\(\sqrt{x^2-4x+4}\)\(\text{=3x-1}\)

⇔\(\text{x-2=3x-1}\)

⇔\(-2x=1\)

⇔\(x=\)\(\dfrac{-1}{2}\)

c)\(\sqrt{x}\)=2-x sai đề bài

d)\(\sqrt{x^2+4}\)=\(\sqrt{3x+8}\)

⇔\(x^2\)\(\text{+4=3x+8}\)

⇔\(x^2\)\(-3x-4=0\)

⇔\(\left(x+1\right)\left(x-4\right)=0\)

⇔\(\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

a) ĐKXĐ: \(-1\leq x\leq 2\)

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

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

Đặt \(\sqrt{2+x-x^2}=t(t\geq 0)\). PT trở thành:

\(t=-3+2t^2\)

\(\Leftrightarrow 2t^2-t-3=0\Leftrightarrow (2t-3)(t+1)=0\)

\(\Rightarrow t=\frac{3}{2}\) (do \(t\geq 0)\)

\(\Rightarrow 2+x-x^2=\frac{9}{4}\Rightarrow x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow (x-\frac{1}{2})^2=0\Rightarrow x=\frac{1}{2}\) (thỏa mãn)

b) ĐK: \(x\geq \frac{1}{3}\)

PT \(\Leftrightarrow \sqrt{(3x-1)+6\sqrt{3x-1}+9}+\sqrt{(3x-1)-6\sqrt{3x-1}+9}=3x+4\)

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

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

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

Nếu \(\sqrt{3x-1}\geq 3\):

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

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

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

\(\Leftrightarrow (\sqrt{3x-1}-1)^2+4=0\) (vô lý)

Nếu \(\sqrt{3x-1}< 3\):

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

\(\Leftrightarrow 3x=2\Rightarrow x=\frac{2}{3}\) (thỏa mãn)

Vậy...........

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )