\(\sqrt{x}\)=6

=>x=\(6^2\)

đúng k nha

√36; k mk nha bn

\(\sqrt{x}=6\)

\(x=6.6\)

\(x=36\)

\(\Rightarrow\sqrt{x}=\sqrt{36}=6\)

ĐKXĐ: \(x>2\)

\(x^2-2\left(m+1\right)x+6m-2=x-2\)

\(\Leftrightarrow x^2-\left(2m+3\right)x+6m=0\) (1)

Pt có nghiệm duy nhất khi và chỉ khi (1) có 2 nghiệm pb thỏa mãn:

\(x_1\le2< x_2\)

\(\Rightarrow\left[{}\begin{matrix}m=1\\f\left(2\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\-2+2m< 0\end{matrix}\right.\) \(\Rightarrow m\le1\)

\(M=\frac{\sqrt{x}}{\sqrt{x}+1}\left(x\ge0\right)\)

Khi \(M=\sqrt{x}-2\)

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

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

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

\(\Leftrightarrow\sqrt{x}=x-\sqrt{x}-2\)

\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(-\sqrt{3}+1\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\1-2\sqrt{3}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4+2\sqrt{3}\\x=4-2\sqrt{3}\end{cases}}\)

Vậy \(x\in\left\{4\pm2\sqrt{3}\right\}\)khi \(M=\sqrt{x}-2\)

1.

\(\lim\dfrac{5\sqrt{3n^2+n}}{2\left(3n+2\right)}=\lim\dfrac{5\sqrt{3+\dfrac{1}{n}}}{2\left(3+\dfrac{2}{n}\right)}=\dfrac{5\sqrt{3}}{6}\Rightarrow a+b=11\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax+b}{x-2}=6\) khi \(x^2+ax+b=0\) có nghiệm \(x=2\)

\(\Rightarrow4+2a+b=0\Rightarrow b=-2a-4\)

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax-2a-4}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)+a\left(x-2\right)}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+a+2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(x+a+2\right)=a+4\Rightarrow a+4=6\Rightarrow a=2\Rightarrow b=-8\)

\(\Rightarrow a+b=-6\)

x=4
x²=8

\(\sqrt{x}=2\Rightarrow x=4\)

⇒\(x^2=4^2=16\)

\(x^2=\left(\sqrt{x}\right)^4=4^2=16\)

x = 4/9

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

=> x = \(\dfrac{2}{3} . \dfrac{2}{3}\)

=> x = \(\dfrac{4}{9}\)

1 quy đồng lên ra được

2 \(A=\dfrac{1}{x-2\sqrt{x-5}+3}\le\dfrac{1}{5-2.0+3}=\dfrac{1}{8}\)

dấu"=" xảy ra<=>x=5

`a)M=(sqrtx/(sqrtx-3)+sqrtx/(sqrtx+3))*(x-9)/sqrt{9x}(x>0,x ne 9)`

`M=((sqrtx(sqrtx+3)+sqrtx(sqrtx-3))/(x-9))*(x-9)/(3sqrtx)`

`M=((x+3sqrtx+x-3sqrtx)/(x-9))*(x-9)/(3sqrtx)`

`M=(2x)/(3sqrtx)=(2sqrtx)/3`

`b)M=6`

`=>2sqrtx=18`

`=>sqrtx=9=>x=81(tmđk)`

a) \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right).\dfrac{x-9}{\sqrt{9x}}\left(x>0,x\ne9\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{3\sqrt{x}}\)

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

b) \(M=6\Rightarrow\dfrac{2\sqrt{x}}{3}=6\Rightarrow2\sqrt{x}=18\Rightarrow\sqrt{x}=9\Rightarrow x=81\)

a) Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}\right)\cdot\dfrac{x-9}{\sqrt{9x}}\)

\(=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{3\sqrt{x}}\)

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

b) Để M=6 thì \(\dfrac{2}{3}\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}=9\)

hay x=81(thỏa ĐK)