a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)

\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)

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

Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)

Bài 1: 

a: Ta có: \(x^2-2\sqrt{5}x+5=0\)

\(\Leftrightarrow x-\sqrt{5}=0\)

hay \(x=\sqrt{5}\)

b: Ta có: \(\sqrt{x+3}=1\)

\(\Leftrightarrow x+3=1\)

hay x=-2

a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)

\(\Leftrightarrow\left|x-3\right|=2\)

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

b) ĐKXĐ: \(x\ge-2\)

Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7(thỏa ĐK)

a) \(\Leftrightarrow\left|x-3\right|=2\)

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

Vậy:.....

b) ĐKXĐ: x ≥ -2

 \(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)

<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)

\(\Leftrightarrow2.\sqrt{x+2}=6\)

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

<=> x + 2 = 9

<=> x = 7

Tham khảo ạ:

a) Ta có: 

⇔[x−3=2x−3=−2⇔[x=5x=1⇔[x−3=2x−3=−2⇔[x=5x=1

b) ĐKXĐ: 

Ta có: ⇔3√x+2−5√x+2+45⋅5√x+2=6⇔3x+2−5x+2+45⋅5x+2=6

hay x=7(thỏa ĐK)

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

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

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

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

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

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

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

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

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

✱ giải pt:

a.\(\sqrt{x^2-4x+4}\)\(=5\)

⇔\(\sqrt{\left(x-2\right)^2}=5\)

⇒\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

vậy....

b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

⇔ \(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

⇔ \(4\sqrt{x+1}=16\)

⇔ \(\sqrt{x+1}=16\)

⇒ \(x+1=256\)

⇔ \(x=255\)

vậy.....

a: \(A=5\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-1\)

\(B=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)

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

b: A=B

=>căn x=-căn x+1

=>căn x=1/2

=>x=1/4

`a)A=[2\sqrt{3}+2-2\sqrt{3}+2]/[(2\sqrt{3}-2)(2\sqrt{3}+2)]`

   `A=4/[12-4]=1/2`

Với `x > 0,x ne 1` có:

`B=[x-2\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]`

`B=[(\sqrt{x}-1)^2]/[\sqrt{x}(\sqrt{x}-1)]=[\sqrt{x}-1]/\sqrt{x}`

`b)B=2/5A`

`=>[\sqrt{x}-1]/\sqrt{x}=2/5 . 1/2`

`<=>5\sqrt{x}-5=\sqrt{x}`

`<=>\sqrt{x}=5/4`

`<=>x=25/16` (t/m)

\(a.\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)( x lớn hơn hoặc =1)
\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}\)+2=0
\(\sqrt{x-1}\left(1+\sqrt{4}-\sqrt{25}\right)=-2\)
\(\sqrt{x-1}\left(1+2-5\right)=-2\)
\(\sqrt{x-1}.\left(-2\right)=-2\)
\(\sqrt{x-1}=-2.2\)
\(\sqrt{x-1}-4\)(ko thỏa mãn)
b)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\dfrac{\sqrt{x-1}}{8}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)\sqrt{x-1}=-17\)
\(7\sqrt{x-1}=-17\)
\(\sqrt{x-1}=-\dfrac{17}{7}\)(ko thỏa mãn căn bậc 2 ko có số âm)

a: Ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

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

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

\(\Leftrightarrow x-1=1\)

hay x=2

Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$

PT $\Leftrightarrow 5x-2=7^2=49$

$\Leftrightarrow 5x=51$

$\Leftrightarrow x=\frac{51}{5}=10,2$

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$

$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$

$\Leftrightarrow 8\sqrt{x-3}=24$

$\Leftrightarrow \sqrt{x-3}=3$

$\Leftrightarrow x-3=9$

$\Leftrightarrow x=12$ (tm)

Bài 1:

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$

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

$\Leftrightarrow (x-3)[(x-2)-\frac{2}{\sqrt{x-2}+1}]=0$

$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$

Nếu $x-3=0$

$\Leftrightarrow x=3$ (tm) 

Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$

$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)

$\Leftrightarrow a^3+a^2-2=0$

$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$

$\Leftrightarrow (a-1)(a^2+2a+2)=0$

Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$

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

Vậy pt có nghiệm duy nhất $x=3$.

Bài 2:

ĐKXĐ: $x\geq 0; x\neq 4$

\(A=\frac{\sqrt{x}(\sqrt{x}-2)-\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)\sqrt{x}-2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-2\sqrt{x}}{\sqrt{x}-2}=\frac{2\sqrt{x}}{2-\sqrt{x}}\)