Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

Yêu cầu?

1.

ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)-4\sqrt{x-1}+4}+\sqrt{(x-1)+6\sqrt{x-1}+9}=5\)

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

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

Ta thấy:

\(\text{VT}=|2-\sqrt{x-1}|+|\sqrt{x-1}+3|\geq |2-\sqrt{x-1}+\sqrt{x-1}+3|=5\)

Dấu "=" xảy ra khi \((2-\sqrt{x-1})(\sqrt{x-1}+3)\geq 0\)

$\Leftrightarrow 2\geq \sqrt{x-1}$

$\Leftrightarrow 5\geq x\geq 1$

2. 

ĐKXĐ: $x\geq \frac{5}{2}$

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

\(\Leftrightarrow \sqrt{(2x-5)-6\sqrt{2x-5}+9}+\sqrt{(2x-5)+2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow \sqrt{(\sqrt{2x-5}-3)^2}+\sqrt{(\sqrt{2x-5}+1)^2}=4\)

\(\Leftrightarrow |\sqrt{2x-5}-3|+|\sqrt{2x-5}+1|=4\)

Thấy rằng:

\(\text{VT}=|3-\sqrt{2x-5}|+|\sqrt{2x-5}+1|\geq |3-\sqrt{2x-5}+\sqrt{2x-5}+1|=4\)

Dấu "=" xảy ra khi $(3-\sqrt{2x-5})(\sqrt{2x-5}+1)\geq 0$

$\Leftrightarrow 3-\sqrt{2x-5}\geq 0$

$\Leftrightarrow 7\geq x\geq \frac{5}{2}$

Vậy........

3. Nhân hai vế với $\sqrt{6}$ và làm tương tự câu 1,2.

\(\sqrt{6+2\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5+4\sqrt{5}+4}\)

= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)

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

= \(-1\)

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)

<=> \(\sqrt{\left(x-10\right)^2}=10\)

<=> \(\left|x-10\right|=10\)

=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)

Vậy S = \(\left\{20;0\right\}\)

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

<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)

<=> \(\left|\sqrt{x}+1\right|=6\)

=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)

Vậy S = \(\left\{25\right\}\)

3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

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

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

<=> \(\left|x-3\right|=\sqrt{3}+1\)

=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)

Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)

4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)

<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)

<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)

<=> \(\left|\sqrt{3x}+1\right|=5\)

=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)

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

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

<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)

<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)

Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)

6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)

<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)

<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)

<=> \(\left|\sqrt{6x}+2\right|=7\)

=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)

=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)

7) \(\sqrt{2x^2-2x\sqrt{6}+3}-\sqrt{5-\sqrt{24}}=0\)

<=> \(\sqrt{2x^2-2x\sqrt{6}+3}=\sqrt{5-\sqrt{24}}\)

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

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

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

<=> \(\left|x\sqrt{2}-\sqrt{3}\right|=\sqrt{3}-\sqrt{2}\)

=> \(\left[{}\begin{matrix}x\sqrt{2}-\sqrt{3}=\sqrt{3}-\sqrt{2}\\x\sqrt{2}-\sqrt{3}=-\left(\sqrt{3}-\sqrt{2}\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\sqrt{2}=2\sqrt{3}-\sqrt{2}\\x\sqrt{2}=\sqrt{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}}\\1\end{matrix}\right.\)

Vậy S = \(\left\{\dfrac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}};1\right\}\)

8) \(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{2}+2}=0\)

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

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

<=> \(\sqrt{2}-1=\left|x-\sqrt{2}\right|\)

=> \(\left[{}\begin{matrix}x-\sqrt{2}=\sqrt{2}-1\\x-\sqrt{2}=-\left(\sqrt{2}-1\right)\end{matrix}\right.=>\left[{}\begin{matrix}x+1=2\sqrt{2}\\x-\sqrt{2}=-\sqrt{2}+1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=1\end{matrix}\right.\) Vậy S = \(\left\{2\sqrt{2}-1;1\right\}\)