1) Ta có: \(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1-21+x=0\)

\(\Leftrightarrow x^2-3x-20=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)

1)\(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow21-x=x^2-2x+1\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow8-x=\left(x-2\right)^2\)

\(\Leftrightarrow8-x=x^2-4x+4\)

\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)

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

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

\(\Leftrightarrow3x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

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

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

Bài 1:

Vì $a\geq 1$ nên:

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

\(\geq 1+\sqrt{4}+0=3\)

Ta có đpcm

Dấu "=" xảy ra khi $a=1$

Bài 2:
ĐKXĐ: x\geq -3$

Xét hàm:

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

\(f'(x)=3x^2-6x+3+\frac{1}{2\sqrt{x+3}}=3(x-1)^2+\frac{1}{2\sqrt{x+3}}>0, \forall x\geq -3\)

Do đó $f(x)$ đồng biến trên TXĐ

\(\Rightarrow f(x)=0\) có nghiệm duy nhất

Dễ thấy pt có nghiệm $x=1$ nên đây chính là nghiệm duy nhất.

\(a,\)

\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

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

Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)

\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :

\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)

\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)

\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)

\(\Leftrightarrow-3\sqrt{x}+11=0\)

\(\Leftrightarrow-3\sqrt{x}=-11\)

\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)

\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)

\(\Leftrightarrow x=\dfrac{121}{9}\)

Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)

Bài 4: 

a, \(\sqrt{3x+4}-\sqrt{2x+1}=\sqrt{x+3}\) (ĐK: \(x\ge\dfrac{-1}{2}\))

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

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

\(\Leftrightarrow\) \(4x+2=2\sqrt{6x^2+11x+4}\)

\(\Leftrightarrow\) \(2x+1=\sqrt{6x^2+11x+4}\)

\(\Rightarrow\) \(4x^2+4x+1=6x^2+11x+4\)

\(\Leftrightarrow\) \(2x^2+7x+3=0\)

\(\Delta=7^2-4.2.3=25\); \(\sqrt{\Delta}=5\)

Vì \(\Delta\) > 0; theo hệ thức Vi-ét ta có:

\(x_1=\dfrac{-7+5}{4}=\dfrac{-1}{2}\)(TM); \(x_2=\dfrac{-7-5}{4}=-3\) (KTM)

Vậy ...

Các phần còn lại bạn làm tương tự nha, phần d bạn chuyển \(-\sqrt{2x+4}\) sang vế trái rồi bình phương 2 vế như bình thường là được

Bài 5: 

a, \(\sqrt{x+4\sqrt{x}+4}=5x+2\)

\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

\(\Rightarrow\) \(\sqrt{x}+2=5x+2\)

\(\Leftrightarrow\) \(5x-\sqrt{x}=0\)

\(\Leftrightarrow\) \(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\sqrt{x}=0\\5\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{25}\end{matrix}\right.\)

Vậy ...

Phần b cũng là hằng đẳng thức thôi nha \(\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}=x-1\); \(\sqrt{x^2+4x+4}=\sqrt{\left(x+2\right)^2}=x+2\) rồi giải như bình thường là xong nha!

VD1:

a, \(\sqrt{2x-1}=\sqrt{2}-1\) (x \(\ge\) \(\dfrac{1}{2}\))

\(\Leftrightarrow\) \(2x-1=\left(\sqrt{2}-1\right)^2\) (Bình phương 2 vế)

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

\(\Leftrightarrow\) \(2x=4-2\sqrt{2}\)

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

Vậy ...

Phần b tương tự nha

c, \(\sqrt{3}x^2-\sqrt{12}=0\)

\(\Leftrightarrow\) \(\sqrt{3}x^2=\sqrt{12}\)

\(\Leftrightarrow\) \(x^2=2\)

\(\Leftrightarrow\) \(x=\pm\sqrt{2}\)

Vậy ...

d, \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

\(\Leftrightarrow\) \(\sqrt{2}\left(x-1\right)=\sqrt{50}\)

\(\Leftrightarrow\) \(x-1=5\)

\(\Leftrightarrow\) \(x=6\)

Vậy ...

VD2: 

Phần a dễ r nha (Bình phương 2 vế rồi tìm x như bình thường)

b, \(\sqrt{x^2-x}=\sqrt{3-x}\) (\(x\le3\); \(x^2\ge x\))

\(\Leftrightarrow\) \(x^2-x=3-x\) (Bình phương 2 vế)

\(\Leftrightarrow\) \(x^2=3\)

\(\Leftrightarrow\) \(x=\pm\sqrt{3}\) (TM)

Vậy ...

c, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\dfrac{\sqrt{3}}{2}\))

\(\Leftrightarrow\) \(2x^2-3=4x-3\) (Bình phương 2 vế)

\(\Leftrightarrow\) \(2x^2-4x=0\)

\(\Leftrightarrow\) \(2x\left(x-2\right)=0\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt! (Có gì không biết cứ hỏi mình nha!)

Bài 1 : 

\(a.\sqrt{x^2-1}\)

\(ĐK:\)

\(x^2-1\ge0\)

\(\Leftrightarrow x^2\ge1\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)

Bài 2 : 

\(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{48}-5\sqrt{50}\)

\(=2\cdot\left|\sqrt{2}-3\right|+4\sqrt{3}-25\sqrt{2}\)

\(=-2\cdot\left(\sqrt{2}-3\right)+4\sqrt{3}-25\sqrt{2}\)

\(=-2\sqrt{2}-6+4\sqrt{3}-25\sqrt{2}\)

\(=-27\sqrt{2}-6+4\sqrt{3}\)