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

Có: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+1\ge1\\\left(x-2\right)^2+4\ge4\\\left(x-2\right)^2+5\ge5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x+2\right)^2+1}\ge1\\\sqrt{\left(x+2\right)^2+4\ge2}\\\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow\sqrt{\left(x+2\right)^2+1}\ge1+\sqrt{\left(x+2\right)^2+4}\ge2+\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}=3+\sqrt{5}\)Đẳng thức xảy ra khi: \(x-2=0\Leftrightarrow x=2\)

Vậy...

1. \(\sqrt{x^2-4x+3}=x-2\)

<=> x² - 4x + 3 = (x - 2)²

<=> x² - 4x + 3 = x² - 4x + 4

<=> x² - x² - 4x + 4x = 1

<=> 0 = 1 (Vô lí)

vậy PT có nghiệm là S = \(\varnothing\)

2. \(\sqrt{4x^2-4x+1}=x-1\)

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

<=> 2x - 1 = x - 1

<=> 2x - x = -1 + 1

<=> x = 0

1: ta có: \(\sqrt{x^2-4x+3}=x-2\)

\(\Leftrightarrow x^2-4x+3=x^2-4x+4\)(vô lý)

2: Ta có: \(\sqrt{4x^2-4x+1}=x-1\)

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

\(\Leftrightarrow x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

1: =>x^2-x=3-x

=>x^2=3

=>x=căn 3 hoặc x=-căn 3

2: =>x^2-4x+3=x^2-4x+4 và x>=2

=>3=4(vô lý)

3: =>2|x-1|=6

=>|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2 hoặc x=4

4: =>|2x-3|=|x-2|

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc x=5/3

5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)

=>x+2=0

=>x=-2

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

\(y=\left[{}\begin{matrix}3x-3\left(\text{với }x\ge2\right)\\3-3x\left(\text{với }x\le\dfrac{1}{2}\right)\\x+1\left(\text{với }\dfrac{1}{2}\le x\le2\right)\end{matrix}\right.\) 

Từ đó ta có đồ thị hàm số như sau:

Từ đồ thị ta thấy phương trình \(\sqrt{4x^2-4x+1}+\sqrt{x^2-4x+4}=m\):

- Có đúng 1 nghiệm khi \(m=\dfrac{3}{2}\)

- Có 2 nghiệm phân biệt khi \(m>\dfrac{3}{2}\)

- Vô nghiệm khi \(m< \dfrac{3}{2}\)

a: TH1: x>=2

A=x+x-2=2x-2

TH2: x<2

A=x+2-x=2

b: TH1: x>=3

A=x-3-x=-3

TH2: x<3

A=3-x-x=-2x+3

c: TH1: x>=1

C=x-x+1=1

TH2: x<1

C=x+x-1=2x-1

d: TH1: m>=3

C=m-3-2m=-3-m

TH2: m<3

C=-m+3-2m=-3m+3

e: TH1: m>=1

E=m-m+1=1

TH2: m<1

E=m+m-1=2m-1

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

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

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

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

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

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

Câu 1:

\(\lim _{x\to +\infty}(2x-1-\sqrt{4x^2-4x-3})=\lim_{x\to +\infty}\frac{(2x-1)^2-(4x^2-4x-3)}{2x-1+\sqrt{4x^2-4x-3}}\) (liên hợp)

\(=\lim_{x\to +\infty}\frac{4}{2x-1+\sqrt{4x^2-4x-3}}=4\lim_{x\to +\infty}\frac{1}{2x-1+\sqrt{4x^2-4x-3}}\)

Ta thấy với \(x\to +\infty\Rightarrow 2x-1+\sqrt{4x^2-4x-3}\to +\infty\)

Do đó: \(\lim_{x\to +\infty}\frac{1}{2x-1+\sqrt{4x^2-4x-3}}=0\) (theo dạng \(\lim _{t\to \infty}\frac{1}{t}=0\) )

\(\Rightarrow \lim _{x\to +\infty}(2x-1-\sqrt{4x^2-4x-3})=0\)

Câu 3:

\(\lim_{x\to 1+} (x^3-1)\sqrt{\frac{x}{x^2-1}}=\lim_{x\to 1+}(x^2+x+1)\sqrt{\frac{x(x-1)^2}{x^2-1}}\)

\(=\lim_{x\to 1+}(x^2+x+1)\sqrt{\frac{x(x-1)}{x+1}}=(1+1+1)\sqrt{\frac{1.0}{1+1}}=0\)

Câu 2:

\(\lim_{x\to 3}\frac{\sqrt{2x^2-2}-\sqrt{4x-3}+2x-7}{9-x^2}=\lim_{x\to 3}\frac{\sqrt{2x^2-2}-4}{9-x^2}-\lim_{x\to 3}\frac{\sqrt{4x-3}-3}{9-x^2}+\lim_{x\to 3}\frac{2x-6}{9-x^2}\)

Ta có:

\(\lim_{x\to 3}\frac{2x^2-2-16}{(\sqrt{2x^2-2}+4)(9-x^2)}=\lim_{x\to 3}\frac{2(x^2-9)}{(\sqrt{2x^2-2}+4)(9-x^2)}=\lim_{x\to 3}\frac{-2}{\sqrt{2x^2-2}+4}=\frac{-1}{4}\) (1)

\(\lim_{x\to 3}\frac{\sqrt{4x-3}-3}{9-x^2}=\lim_{x\to 3}\frac{4x-3-9}{(\sqrt{4x-3}+3)(9-x^2)}=\lim_{x\to 3}\frac{4(x-3)}{(\sqrt{4x-3}+3)(9-x^2)}\)

\(=\lim_{x\to 3}\frac{-4}{(\sqrt{4x-3}+3)(3+x)}=-\frac{1}{9}\) (2)

\(\lim _{x\to 3}\frac{2x-6}{9-x^2}=\lim_{x\to 3}\frac{2(x-3)}{9-x^2}=\lim_{x\to 3}\frac{-2}{x+3}=\frac{-1}{3}\) (3)

Từ \((1); (2); (3)\Rightarrow \lim_{x\to 3}\frac{\sqrt{2x^2-2}-\sqrt{4x-3}+2x-7}{9-x^2}=\frac{-1}{4}+\frac{1}{9}-\frac{1}{3}=\frac{-17}{36}\)

$x$ tiến tới mấy hả bạn?

\(1-\sqrt{2}x\) nha

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

\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)

\(\Leftrightarrow4x^2+4x-1=0\)

\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)

\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)

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

\(M=1+8+1=10\)