1.

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+2}-\sqrt{2-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{2x}{x\left(\sqrt{x+2}+\sqrt{2-x}\right)}=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{x+2}+\sqrt{2-x}}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

Vậy cần bổ sung \(f\left(0\right)=\dfrac{\sqrt{2}}{2}\) để hàm liên tục tại \(x=0\)

2.

a. \(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(x+\dfrac{3}{2}\right)=\dfrac{3}{2}\)

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{1+x}-1}=\lim\limits_{x\rightarrow0^+}\dfrac{x\left(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1\right)}{x\left(\sqrt[]{x+1}+1\right)}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}{\sqrt[]{x+1}+1}=\dfrac{3}{2}\)

\(\Rightarrow f\left(0\right)=\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)\) nên hàm liên tục tại \(x=0\)

2b.

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{x^3-x^2+2x-2}{x-1}=\lim\limits_{x\rightarrow1^-}\dfrac{x^2\left(x-1\right)+2\left(x-1\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1^-}\dfrac{\left(x^2+2\right)\left(x-1\right)}{x-1}=\lim\limits_{x\rightarrow1^-}\left(x^2+2\right)=3\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=f\left(1\right)=\lim\limits_{x\rightarrow1^+}\left(3x+a\right)=a+3\)

- Nếu \(a=0\Rightarrow f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^+}f\left(x\right)\) hàm liên tục tại \(x=1\)

- Nếu \(a\ne0\Rightarrow\lim\limits_{x\rightarrow1^-}f\left(x\right)\ne\lim\limits_{x\rightarrow1^+}f\left(x\right)\Rightarrow\) hàm không liên tục tại \(x=1\)

Hai biểu thức bằng nhau thì:

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

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

\(\Leftrightarrow x^2+2x\left(1+\sqrt{3}\right)+2\sqrt{3}=0\)

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

pt có hai hai nghiệm phân biệt là :

\(x_1=-\left(1+\sqrt{3}\right)+\sqrt{4}=1-\sqrt{3}\)

\(x_2=-\left(1+\sqrt{3}\right)-\sqrt{4}=-3-\sqrt{3}\)

a: Khi x=16 thì B=1/(4-3)=1

b: P=A-B

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

ĐK: \(x\ge0;x\ne9\)

a) Khi \(x=16\) TMĐKXĐ thì \(B=\dfrac{1}{\sqrt{16}-3}=1\)

b) \(P=A-B\)

\(P=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

\(=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

c) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

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

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

\(\Leftrightarrow4=3\) (Sai)

Vậy \(x\in\varnothing\)

\(a,x=16\Rightarrow B=\dfrac{1}{\sqrt{16}-3}=\dfrac{1}{4-3}=1\)

\(b,\) Rút gọn : \(A=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}\left(dkxd:x\ne9,x\ge0\right)\)

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

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

\(=\dfrac{x+3+2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+2\sqrt{x}-3}{x-9}\)

  Rút gọn \(P=A-B=\dfrac{x+2\sqrt{x}-3}{x-9}-\dfrac{1}{\sqrt{x}-3}\left(dkxd:x\ge0,x\ne9\right)\)

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

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

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

\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

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

\(c,P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(dkxd:x\ne9,x\ne4,x\ge0\right)\)

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

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

\(\Leftrightarrow\dfrac{x-3\sqrt{x}+\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=0\)

\(\Leftrightarrow-2\sqrt{x}+1=0\) ( Mất mẫu là bạn lấy mẫu nhân ngược vào 0 bên vế phải nha. )

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

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

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy khi \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(x=\dfrac{1}{4}\)

x=4
x²=8

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

⇒\(x^2=4^2=16\)

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

Làm sai kìa !

Cái chỗ \(\left|\sqrt{x-2}-5+3-\sqrt{x-2}\right|\ge2\) chứ  ? Trị tuyệt đối luôn dương mà

Cái trên là vừa phát hiện trong khi giải cái dưới

Vấn đề là giá trị của x cơ

Kiến thức cơ bản r 

\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(ab\ge0\)

\(C\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(\sqrt{x-2}-5\right)\left(3-\sqrt{x-2}\right)\ge0\)

TH1: \(\hept{\begin{cases}\sqrt{x-2}-5\ge0\\3-\sqrt{x-2}\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}\ge5\\\sqrt{x-2}\le3\end{cases}}}\) ( loại ) 

TH2 : \(\hept{\begin{cases}\sqrt{x-2}-5\le0\\3-\sqrt{x-2}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}\le5\\\sqrt{x-2}\ge3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\le25\\x-2\ge9\end{cases}\Leftrightarrow}\hept{\begin{cases}x\le27\\x\ge11\end{cases}\Leftrightarrow}11\le x\le27}\) ( nhận ) 

Vậy GTNN của \(C\) là \(2\) khi \(11\le x\le27\)

cái = 0 của pt 2 ý,,,,bạn thấy nha,,,do x>0 ( ĐKXĐ) ta có \(\frac{5\left(x+49\right)}{\sqrt{5x^2+4x}+21}\ge\frac{x+6}{\sqrt{x^2-3x-18}+6}\)

Từ đó dẫn đến vô lí

b)\(\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}\)

Đk:....

\(\Leftrightarrow\sqrt{5x^2+4x}-21-\left(\sqrt{x^2-3x-18}-6\right)-\left(5\sqrt{x}-15\right)=0\)

\(\Leftrightarrow\frac{5x^2+4x-441}{\sqrt{5x^2+4}+21}-\frac{x^2-3x-18-36}{\sqrt{x^2-3x-18}+6}-\frac{25x-225}{5\sqrt{x}+15}=0\)

\(\Leftrightarrow\frac{\left(x-9\right)\left(5x+49\right)}{\sqrt{5x^2+4}+21}-\frac{\left(x-9\right)\left(x+6\right)}{\sqrt{x^2-3x-18}+6}-\frac{25\left(x-9\right)}{5\sqrt{x}+15}=0\)

\(\Leftrightarrow\left(x-9\right)\left(\frac{5x+49}{\sqrt{5x^2+4}+21}-\frac{x+6}{\sqrt{x^2-3x-18}+6}-\frac{25}{5\sqrt{x}+15}\right)=0\)

chịu cái trong ngoặc r` bình phương đi :v

ừ nhỉ,,tui lm thiếu r

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:

\(x_1\le2< x_2\)

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

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