hình như đề bài bị sai số thì phải bạn ạ

mình giải cứ bị lệch số ấy

1.

\(\lim\limits_{x\to +\infty}(x^3+3x^2+2)=+\infty\)

2. 

\(\lim\limits_{x\to -\infty}\sqrt{4x^2-x+5}=\lim\limits_{x\to -\infty}-x.\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}=+\infty\) do $-x\to +\infty$ và $\lim\limits_{x\to -\infty}\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}=4>0$

3.

\(\lim\limits_{x\to +\infty}(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3})=\lim\limits_{x\to +\infty}\frac{x^2-2x-1-(x^2-7x+3)}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}\)

\(=\lim\limits_{x\to +\infty}\frac{5x-4}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}=\lim\limits_{x\to +\infty}\frac{5-\frac{4}{x}}{\sqrt{1-\frac{2}{x}-\frac{1}{x^2}}+\sqrt{1-\frac{7}{x}+\frac{3}{x^2}}}\)

\(=\frac{5}{1+1}=\frac{5}{2}\)

a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

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

c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)

\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)

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

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x\ge0\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)

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

Để A<0 thì \(2< \sqrt{x}< 3\)

=>4<x<9

`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`

`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`

`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`

`=(3x-sqrtx-20)/

Xét \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Phương trình trên tương đương:
x³+4x+5=0
<=>x(x²-1)+5(x+1)=0
<=>x(x-1)(x+1)+5(x+1)=0
<=>(x+1)(x²-x+5)=0
<=>x+1=0 hoặc x²-x+5=0(vô nghiệm)
<=>x=-1
Vậy pt trên có nghiệm là x=-1

x=-1 nha 

Bài này đi thi vio mk cũng gặp ..

    bằng 1 ak
 

1) ĐKXĐ: \(x\ge0\)

\(\sqrt{x}=2\sqrt{2}\Rightarrow x=8\left(tmđkxđ\right)\)

2) ĐKXĐ: \(x\ge-1\)

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

\(\Leftrightarrow\frac{x+1}{2}=\frac{5}{4}\)

\(\Leftrightarrow2x+2=5\Leftrightarrow x=\frac{3}{2}\left(TMĐKXĐ\right)\)

1, 

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

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

=> \(x=8\)

2.

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

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

=>  \(\frac{x+1}{2}=\frac{5}{4}\)

=> 4 ( x + 1 ) = 5.2

=> 4x + 4 = 10

=> 4x = 6

=. x = \(\frac{3}{2}\)