ĐKXĐ: \(x\ge0;x\ne4\)

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

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

b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)

c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

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

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

\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)

a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

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

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

b: Thay x=36 vào A, ta được:

\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)

\(\Leftrightarrow4\sqrt{x}=2\)

hay \(x=\dfrac{1}{4}\)

ko cần làm câu a nha các bạn

a) \(A\left(x\right)=x^2-10x+25\)

\(\Rightarrow A\left(x\right)=\left(x-5\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}A\left(0\right)=\left(0-5\right)^2=25\\A\left(-1\right)=\left(-1-5\right)^2=36\end{matrix}\right.\)

b) \(A\left(x\right)+B\left(x\right)=6x^2-5x+25\)

\(\Rightarrow B\left(x\right)=6x^2-5x+25-A\left(x\right)\)

\(\Rightarrow B\left(x\right)=6x^2-5x+25-\left(x^2-10x+25\right)\)

\(\Rightarrow B\left(x\right)=6x^2-5x+25-x^2+10x-25\)

\(\Rightarrow B\left(x\right)=5x^2+5x\)

\(\Rightarrow B\left(x\right)=5x\left(x+1\right)\)

c) \(A\left(x\right)=\left(x-5\right)C\left(x\right)\)

\(\Rightarrow C\left(x\right)=\dfrac{\left(x-5\right)^2}{x-5}=x-5\left(x\ne5\right)\)

d) Nghiệm của B(x)

\(\Leftrightarrow B=0\)

\(\Leftrightarrow5x\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của B(x)

một bể cá dạng hình hộp chữ nhật có chiều dài 1 2 m chiều rộng 0 5 m bên trong có một hòn non bộ  đặc có thể thích bằng 0,09m3 nếu đổ vào 150l nước thì hòn non bộ ngập hoàn toàn trong nước . hỏi chiều cao mực nước ở trong bẻ là bao nhiêu

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

a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

\(\left(1\right)\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x+1}\)

b) Để \(A=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x^2}{x+1}=-\dfrac{1}{2}\left(x\ne-1\right)\)

\(\Leftrightarrow2x^2=-\left(x+1\right)\)

\(\Leftrightarrow2x^2+x+1=0\)

\(\Delta=1-8=-7< 0\)

Nên phương trình trên vô nghiệm \(\left(x\in\varnothing\right)\)

c) Để \(A< 1\) 

\(\Leftrightarrow\dfrac{x^2}{x+1}< 1\)

\(\Leftrightarrow x^2< x+1\left(x\ne-1\right)\)

\(\Leftrightarrow x^2-x-1< 0\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2< \dfrac{5}{4}\)

\(\Leftrightarrow-\dfrac{\sqrt[]{5}}{2}< x-\dfrac{1}{2}< \dfrac{\sqrt[]{5}}{2}\)

\(\Leftrightarrow\dfrac{-\sqrt[]{5}+1}{2}< x< \dfrac{\sqrt[]{5}+1}{2}\)

d) Để A nguyên

\(\Leftrightarrow\dfrac{x^2}{x+1}\in Z\)

\(\Leftrightarrow x^2⋮x+1\)

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

\(\Leftrightarrow x^2-x^2+x⋮x+1\)

\(\Leftrightarrow x⋮x+1\)

\(\Leftrightarrow x-x-1⋮x+1\)

\(\Leftrightarrow-1⋮x+1\)

\(\Leftrightarrow x+1\in\left\{-1;1\right\}\)

\(\Leftrightarrow x\in\left\{-2;0\right\}\left(x\in Z\right)\)

!ERROR 404!

1: Để A>0 thì x-1<0

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

1) Để A > 0 thì:

\(x-1< 0\Leftrightarrow x< 1\)

\(\Rightarrow0\le x< 1\) và \(x\ne1\)

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

Để A<1 thì \(\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Rightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Mà x\(\ge0,x\ne1\)

\(\Rightarrow0\le x< 1\)

1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

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

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

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

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

\(\Leftrightarrow x=1\left(nhận\right)\)

2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)

a) đk: x khác 0;1

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

= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)

= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)

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

b) Để \(\left|2x-5\right|=3\)

<=>  \(\left[{}\begin{matrix}2x-5=3< =>2x=8< =>x=4\left(c\right)\\2x-5=-3< =>2x=2< =>x=1\left(l\right)\end{matrix}\right.\)

Thay x = 4 vào A, ta có: 

\(A=\dfrac{4^2}{4-1}=\dfrac{16}{3}\)

c) Để A = 4

<=> \(\dfrac{x^2}{x-1}=4\)

<=> \(\dfrac{x^2}{x-1}-4=0< =>\dfrac{x^2-4x+4}{x-1}=0\)

<=> \(\left(x-2\right)^2=0\)

<=> x = 2 (T/m)

d) Để A < 2

<=> \(\dfrac{x^2}{x-1}< 2< =>\dfrac{x^2}{x-1}-2< 0< =>\dfrac{x^2-2x+2}{x-1}< 0\)

<=> \(\dfrac{\left(x-1\right)^2+1}{x-1}< 0\)

Mà \(\left(x-1\right)^2+1>0\)

<=> x - 1 < 0 <=> x < 1

KHĐK: x < 1 ( x khác 0)

e) Để A thuộc Z

<=> \(\dfrac{x^2}{x-1}\in Z\)

<=> \(x^2⋮x-1\)

<=> \(x^2-x\left(x-1\right)-\left(x-1\right)⋮x-1\) 

<=> \(1⋮x-1\)

Ta có bảng: 

KL: Để A thuộc Z <=> \(x\in\left\{2;0\right\}\) 

f) Để A thuộc N <=> \(x\in\left\{2;0\right\}\) 

b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

Giúp mình những câu mik in đậm nhé

Thanks mọi người

a) ĐKXĐ: \(x\ge0,x\ne9\)

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

b) \(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{64}+3}=\dfrac{3}{8+3}=\dfrac{3}{11}\)

c) \(2A=\dfrac{6}{\sqrt{x}+3}=1\Rightarrow\sqrt{x}+3=6\Rightarrow x=9\left(tm\right)\)

g) \(A=\dfrac{3}{\sqrt{x}+3}\in Z\)

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

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

h) \(A=\dfrac{3}{\sqrt{x}+3}\in Z\)

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

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

k) \(2A=\dfrac{6}{\sqrt{x}+3}=m\)