a)\(ĐKXĐ\Leftrightarrow\begin{cases}\sqrt{x}\ge0\\\sqrt{x}-1\ne0\end{cases}\Leftrightarrow\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(A=\frac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)+1\cdot\left(\sqrt{x}-1\right)-3\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)

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

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

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

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

b)\(S=A\cdot B\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}+3}{\sqrt{x}+2}\)

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

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

Để S đạt GTLN thì \(\frac{1}{\sqrt{x}+2}\)  đạt GTLN 

\(\frac{1}{\sqrt{x}+2}\) đạt GTLN \(\Leftrightarrow\sqrt{x}+2\) đạt GTNN 

GTNN \(\sqrt{x}+2\) là 2 \(\Leftrightarrow x=0\)

Vậy GTLN của S là \(\frac{3}{2}\Leftrightarrow x=0\)

ĐKXĐ \(\Leftrightarrow\)\(\sqrt{x}\ge0\) và \(\sqrt{x}-1\ne0\)

\(\Leftrightarrow x\ge0\) và \(x\ne1\)

SAO KHÔNG XEM ĐƯỢC VẬY TOÀN LEFT RIGHT FRAC CÁI GÌ CHẢ HIỂU NỔI 

a/ \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)   \(\left(ĐK:x\ge0;x\ne1\right)\)

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

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

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

1. a, \(2^{x+2}.3^{x+1}.5^x=10800\)

\(2^x.2^2.3^x.3.5^x=10800\)

\(\Rightarrow\left(2.3.5\right)^x.12=10800\)

\(\Rightarrow30^x=\frac{10800}{12}=900\)

\(\Rightarrow30^x=30^2\)

\(\Rightarrow x=2\)

b,\(3^{x+2}-3^x=24\)

\(\Rightarrow3^x\left(3^2-1\right)=24\)

\(\Rightarrow3^x.8=24\)\(\Rightarrow3^x=3^1\Rightarrow x=1\)

2, c, Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Dấu bằng xảy ra khi \(ab\ge0\)

Ta có: \(\left|x-2017\right|=\left|2017-x\right|\)

 \(\Rightarrow\left|x-1\right|+\left|2017-x\right|\ge\left|x-1+2017-x\right|\)\(=\left|2016\right|=2016\)

Dấu bằng xảy ra khi \(\left(x-1\right)\left(2017-x\right)\ge0\)\(\Rightarrow2017\ge x\ge1\)

Vậy \(Min_{BT}=2016\)khi \(2017\ge x\ge1\)

d, Áp dụng BĐT \(\left|a\right|-\left|b\right|\le\left|a-b\right|\forall a,b\inℝ\)

Dấu bằng xảy ra khi \(b\left(a-b\right)\ge0\)

Ta có \(B=\left|x-2018\right|-\left|x-2017\right|\le\left|x-2018-x+2017\right|\)

\(\Rightarrow B\le1\)

Dấu bằng xảy ra khi \(\left(x-2017\right)\left[\left(x-2018\right)-\left(x-2017\right)\right]\ge0\)

\(\Rightarrow x\le2017\)

Vậy \(Max_B=1\) khi \(x\le2017\)

để BT \(\frac{5}{\sqrt{2x+1}+2}\) nguyên thì \(\sqrt{2x+1}+2\inƯ\left(5\right)\)

suy ra \(\sqrt{2x+1}+2\in\left\{-5;-1;1;5\right\}\)

\(\Rightarrow\sqrt{2x+1}\in\left\{-7;-3;-1;3\right\}\)

Mà \(\sqrt{2x+1}\ge0\) nên \(\sqrt{2x+1}\)chỉ có thể bằng 3

\(\Rightarrow2x+1=9\Rightarrow x=4\)( thỏa mãn điều kiện \(x\ge-\frac{1}{2}\))

Đây là cách lớp 9. Mk đang phân vân ko biết giải theo cách lớp 7 thế nào!!!!

Bài 1 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

b) Để \(A< -1\)

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

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

\(\Leftrightarrow2\sqrt{x}< 1\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)

\(\Leftrightarrow x< \frac{1}{4}\)

Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)

\(A=\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)

\(A=\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{\left(\sqrt{x}+1\right)}{x\sqrt{x}+1}\)

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

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

Giúp mình câu b vớiiii. Câu a mình làm được A= \(\frac{\sqrt{x}}{x-\sqrt{x}+1}\)