Trả lời:

a, \(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\left(ĐK:x\ge0;x\ne25\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

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

b, Thay x = 9 vào A, ta được:

\(A=\frac{\sqrt{9}-5}{\sqrt{9}+5}=\frac{3-5}{3+5}=\frac{-2}{8}=-\frac{1}{4}\)

c, \(A< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}-\frac{1}{3}< 0\)

\(\Leftrightarrow\frac{3\left(\sqrt{x}-5\right)}{3\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\frac{3\sqrt{x}-15-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\frac{2\sqrt{x}-20}{3\left(\sqrt{x}+5\right)}< 0\) 

\(\Rightarrow2\sqrt{x}-20< 0\) (vì \(3\left(\sqrt{x}+5\right)>0\) )

\(\Leftrightarrow2\sqrt{x}< 20\)

\(\Leftrightarrow\sqrt{x}< 10\)

\(\Leftrightarrow x< 100\)

Vậy \(0\le x< 100\)và \(x\ne25\) là giá trị cần tìm.

Trả lời:

 \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(ĐK:x\ge0;x\ne16\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)

Trả lời:

b, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x>0\right)\)

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

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

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

c, \(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}>\frac{3}{2}\) \(\left(ĐK:x>0\right)\)

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

\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)

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

\(\Rightarrow2\sqrt{x}+1-3\sqrt{x}>0\Leftrightarrow1-\sqrt{x}>0\)

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

Vậy \(0< x< 1\) là giá trị cần tìm.

Trả lời:

a, \(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)

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

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

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

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

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

b, \(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\) \(\left(ĐK:x>0\right)\)

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

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

\(\Rightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2x+3\sqrt{x}-2=0\)

\(\Leftrightarrow2x+4\sqrt{x}-\sqrt{x}-2=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=0\\2\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-2\left(voli\right)\\2\sqrt{x}=1\end{cases}\Leftrightarrow}\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)}\)

Vậy x = 1/4 là giá trị cần tìm.

Trên cạnh AE nhé các bạn. Đề lộn.

Làm như vậy không ổn lắm bởi vì còn phải xét trường hợp \(x=0\)và \(x< 0\)nữa, rất mất thời gian. Bạn cứ làm theo cách thông thường đưa về phương trình tích là được rồi.

a, \(Na_2O_2+2KO_2+2CO_2\rightarrow O_2\uparrow+K_2CO_3+Na_2CO_3\)

b, Tỉ lệ 2:1

a) ĐKXĐ : \(x\ge5\)

Đặt \(\sqrt{x-5}=a;\sqrt[3]{3-x}=b\)(a \(\ge0\))

Khi đó phương trình thành a + b = 2

Lại có \(b^3+a^2=-2\)

=> HPT : \(\hept{\begin{cases}a+b=2\\b^3+a^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+\left(2-b\right)^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+b^2-4b+6=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2-b\\\left(b+3\right)\left(b^2-2b+2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5\\b=-3\end{cases}}\)(tm)

a = 5 => x = 30 (tm) 

Vậy x = 30 là nghiệm phương trình 

d) Ta có \(\sqrt{25x^2-20x+4}+\sqrt{25x^2-40x+16}=0\)

<=> \(\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-4\right)^2}=2\)

<=> |5x - 2| + |5x - 4| = 2

Lại có |5x - 2| + |5x - 4| = |5x - 2| + |4 - 5x| \(\ge\left|5x-2+4-5x\right|=2\)

Dấu "=" xảy ra <=> \(\left(5x-2\right)\left(4-5x\right)\ge0\Leftrightarrow\frac{2}{5}\le x\le\frac{4}{5}\)

Vậy \(\frac{2}{5}\le x\le\frac{4}{5}\)là nghiệm phương trình 

đvdt: đơn vị diện tích

k cho mình nhé:))

EM NHÉ