ĐKXĐ: \(x\ge0;x\ne4.\)

\(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}.\)

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

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

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

b) Để \(A=\frac{5}{4}\)\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\Leftrightarrow\frac{4\sqrt{x}}{4\left(\sqrt{x}-2\right)}-\frac{5\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}=0\)

\(\Leftrightarrow\frac{4\sqrt{x}-5\sqrt{x}+10}{4\left(\sqrt{x}-2\right)}=0\Leftrightarrow-\sqrt{x}+10=0\)

\(\Leftrightarrow\sqrt{x}=10\Leftrightarrow x=100\left(tmđk\right).\)

Vậy để A=5/4 thì x=100

Tự tìm ĐK nha

a) \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

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

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

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

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

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

b) \(A=\frac{5}{4}\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\)

\(\Leftrightarrow4\sqrt{x}=5\left(\sqrt{x}-2\right)\)

\(\Leftrightarrow4\sqrt{x}=5\sqrt{x}-10\)

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

\(\Leftrightarrow x=100\)( thỏa mãn )

Vậy...

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

có điêu kiện j nữa ko ?

Có \(x,y\in Z\)

\(\frac{7}{x}-\frac{2}{y}=5\)

\(7y-5x-5xy=0\)

hết

\(\frac{x+5}{4}=\frac{x-3}{5}\)

\(\Rightarrow x+5=\left(x-3\right).\frac{4}{5}\)

\(\Rightarrow x+5=\frac{4}{5}x-\frac{12}{5}\)

\(\Rightarrow\frac{1}{5}x=-\frac{37}{5}\)

\(\Rightarrow x=-37\)

Vậy  x = -37 

\(\frac{x+5}{4}=\frac{x-3}{5}\Leftrightarrow5\left(x+5\right)=4\left(x-3\right)\Leftrightarrow5x+25=4x-13\Leftrightarrow x=-38\) 

nhầm mình nhân sai

\(\Rightarrow\frac{x+5}{2015}+1+\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2015}{5}+1+\frac{x+2016}{4}+1+\frac{x+2017}{3}+1\)

\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}\)

\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Rightarrow x=-2020\)

thanks

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+6-4}=\frac{2x-2+3y-6-z+3}{4+6-4}\)

\(=\frac{\left(2x+3y-z\right)+\left(-2+6+3\right)}{6}=\frac{50+\left(-5\right)}{6}=\frac{45}{6}=7,5\)

\(\frac{x-1}{2}=7,5\Rightarrow x-1=15\Rightarrow x=16\)

\(\frac{y-2}{3}=7,5\Rightarrow y-2=24,5\Rightarrow y=20,5\)

\(\frac{z-3}{4}=7,5\Rightarrow z-3=30\Rightarrow z=33\)

a,\(\frac{17}{12}\)

b,\(\frac{-134}{81}\)

c,x=2

d,x=7