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1) Thay x=16 vào A ta có:
A=\(\frac{16+\sqrt{16}+1}{\sqrt{16}+2}\)
A=\(\frac{16+4+1}{4+2}\)
A=\(\frac{21}{6}=\frac{7}{2}\)
\(2,\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{x-\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x-x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}\)\(\left(đpcm\right)\)
\(3,P=A.B=\frac{x+\sqrt{x}+1}{\sqrt{x}+2}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
Ta thấy \(\left(\sqrt{x}-1\right)^2>0\Rightarrow x-2\sqrt{x}+1>0\)
\(\Rightarrow x+\sqrt{x}+1>3\sqrt{x}\)
\(\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>\frac{3\sqrt{x}}{\sqrt{x}}\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>3\left(đpcm\right)\)
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
phamthiminhtrang
\(a,x=\frac{3}{6}-\frac{8}{16}\)
\(\Rightarrow x=0\)
\(b,\frac{12}{16}:x=\frac{32}{64}\)
\(x=\frac{12}{16}:\frac{32}{64}\)
\(x=\frac{12}{16}\cdot\frac{64}{32}\)
\(x=\frac{3}{8}\)
\(a,\)\(x\)\(=\frac{3}{6}-\frac{8}{16}=\frac{1}{2}-\frac{1}{2}=0\)
\(b,\)\(\frac{12}{16}\)\(:\)\(x\)\(=\frac{32}{64}\)
\(=>\) \(x\)\(=\)\(\frac{12}{16}:\frac{32}{64}\)
\(x\) \(=\)\(\frac{12}{16}.\frac{64}{32}\)
\(x\)\(=\)\(\frac{3}{4}.2\)
\(x\)\(=\)\(\frac{6}{4}=\frac{3}{2}\)
b) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
=> 3.(x + 2) = (x - 5). (-4)
= 3x + 6 = -4x + 20
= 3x - 4x = 26
=> -x = 26
=> x = -26
c) \(\frac{x}{-2}=\frac{-8}{x}\)
=> x.x = -2 . (-8)
=> x2 = 16
=> x = 4
t i c k nha!!! 645646677778879078452352543546456457567564546567
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1