\(\frac{x-3}{\sqrt{x-3}}\) = 5
Vay x bang bao nhieu?
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Trả lời:
\(a.-\frac{3}{5}=-\frac{6}{10}=-\frac{9}{15}=-\frac{12}{20}\)
_Su_
\(ĐKXĐ:x\ge0\)
Đề sai???
Sửa lại
\(a,P=\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+2-x+\sqrt{x}-1+x-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
Ta có : -3/4=x/5
➜\(\dfrac{-3}{x}\) = \(\dfrac{4}{5}\)
➜-3.5 = x.4
➞-15 =x.4
➜x = \(\dfrac{-15}{4}\)
Theo bài ra ta có :
\(\dfrac{-3}{4}=\dfrac{x}{5}\Rightarrow\left(-3\right)\cdot5=4\cdot x\)
\(\Rightarrow-15=4x\)
\(\Rightarrow4x=-15\)
\(\Rightarrow x=\left(-15\right):4=-\dfrac{15}{4}\)
Vậy x = \(\dfrac{-15}{4}\)
xyz=46656
\(\Leftrightarrow x.xk.xk^2=46656\Leftrightarrow x^3.k^3=46656\Leftrightarrow\left(xk\right)^3=46656\Rightarrow xk=36\)Ta có xk=36=> y=36
Vậy \(x+z=114-y=114-36=78\)
\(\frac{x-3}{\sqrt{x-3}}=5\)
=> \(\sqrt{x-3}=5\)
=> \(x-3=25\)
=> \(x=28\)
\(\frac{x-3}{\sqrt{x-3}}=5\Rightarrow x-3=5\sqrt{x-3}\)ĐK : x >= 3
\(\Leftrightarrow x^2-6x+9=25\left(x-3\right)\Leftrightarrow x^2-6x+9=25x-75\)
\(\Leftrightarrow x^2-31x+84=0\Leftrightarrow x=28;x=3\)