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ĐK:\(x\ge0\)
\(\left(x^2-1\right)\sqrt{x}=0\Leftrightarrow\left(x-1\right)\left(x+1\right)\sqrt{x}=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\\x=0\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Ủa lớp 7 sao học căn r nè
a) \(\left|x\right|+x\)
Vì \(\left|x\right|\ge0\) nên ta có 3TH:
TH1: \(x>0\)
\(\Rightarrow\left|x\right|+x=2x\)
TH2: \(x=0\)
\(\Rightarrow\left|x\right|+x=0\)
TH3: \(x< 0\)
\(\Rightarrow\left|x\right|+x=0\)
\(x:0,16=9:x\)
\(\frac{x}{0,16}=\frac{9}{x}\)
\(\Rightarrow x^2=0,16.9\)
\(\Rightarrow x^2=1,44\)
\(\Rightarrow x=\sqrt{1,44}\)
\(\Rightarrow x=1,2\)
\(x:\left[\dfrac{8}{5}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{5}\right]=\dfrac{15}{7}+\dfrac{6}{5}\left[\left(2\dfrac{1}{7}\right)^2-\dfrac{50}{49}\right]\)
\(\Leftrightarrow x:\left[\dfrac{32}{45}-\dfrac{18}{45}\right]=\dfrac{15}{7}+\dfrac{6}{5}\cdot\left(\dfrac{225}{49}-\dfrac{50}{49}\right)\)
\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{15}{7}+\dfrac{6}{5}\cdot\dfrac{25}{7}\)
\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{45}{7}\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow x^2=1.44\)
hay \(x\in\left\{1.2;-1.2\right\}\)
\(\frac{x}{0,16}=\frac{9}{x}\)
\(\Rightarrow x^2=0,16.9=\frac{36}{25}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{9}{5}\\x=-\frac{9}{5}\end{array}\right.\)
\(\frac{x}{0,16}=\frac{9}{x}\)
\(\Leftrightarrow x.x=0,16.9\)
\(\Leftrightarrow x^2=1,44=\left(\pm1,2\right)^2\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1,2\\x=-1,2\end{array}\right.\)
Vậy \(x\in\left\{1,2;-1,2\right\}\)
Chúc bạn học tốt!
\(\dfrac{-x}{0.45}=\dfrac{5.7}{0.35}\)
\(\Leftrightarrow x=\dfrac{-5.7\cdot0.45}{0.35}=-\dfrac{513}{70}\)
x:0,16 = 9:x
x:0,16.x=9
x2:0,16=9
x2 =9.0,16
x2 =1,44
x2 = (6/5)2
=> x = 6/5
Vậy x =6/5
Ủng hộ mk nha