\(\left(\left|x\right|-\dfrac{3}{2}\right)^2=\dfrac{9}{25}\Rightarrow\left[{}\begin{matrix}\left|x\right|-\dfrac{3}{2}=\dfrac{3}{5}\\\left|x\right|-\dfrac{3}{2}=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x\right|=\dfrac{21}{10}\\\left|x\right|=\dfrac{9}{10}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{21}{10}\\x=-\dfrac{21}{10}\\x=\dfrac{9}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

a) 5 giá trị của x: \(0,21;0,22;0,23;0,24;0,25\)

b) \(\Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{25}{12}=\dfrac{73}{12}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{73}{12}-\dfrac{25}{12}=4\Rightarrow x=4-\dfrac{1}{3}=\dfrac{11}{3}\)

\(\dfrac{R_1}{R_2}=\dfrac{S_2}{S_1}\Rightarrow R_2=\dfrac{R_1.S_1}{S_2}=\dfrac{60.4}{2,4}=100\left(\Omega\right)\)

Huhu nhìn lộn đề =((

\(18ph=\dfrac{3}{10}h,12ph=\dfrac{1}{5}h\)

\(t_{tổng}=t_1+t_2=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{1}{2}\left(h\right)\)

\(S_{AB}=v.t=\dfrac{1}{2}.20=10\left(km\right)\)

Gọi độ dài 3 cạnh tam giác theo thứ tự tăng dần lần lượt là a,b,c(m)(a,b,c>0)

Áp dụng t/c dtsbn:

\(\dfrac{a}{7}=\dfrac{b}{5}=\dfrac{c}{3}=\dfrac{a-c}{7-3}=\dfrac{8}{4}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a=2.7=14\\b=2.5=10\\c=2.3=6\end{matrix}\right.\)(nhận)

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