Bài 11: Số vô tỉ. Khái niệm về căn bậc hai

\(=18+9^2+1=19+81=100\)

gì vậy

câu 3:

a) \(\sqrt{6}< 4\)

b) \(\sqrt{11.3}< \sqrt{44}\)

câu 4:

a)\(\Leftrightarrow x^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

b) ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\left|\sqrt{x}-3\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=-6\\\sqrt{x}-3=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(vôlí\right)\\\sqrt{x}=9\end{matrix}\right.\\ \Leftrightarrow x=81\)

\(3,\\ a,\sqrt{6}< \sqrt{16}=4\\ b,\sqrt{11\cdot3}=\sqrt{33}< \sqrt{44}\\ 4,\\ a,\Leftrightarrow x^2=9\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left|\sqrt{x}-3\right|=6\\ \Leftrightarrow\sqrt{x}-3=6\left(\sqrt{x}-3\ge-3\right)\\ \Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

\(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{2x}{4}=\dfrac{3y}{15}=\dfrac{2x-3y}{4-15}=\dfrac{44}{-11}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).2=-8\\y=\left(-4\right).5=-20\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{2x-3y}{4-15}=\dfrac{44}{-11}=-4\)

\(\dfrac{x}{2}=-4\Rightarrow x=-8\\ \dfrac{y}{5}=-4\Rightarrow y=-20\)

a, Áp dụng t/c dtsbn:

\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{2}{-2}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-7\\y=-5\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{y}=\dfrac{7}{2}\Rightarrow\dfrac{x}{7}=\dfrac{y}{2}=\dfrac{x+y}{7+2}=\dfrac{-27}{9}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-21\\y=-6\end{matrix}\right.\)

c, \(\dfrac{x}{32}=\dfrac{2}{x}\Rightarrow x^2=2\cdot32=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

d, \(\left|x+\dfrac{1}{3}\right|-2=\dfrac{1}{2}\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{2}\\x+\dfrac{1}{3}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{17}{6}\end{matrix}\right.\)

\(3\sqrt{x}-5=7\left(x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=12\Leftrightarrow\sqrt{x}=4\\ \Leftrightarrow x=16\left(tm\right)\)

\(5^x+5^{x+1}=150\)

\(\Leftrightarrow5^x=25\)

hay x=2

\(=\dfrac{5}{9}-\dfrac{4}{45}=\dfrac{7}{15}\)

0,(5) - 0,0(8)

= \(\dfrac{5}{9}-\dfrac{4}{45}\)

= \(\dfrac{25}{45}-\dfrac{4}{45}\)

= \(\dfrac{21}{45}=\dfrac{7}{15}=0,4\left(6\right)\)

Hình đâu thế bạn!!!

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\(a,\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow x^2=-1\left(vô.lí\right)\Rightarrow x\in\varnothing\\ c,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\ d,\Rightarrow x^2=3\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

a) \(\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

b) \(x^2+1=0\)

\(\Rightarrow x^2=-1\left(VLý.do.x^2\ge0\forall x\right)\)

Vậy \(S=\varnothing\)

c) \(\Rightarrow x=\pm\sqrt{2}\)

d) \(\Rightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)

\(\sqrt{\dfrac{0.3+1.2}{0.7}}:\sqrt{\dfrac{0.4}{1.2}}=\sqrt{\dfrac{1.5}{0.7}\cdot\dfrac{1.2}{0.4}}=\dfrac{3\sqrt{35}}{7}\)