Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bai 1
a) \(\sqrt{0,36}+\sqrt{0,49}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\frac{2}{3}-\frac{5}{6}\)
=\(-\frac{1}{6}\)
Bài 2
a)\(x^2=81\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
b) \(\left(x-1\right)^2=\frac{9}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{3}{4}\\x-1=\frac{-3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
c) \(x-2\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
Ta có:
\(\sqrt{0,49}+\sqrt{\frac{25}{36}}=\sqrt{0,7^2}+\sqrt{\left(\frac{5}{6}\right)^2}=0,7+\frac{5}{6}=\frac{23}{15}\)
a) \(C=3\cdot\sqrt{25}-3\cdot\sqrt{\frac{1}{9}}\)
\(C=3\cdot5-3\cdot\frac{1}{3}\)
\(C=15-1=14\)
b) \(D=-4\sqrt{\frac{4}{25}}+3\sqrt{0,16}-2\sqrt{0,04}\)
\(D=-4\cdot\frac{2}{5}+3\cdot\frac{2}{5}-2\cdot\frac{1}{5}\)
\(D=\frac{1}{5}\cdot\left(-8+6-2\right)\)
\(D=\frac{1}{5}\cdot\left(-4\right)=-\frac{4}{5}\)
1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)
b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)
c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)
d) đề baì có sai ko ban?
= 0,6 : 5/4 + 1/4 + 2/9 : 5/9 - 1/4
= 3/5 . 4/5 + 2/9 . 9/5
= 12/25 + 2/5
= 22/25
a)\(\sqrt{0,36}\)+\(\sqrt{0,49}\)=0,6+0,7=1,3
b)\(\sqrt{\frac{4}{9}}\)-\(\sqrt{\frac{25}{36}}\)=2/3-5/6=4/6-5/6=-1/6
a) \(\sqrt{0,36}+\sqrt{0,49}=\sqrt{\left(0,6\right)^2}+\sqrt{\left(0,7\right)^2}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{\left(\frac{5}{6}\right)^2}=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}\)