`@` `\text {Ans}`

`\downarrow`

\(S=\sqrt{0,49}+\sqrt{\dfrac{1}{9}}-\sqrt{\dfrac{25}{4}}\)

`S=0,7 + 1/3 - 5/2`

`S=31/30 - 5/2 = -22/15`

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)\(\sqrt{0,36}\)+\(\sqrt{0,49}\)=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) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{\left(\frac{5}{6}\right)^2}=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}\)

ẹc, mình chưa học tới bài căn bậc :V

\(\sqrt{\frac{9}{16}:\frac{0,49}{0,64}}=\frac{6}{7}\)

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) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\sqrt{\left(-5\right)^2}=\sqrt{25}=5\)

\(\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)

\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=5+9-5\cdot0,7=10,5\)

a)\(\sqrt{\left(-5\right)^2}=\sqrt{25}=5=-5\)

b)\(\sqrt{\left(3-\sqrt{5}\right)^2}=\sqrt{\left(3^2-2.3.\sqrt{5}+\sqrt{5}^2\right)}=\sqrt{\left(9-6.\sqrt{5}+5\right)}=\sqrt{14-\sqrt{5}}\)

c)\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=5+9-5.0,7=14-3,5=10,5\)

a)\(\sqrt{\left(-5\right)^2}=\sqrt{25}=\pm5\)

b)\(\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)

c)\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=\sqrt{5^2}+\sqrt{9^2}-5\sqrt{\left(0,7\right)^2}=5+9-5\cdot0,7=10,5\)

A, D là đúng. còn đâu sai

\(\begin{array}{l}a)\sqrt {0,49}  + \sqrt {0,64}  = 0,7 + 0,8 = 1,5;\\b)\sqrt {0,36}  - \sqrt {0,81}  = 0,6 - 0,9 =  - 0,3;\\c)8.\sqrt 9  - \sqrt {64}  = 8.3 - 8 = 24 - 8 = 16;\\d)0,1.\sqrt {400}  + 0,2.\sqrt {1600}  = 0,1.20 + 0,2.40 = 2 + 8 = 10\end{array}\)