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

a) \(\sqrt{0,16}+\sqrt{0,04}-\sqrt{0,25}\)

= 0,4 + 0,2 - 0,5 

= 0,1

b) \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)

= \(\sqrt{\left(85-84\right)\left(85+84\right)}\) - \(\sqrt{\left(26-24\right)\left(26+24\right)}\)

= \(\sqrt{169}\) - \(\sqrt{2.50}\)

= 13 - 10

= 3 

 Chúc bạn học tốt

a) Ta có: \(\sqrt{0.16}+\sqrt{0.04}-\sqrt{0.25}\)

\(=0,4+0,2-0,5\)

=0,1

b) Ta có: \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)

=13-10

=3

a: \(A=5\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-1\)

\(B=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)

\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

b: A=B

=>căn x=-căn x+1

=>căn x=1/2

=>x=1/4

a) \(A=\dfrac{1}{\sqrt{25}}+\dfrac{\sqrt{49}}{\sqrt{36}}-\dfrac{2}{\sqrt{100}}.\)

\(=\dfrac{1}{5}+\dfrac{7}{6}-\dfrac{1}{5}.\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\dfrac{7}{6}.\)

\(=0+\dfrac{7}{6}=\dfrac{7}{6}.\)

Vậy \(A=\dfrac{7}{6}.\)

b) \(B=\sqrt{\dfrac{0,01}{1,21}}+3.\dfrac{2}{\sqrt{10^2}+2^2+40}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+3.\dfrac{2}{10+4+40}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+3.\dfrac{1}{37}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+\dfrac{1}{9}-\dfrac{3}{4}.\)

\(=\dfrac{36}{396}+\dfrac{44}{396}-\dfrac{297}{296}.\)

\(=-\dfrac{217}{396}.\)

Vậy \(B=-\dfrac{217}{396}.\)

a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=-\dfrac{2}{\sqrt{x}+1}\)

c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)

d: |B|=A

=>\(\left|-\dfrac{2}{\sqrt{x}+1}\right|=4\)

=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)

=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)

=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)

a)\(\sqrt{0,01}-\sqrt{0,25}\)

=\(\sqrt{\left(0,1\right)^2}-\sqrt{\left(0,5\right)^2}\)

= 0,1 - 0,5 = - 0,4

b)\(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)

= 5 - 0,5

= 4,5.

a) 0,1 - 0,5 = -0,4

b)0,5 . 10 - 0,5 = 5 - 0,5 = 4,5

thầy thông cảm máy em không có dấu căn.

\(\)a)

\(\sqrt{0,01}-\sqrt{0,25}=\sqrt{\frac{1}{100}}-\sqrt{\frac{1}{4}}=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{-4}{10}=\frac{-2}{5}\)

b)

\(0,5.\sqrt{100}-\sqrt{\frac{1}{4}}=0,5.10-\frac{1}{2}=5-\frac{1}{2}=\frac{10}{2}-\frac{1}{2}=\frac{9}{2}\)

ai k mình mình k lại

a) \(\sqrt{0,01}=0,1;\sqrt{0,25}=0,5\)= 0,1-0,5 = -0,4 

b)  = 0,5 x 10 - \(\frac{1}{2}\)= 4,5 . ( Đơn giản nhỉ :) )