0,5*\(\sqrt{100}\)-\(\sqrt{25}\)
so sánh các số sau: a,\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}và\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\)
\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5.10-\frac{\sqrt{4}}{\sqrt{25}}=5-\frac{2}{5}=\frac{23}{5}=\frac{138}{30}\)
\(\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5=\left(\sqrt{\frac{10}{9}-\frac{3}{4}}\right):5=\sqrt{\frac{13}{36}}:5=\frac{\sqrt{13}}{6}:5=\frac{\sqrt{13}}{30}\)
Vì 13 < 138 nên \(\sqrt{13}< 138\Rightarrow\frac{\sqrt{13}}{30}< \frac{138}{30}\)
Vậy \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}>\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\).
So sánh các số sau:
a) \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}v\text{à}\left(\sqrt{1\frac{1}{9}}-\sqrt{\frac{9}{16}}\right):5\)
b) \(\sqrt{25+9}v\text{à}\sqrt{25}+\sqrt{9}\)
Bài 1 : Tính giá trị biểu thức :
1/ 0,2.\(\sqrt{100}\) -\(\sqrt{\dfrac{16}{25}}\)
2/ \(\dfrac{2^7.9^{3^{ }}}{6^5.8^2}\)
3/\(\sqrt{0,01}\) - \(\sqrt{0,25}\)
4/ 0,5 . \(\sqrt{100}\) - \(\sqrt{\dfrac{1}{4}}\)
5/ 7. \(\sqrt{0,01}\) + 2.\(\sqrt{0,25}\)
6/ 0,5.\(\sqrt{100}\) - \(\sqrt{\dfrac{1}{25}}\)
1.
0,2 . \(\sqrt{100}\) - \(\sqrt{\dfrac{16}{25}}\)
= 0,2 . 10 - \(\dfrac{4}{5}\)
= 2 - \(\dfrac{4}{5}\)
= \(\dfrac{6}{5}\)
1/ \(0,2.\sqrt{100}-\sqrt{\dfrac{16}{25}}\)
\(=0,2.10-0,8\)
\(=2-0,8=1,2\)
2/ \(\dfrac{2^7.9^3}{6^5.8^2}\)
\(=\dfrac{93312}{497664}=\dfrac{3}{16}=0,1875\)
3/ \(\sqrt{0,01}-\sqrt{0,25}\)
\(=0,1-0,5\)
\(=-0,4\)
4/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)
\(=0,5.10-0,5\)
\(=5-0,5=4,5\)
5/ \(7.\sqrt{0,01}+2.\sqrt{0,25}\)
\(=7.0,1+2.0,5\)
\(=0,7+1=1,7\)
6/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{25}}\)
\(=0,5.10-0,2\)
\(=5-0,2=4,8\)
So sánh \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}\) và \((\sqrt{\frac{11}{9}-\sqrt{\frac{9}{16}}})\div5\)
\(0,5\sqrt{100}-\sqrt[]{\frac{4}{25}}va\left(\sqrt{\frac{10}{9}}-\sqrt{\frac{9}{16}}\right):5\)
so sanh
(1+1_1/4+1_1/2+1_3/4+2+2_1/4+2_1/2+2_3/4+......+4_3/4) : 23 tính nhanh !!!!! Ai nhanh mình kêu anh chị mình vào nha !! Mình cần gấp lắm !! Cảm ơn mọi người
1)so sánh các số sau:
a)0,5\(\sqrt{100}\)-\(\sqrt{\dfrac{4}{25}}\) và (\(\sqrt{1\dfrac{1}{9}}\)-\(\sqrt{\dfrac{9}{16}}\)):5
b)\(\sqrt{25+9}\) và \(\sqrt{25}+\sqrt{9}\)
2) CMR: Với a,b dương thì \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
1.
a. \(0,5\sqrt{100}-\sqrt{\dfrac{4}{25}}=5-\dfrac{2}{5}=\dfrac{23}{5}>1\)
\(\dfrac{\left(\sqrt{1\dfrac{1}{9}}-\sqrt{\dfrac{9}{16}}\right)}{5}=\dfrac{\dfrac{\sqrt{10}}{3}-\dfrac{3}{4}}{5}=\dfrac{-9+4\sqrt{10}}{60}\approx0,06< 1\)
\(\Rightarrow0,5\sqrt{100}-\sqrt{\dfrac{4}{25}}>\dfrac{\left(\sqrt{1\dfrac{1}{9}}-\sqrt{\dfrac{9}{16}}\right)}{5}\)
2.
Ta có:
\(\left(\sqrt{a+b}\right)^2=a+b\)
\(\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2+2\sqrt{ab}+\left(\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
=> \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
1b.
Áp dụng công thức trên
=> \(\sqrt{25+9}< \sqrt{25}+\sqrt{9}\)
2.
\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\\ \Rightarrow a+b< a+2\sqrt{ab}+b\\ \Rightarrow2\sqrt{ab}>0\\ \Rightarrow\sqrt{ab}>0\)
Luôn đúng với mọi a;b dươn g
=> đpcm
a) \(\dfrac{2}{5}\sqrt{25}\) -\(\dfrac{1}{2}\sqrt{4}\) b)0,5\(\sqrt{0,09}\) +5\(\sqrt{0,81}\) c)\(\dfrac{2}{5}\sqrt{\dfrac{25}{36}}\) -\(\dfrac{5}{2}\sqrt{\dfrac{4}{25}}\)
d)-2\(\sqrt{\dfrac{-36}{-16}}\) + 5\(\sqrt{\dfrac{-81}{-25}}\)
`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
100 x \(\sqrt{0,01}\) - 0,5 x \(\sqrt{400}\) + \(\sqrt{0,09}\)
\(100\times\sqrt{0,01}-0,5\times\sqrt{400}+\sqrt{0,09}=100\times0,1-0,5\times20+0,3\)
\(=10-10+0,3=0+0,3=0,3\)
a) \(\sqrt{0,5}-\sqrt{0,25}\)
b) \(0,5.\sqrt{100}-\sqrt{\frac{1}{4}}\)
a, = \(\sqrt{2}\)/2 - 1/2 = \(\sqrt{2}-1\)/2
b, = 0,5 . 10 - 1/2 = 5 - 1/2 =9/2
Đáp án là:
a) = \(\frac{-1+\sqrt{2}}{2}\) .
b) = \(\frac{9}{2}\) .