Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Chú ý: Trong các kết quả trên, hai kết quả đầu là căn bậc hai đúng, hai kết quả cuối là căn bậc hai gần đúng chính xác đến 6 chữ số thập phân (được làm tròn đến chữ số thập phân thứ sáu)
a: \(C=25\cdot\dfrac{-1}{125}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-1}{5}+\dfrac{1}{5}-\dfrac{1}{2}-\dfrac{1}{2}\)
=-1
b: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
\(=20-12+5-6=7\)
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/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{25}}\)
\(=0,5.10-0,2\)
\(=5-0,2=4,8\)
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~
\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}\)
\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\.............\\\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\end{matrix}\right.\)
Suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{99}{\sqrt{100}}\)
\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{99}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}\)
\(\)\(linh>10\left(đpcm\right)\)
Bài này ko phải 100 nhé
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ 
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
.............................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.........+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+.....+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
\(=10.\dfrac{1}{10}.\dfrac{4}{3}+3.7-\dfrac{1}{6}.2\)
\(=1.\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{28}{21}+\dfrac{441}{21}+\dfrac{7}{21}\)
\(=\dfrac{469}{21}+\dfrac{7}{21}\)
\(=\dfrac{68}{3}\)
\(10.\sqrt{0,01}.\sqrt{\dfrac{16}{9}}+3.\sqrt{49}-\dfrac{1}{6}.\sqrt{4}\)
= 10 . 0,1 . \(\dfrac{4}{3}\) + 3. 7 - \(\dfrac{1}{6}.2\)
= 1 . \(\dfrac{4}{3}\) + 21 - \(\dfrac{1}{3}\)
= \(\dfrac{4}{3}+21-\dfrac{1}{3}\)
= 22
\(\sqrt{3783025}=1945\)
\(\sqrt{1125\cdot45}=\sqrt{50625}=225\)
\(\sqrt{\dfrac{0,3+1,2}{0,7}}=\sqrt{\dfrac{15}{7}}=\dfrac{\sqrt{105}}{7}\)
\(\sqrt{\dfrac{6,4}{1,2}}=\sqrt{\dfrac{16}{3}}=\dfrac{4\sqrt{3}}{3}\)
\(\sqrt{3783025}=1945\) \(\sqrt{1125.45}=\sqrt{50625}=225\) \(\sqrt{\dfrac{0,3+1,2}{0,7}}=\sqrt{\dfrac{1,5}{0,7}}=\sqrt{\dfrac{105}{7}}=1,4638...\) \(\sqrt{\dfrac{6,4}{1,2}}=\sqrt{\dfrac{16}{3}}=5,\left(3\right)\)