Tính
1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
1: =1-1+5-5+7-7+8-8=0
2: =14-23+5+14-5+23+17
=28+17=45
3: =12-12+9-9+14-44-3=-33
4: =22-8-8-12+4
=22-16-8
=-2
Bài \(1\)
\(1)\) \(1-5+7-8+4-1+5-7+8\)
\(=(1-1)+(5-5)+(7-7)+(8-8)\)
\(=0+0+0+0\)
\(=0\)
\(2)\) \(14-23+(5+14)-(5-23)+17\)
\(=14-23+5+14-5+23+17\)
\(=(14+14)+(23-23)+(5-5)+17\)
\(=28+17\)
\(=45\)
\(3)\) \(12-44+9-3+14-19-9-12\)
\(=(12-12)+(9-9)+(14-44)+3\)
\(=-30+3\)
\(=-33\)
\(4)\) \(22-(4-8+12)+(-8-12+4)\)
\(=22-4+8-12-8-12+4\)
\(=22+(4+4)+(8-8)+(-12-12)\)
\(=22-24\)
\(=-2\)
1: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)
\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)
2: \(\sqrt{28}-4\cdot\sqrt{63}+7\cdot\sqrt{112}\)
\(=2\sqrt7-4\cdot3\sqrt7+7\cdot4\sqrt7\)
\(=2\sqrt7-12\sqrt7+28\sqrt7=18\sqrt7\)
3: \(\sqrt{49}-5\cdot\sqrt{28}+\frac12\cdot\sqrt{63}\)
\(=7-5\cdot2\sqrt7+\frac12\cdot3\sqrt7=7-10\sqrt7+1,5\sqrt7=7-8,5\cdot\sqrt7\)
4: \(\left(2\sqrt6-4\sqrt3-\frac14\sqrt8\right)\cdot3\sqrt6\)
\(=6\cdot\sqrt{36}-12\sqrt{18}-\frac14\cdot2\cdot\sqrt2\cdot3\sqrt6\)
\(=36-36\sqrt2-\frac32\sqrt{12}=36-36\sqrt2-3\sqrt3\)
6: \(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt3=\sqrt{\frac{48}{3}}-3\cdot\sqrt{\frac{27}{3}}-\sqrt{\frac{147}{3}}\)
\(=\sqrt{16}-3\cdot\sqrt9-\sqrt{49}=4-3\cdot3-7\)
=-3-9
=-12
5: \(\left(2\cdot\sqrt{1\frac{9}{16}}-5\cdot\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)
\(=\left(2\cdot\sqrt{\frac{25}{16}}-5\cdot\sqrt{\frac{81}{16}}\right):\sqrt{16}\)
\(=\left(2\cdot\frac54-5\cdot\frac94\right):4=\frac{10-45}{4\cdot4}=\frac{-35}{16}\)
7: \(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt2-\sqrt{162}:\sqrt2\)
\(=\left(5\sqrt2-21\right):\sqrt2-9\sqrt2:\sqrt2\)
\(=5-\frac{21}{\sqrt2}-9=-4-\frac{21\sqrt2}{2}=\frac{-8-21\sqrt2}{2}\)
8: \(\left(2\cdot\sqrt{1\frac{9}{10}}-\sqrt{5\frac{1}{10}}\right):\sqrt{10}=\left(2\cdot\sqrt{\frac{19}{10}}-\sqrt{\frac{51}{10}}:\sqrt{10}\right)\)
\(=2\cdot\frac{\sqrt{19}}{10}-\frac{\sqrt{51}}{10}=\frac{2\sqrt{19}-\sqrt{51}}{10}\)
9: \(2\cdot\sqrt{\frac{16}{3}}-3\cdot\sqrt{\frac{1}{27}}-6\cdot\sqrt{\frac{4}{75}}\)
\(=2\cdot\frac{4}{\sqrt3}-3\cdot\frac{1}{3\sqrt3}-6\cdot\frac{2}{5\sqrt3}\)
\(=\frac{8}{\sqrt3}-\frac{1}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7\sqrt3}{3}-\frac{12\sqrt3}{15}=\frac{35\sqrt3-12\sqrt3}{15}=\frac{23\sqrt3}{15}\)
10: \(2\sqrt{27}-6\cdot\sqrt{\frac43}+\frac35\cdot\sqrt{75}\)
\(=2\cdot3\sqrt3-6\cdot\frac{2}{\sqrt3}+\frac35\cdot5\sqrt3\)
\(=6\sqrt3-4\sqrt3+3\sqrt3=5\sqrt3\)
11: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)
14dm5cm=14,5dm;3dm7cm=3,7dm
chu vi hình chữ nhật đó là:
(14,5+3,7)x2=36,4(dm)
ĐS:36,4dm
14 dm 5 cm = 14,5 dm
3 dm 7 cm = 3,7 dm
Chiều rộng HCN là :
14,5 - 3,7 = 10,8 ( dm )
chu vi HCN là :
( 14,5 + 10,8 ) x 2 = 50,6 ( dm )
ĐS:..
2 = 1 + 1 6 = 2 + 4 8 = 5 + 3 10 = 8 + 2
3 = 1 + 2 6 = 3 + 3 8 = 4 + 4 10 = 7 + 3
4 = 3 + 1 7 = 1 + 6 9 = 8 + 1 10 = 6 + 4
4 = 2 + 2 7 = 5 + 2 9 = 6+ 3 10 = 5 + 5
5 = 4 + 1 7 = 4 + 3 9 = 7 + 2 10 = 10 + 0
5 = 3 + 2 8 = 7 + 1 9 = 5 + 4 10 = 0 + 10
6 = 5 + 1 8 = 6 + 2 10 = 9 + 1 1 = 1 + 0
1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9
= 1/3 - 1/9
= bam may tinh di hihi :3
#Giải :
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{3}-\frac{1}{9}=\frac{9}{27}-\frac{3}{27}=\frac{6}{27}\)
P/S : Bước 1 có thể viết hết,xin lỗi vì không viết đầu bài.
#By_Ami.