Tính
1+1/3+/19+1/27+1/81
37,52+4,7×2,3_9,8
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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 \(1\)
\(1)\) \(1-5+7-8+4-1+5-7+8\)
\(=(1-1)+(5-5)+(7-7)+(8-8)\)
\(=0+0+0+0\)
\(=0\)
\(2)\) \(14-23+(5+14)-(5-23)+17\)
\(=14-23+5+14-5+23+17\)
\(=(14+14)+(23-23)+(5-5)+17\)
\(=28+17\)
\(=45\)
\(3)\) \(12-44+9-3+14-19-9-12\)
\(=(12-12)+(9-9)+(14-44)+3\)
\(=-30+3\)
\(=-33\)
\(4)\) \(22-(4-8+12)+(-8-12+4)\)
\(=22-4+8-12-8-12+4\)
\(=22+(4+4)+(8-8)+(-12-12)\)
\(=22-24\)
\(=-2\)
A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
A = \(\frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\)
A = \(\frac{81+27+9+3+1}{81}\)
A = \(\frac{121}{81}\)
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: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)
\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)
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: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)
1 + (-2) + 3 + (-4) + ... + 19 + (-20)
= (1 - 2) + (3 - 4) + ... + (19 - 20)
= -1 - 1 - ... - 1 (10 chữ số 1)
= -10
đặt A=1+(-2)+3+(-4)+...+19+(-20)
A=(1+3+..+19)-(2+4+..+20) (1)
Đặt B=1+3+...+19
số các số hạng của tôngB là
(19-1) :2+1=10 (số hạng)
tổng B là
B=(19+1).10 / 2
B=100 (2)
Đặt C=2+4+...+20
số các số hạng của tổng C là
(20-2) :2+1=10 (số hạng)
tổng C là
C= (20+2).10 / 2
C= 110 (3)
thay (2) và (3) vào (1) ta được:
A= 100-110
A=-10
Vậy 1 + (-2) + 3 + (-4) +....+19 +(-20)=-10
CHÚC BN HỌC TỐT :>>
Đáp án :
`-1/2`
Giải thích các bước giải :
`1/2` - (`1/3` + `2/3`)
= `1/2` - 1
= `-1/2`
________________________
Đáp án :
`7/9`
Giải thích các bước giải :
`4/9` + `1/2` . `2/3`
= `4/9` + `1/3`
= `12/27` + `9/27`
= `21/27` = `7/9`
\(1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6}{30}-\dfrac{5}{30}=\dfrac{1}{30}\)
\(\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{7}{42}-\dfrac{6}{42}=\dfrac{1}{42}\)
`@mt`
\(1,=\dfrac{1}{12}-\dfrac{4}{12}=\dfrac{3}{12}\\ 2,=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\\ 3,\dfrac{-1}{6}-\dfrac{2}{5}=\dfrac{-5}{30}-\dfrac{12}{30}=\dfrac{-17}{30}\\ 4,=\dfrac{15}{20}-\dfrac{1}{20}=\dfrac{14}{20}\)
1,=112−412=3122,=636−136=5363,−16−25=−530−1230=−17304,=1520−120=1420
\(\dfrac{11}{2}.\dfrac{21}{3}+\dfrac{11}{3}.\dfrac{1}{2}\)
\(=\dfrac{77}{2}+\dfrac{11}{6}=\dfrac{121}{3}\)
bn ơi phép thứ nhất /19 là sao mình ko hiểu