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

mình cảm ơn ạ

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) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)

2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)

3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)

4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)

5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)

\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)

\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)

\(=0\cdot\sqrt{6}=0\)

.

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` 

a: =7/4*4/7=1

b: =5/2*2/9*5/3=50/54=25/27

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: Ta có: \(S_1=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2020\right)+2021\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)

\(=-1\cdot1010+2021\)

\(=-1010+2021=1011\)

2) Ta có: \(S_2=\left(-2\right)+4+\left(-6\right)+8+...+\left(-2014\right)+2016\)

\(=\left(-2+4\right)+\left(-6+8\right)+...+\left(-2014+2016\right)\)

\(=2+2+...+2\)

\(=2\cdot504=1008\)

Cho mình cảm ơn bạn NGUYỄN LÊ PHƯỚC THỊNH nhé 

`1/2 -1/6- 1/9`

`= 3/6 -1/6 -1/9`

`= 2/6 -1/9`

`= 1/3-1/9`

`=3/9-1/9`

`=2/9`

\(\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{9}{18}-\dfrac{3}{18}-\dfrac{2}{18}=\dfrac{4}{18}=\dfrac{2}{9}\)

1: \(23+\left(-13\right)+\left(-50\right)\)

\(=23-13-50\)

=10-50

=-40

2: \(-5+15+\left(-123\right)\)

\(=\left(-5+15\right)-123\)

=10-123

=-113

3: \(5871:\left\{928-\left[\left(-82+247\right)\right]\right\}\cdot5\)

\(=5871-\left\{928+82-247\right\}\cdot5\)

\(=5871-763\cdot5=5871-3815=2056\)

4: \(40-\left(4\cdot5^2-3\cdot2^3\right)\)

\(=40-4\cdot5^2+3\cdot2^3\)

\(=40-4\cdot25+3\cdot8\)

=40-100+24

=64-100

=-36

5: \(6^2\cdot5-7^2+149\)

\(=36\cdot5-49+149\)

\(=180+149-49\)

=180+100

=280

6: \(-210:\left[16+3\cdot\left(6+3\cdot2^2\right)\right]+\left(-2022\right)\)

\(=-210:\left[16+3\cdot\left(6+3\cdot4\right)\right]+\left(-2022\right)\)

\(=-210:\left[16+3\cdot18\right]+\left(-2022\right)\)

\(=-210:70-2022\)

=-3-2022

=-2025

7: \(5\cdot2^3+7^{11}:7^9-2023^0\cdot1^8\)

\(=5\cdot8+7^2-1\)

=40+49-1

=39+49

=88

8: \(400:\left\{5\cdot\left[360-\left(290+2\cdot5^2\right)\right]\right\}\)

\(=400:\left\{5\cdot\left[360-290-2\cdot25\right]\right\}\)

\(=400:\left\{5\cdot20\right\}\)

\(=\dfrac{400}{100}=4\)

9: \(75-\left(3\cdot5^2\right)-4\cdot5^3\)

\(=75-3\cdot25-4\cdot5^3\)

=-4*125

=-500

`@` `\text {Ans}`

`\downarrow`

`1.`

\(\left(-4xy\right)\cdot\left(2xy^2-3x^2y\right)\)

`=`\(\left(-4xy\right)\left(2xy^2\right)+\left(-4xy\right)\left(-3x^2y\right)\)

`=`\(-8\left(x\cdot x\right)\left(y\cdot y^2\right)+12\left(x\cdot x^2\right)\left(y\cdot y\right)\)

`=`\(-8x^2y^3+12x^3y^2\)

`2.`

\(\left(-5x\right)\left(3x^3+7x^2-x\right)\)

`=`\(\left(-5x\right)\left(3x^3\right)+\left(-5x\right)\left(7x^2\right)+\left(-5x\right)\left(-x\right)\)

`=`\(-15x^4-35x^3+5x^2\)

`3.`

\(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)

`=`\(3x\left(4x+5\right)-2\left(4x+5\right)-12x^2+6x\)

`=`\(12x^2+15x-8x-10-12x^2+6x\)

`=`\(\left(12x^2-12x^2\right)+\left(15x-8x+6x\right)-10\)

`=`\(13x-10\)

`4.`

\(2x^2\left(x^2-7x+9\right)\)

`=`\(2x^2\cdot x^2+2x^2\cdot\left(-7x\right)+2x^2\cdot9\)

`=`\(2x^4-14x^3+18x^2\)

`5.`

\(\left(3x-5\right)\left(x^2-5x+7\right)\)

`=`\(3x\left(x^2-5x+7\right)-5\left(x^2-5x+7\right)\)

`=`\(3x^3-15x^2+21x-5x^2+25x-35\)

`=`\(3x^3-20x^2+46x-35\)

C xem lại bài cuối ạ.