1800 nha

đúng k nhỉ

\(1+2+3+4+5\times1\times2\times3\times4\times5\)

=\(1+2+3+4+\left(5\times1\times2\times3\times4\times5\right)\)

\(=10+600\)

\(=610\)

ban sai rùi nha

bang 450

câu 1 bỏ dấu ngoặc rồi tính

( 36 + 79 ) + ( 145 _ 79 _ 36 )

\(=36+79+145-79-36\)

\(=\left(36-36\right)+\left(79-79\right)+145\)\

\(=0+0+145=145\)

10 _ [ 12 _ ( -9 _ 1 ) ]

\(=10-12-10\)

\(=10-10-12\)

\(=0-12=-12\)

( 38 _ 29 + 43) _ ( 43 + 38 )

\(=38-29+43-43-38\)

\(=\left(38-38\right)+\left(43-43\right)-29\)

\(=0+0-29=-29\)

271 _ [ ( -43 ) + 271 _ ( -17 ) ]

\(=271+43-271-17\)

\(=\left(271-271\right)+\left(43-17\right)\)

\(=0+26=26\)

- 144 _ [ 29 _ ( + 144 ) _ ( + 144 )]

\(=-144-19+144+144\)

\(=\left(-144+144+144\right)-19\)

\(=144-19=125\)

đợi mk lm tiếp câu 2 nha .

bài 2 tính tổng các số nguyên

- 18 < hoặc bằng x < hoặc bằng 17

\(\Rightarrow x\in\left\{-18;-17;-16;....;17\right\}\)

tổng \(x=-18+\left(-17\right)+\left(-16\right)+...+17=-18\)

- 27 < hoặc bằng x < hoặc bằng 27

\(\Rightarrow x\in\left\{-27;-26;-25;..;27\right\}\)

Tổng \(x=-27+\left(-26\right)+\left(-25\right)+...+27=0\)

câu 3 tìm x

- 16 + 23 + x = -16

\(x=-16-23+16\)

\(x=-23\)

2x _ 35 = 15

\(2x=15+35\)

\(2x=50\)

\(x=25\)

3x + 17 = 12

\(3x=29\)

\(x=\frac{29}{3}\)

dấu giá trị tuyệt đối x - 1 dấu giá trị tuyệt đối = 0

\(\Rightarrow x-1=0\)

\(x=1\)

- 13 nhân dấu giá trị tuyệt đối  x dấu giá trị tuyệt đối = -26

\(|x|=-26:\left(-13\right)\)

\(|x|=2\)

\(\Rightarrow x\in\left\{\pm2\right\}\)

câu 4 tính hợp lí

35 nhân 18 _ 5 nhân 7 nhân 28

\(=35.18-35.28\)

\(=35.\left(18-28\right)\)

\(=35.\left(-10\right)\)

\(=-350\)

45 _ 5 nhân ( 12 + 9 )

\(=45-5.21\)

\(=45-105=-60\)

24 nhân ( 16 _ 5 ) _ 16 nhân ( 24 _ 5 )

\(=24.16-24.5-16.24+16.5\)

\(=24.\left(16-16\right)+5.\left(-24+16\right)\)

\(=24.0+5.\left(-8\right)\)

\(=0-40=-40\)

29 nhân ( 19 _ 13 ) _ 19 nhân ( 29 _ 13 )

\(=29.19-29.13-19.29+19.13\)

\(=19.\left(29-29\right)+13.\left(-29+19\right)\)

\(=19.0+13.\left(-10\right)\)

\(=0+\left(-130\right)=-130\)

31 nhân ( -18 ) + 31 nhân ( -81 ) _ 31

\(=31.\left[\left(-18\right)+\left(-81\right)-1\right]\)

\(=31.\left(-100\right)=-3100\)

( - 12) nhân 47 + ( -12 ) nhân 52 + ( -12 )

\(=\left(-12\right).\left(47+52+1\right)\)

\(=\left(-12\right).100=-1200\)

13 nhân ( 23 + 22 ) _ 3 nhân ( 17 + 28 )

\(=13.45-3.45\)

\(=45.\left(13-3\right)\)

\(=45.10=450\)

- 48 + 48 nhân ( -78 ) + 48 nhân ( -21)

\(=48.\left(-1-78-21\right)\)

\(=48.\left(-100\right)\)

\(=-4800\)

chu so

tan cung

la 0

     \(\frac{3}{4}=\frac{6}{8}\)                      \(\frac{2}{3}=\frac{6}{9}\)

Ta có sơ đồ :

Tổ 1 : |----|----|----|----|----|----|----|----|----|

Tổ 2 : |----|----|----|----|----|----|----|----|

Tổ 1 có :

5: ( 9 - 8 ) x 9 = 45 ( người )

Tổ 2 có :

45 - 5 = 40 người

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

\(x=\dfrac{9}{12}-\dfrac{2}{12}\)

\(x=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)

\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)

\(x=\dfrac{1009}{2020}\)

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)

\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)

a. \(2^4\cdot3^2\)
b. \(x^5\)

c. \(3^4\cdot4^2\cdot2^2\)