\(\dfrac{1}{12}\times\dfrac{4}{5}=\dfrac{4}{60}=\dfrac{1}{15}\\ \dfrac{9}{5}:\dfrac{4}{7}=\dfrac{9}{5}\times\dfrac{7}{4}=\dfrac{63}{20}\\ 4\times\dfrac{3}{7}=\dfrac{4\times3}{7}=\dfrac{12}{7}\\ \dfrac{1}{2}:5=\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{1}{10}\)

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

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

35 : 7 = 5

= 5

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

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

\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)

\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)

\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6}{30}-\dfrac{5}{30}=\dfrac{1}{30}\)

\(\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{7}{42}-\dfrac{6}{42}=\dfrac{1}{42}\)

`@mt`

`@` `\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.`

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

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

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

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

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 + (-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 :>>

1 + (-2) + 3 + (-4) + ... + 19 + (-20)

= (1 - 2) + (3 - 4) + ... + (19 - 20)

= -1 - 1 - ... - 1

= -10

1) \(-6x^4+4x^3-2x^2\)

2) \(=x^2+4x-21-x^2-4x+5=-16\)

3) \(=6x^2-4x-x^2-4x-4=5x^2-8x-4\)

4) \(=2x^3-4x^2-8x-3x^2+6x+12=2x^3-7x^2-2x+12\)

giúp mình với

1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)

\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)

Bài 2,3 tương tự, bạn tự làm nhé!

1: \(=\dfrac{1}{3}\cdot\dfrac{3}{7}\cdot\dfrac{1}{2}=\dfrac{1}{7\cdot2}=\dfrac{1}{14}\)

2: =5/12+1/7

=35/84+12/84=47/84

3: =8(8/9-6/9)

=8*2/9=16/9

4: \(=\dfrac{5}{12}\cdot\dfrac{12}{5}=1\)

5: =16/5+6

=16/5+30/5=46/5

6: =10*1/2-10*1/5

=5-2=3