:((

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2² + (x - 3) = 5²

4 + (x - 3) = 25

x - 3 = 25 - 4

x - 3 = 21

x = 21 + 3

x = 24

2² + ( x - 3 ) = 5²

<=> 4 + ( x - 3 ) = 25

<=> x - 3 = 21

<=> x = 24

9x - 2 . 3² = 3⁴

<=>9x - 2 = 3⁴ : 3²

<=>9x - 2 = 9

<=>9x = 11

<=> x = 11/9

10x + 2² . 5 = 10²

<=> 10x + 4 . 5 = 100

<=> 10x + 4 = 20

<=> 10x = 16

<=> x = 1,6

diễn giải ra nhé

X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000

Câu 1:

a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$

b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)

c.

\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)

d.

\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)

Câu 2:

a. $x\times \frac{2}{7}=\frac{6}{11}$

$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$

b. $x: \frac{3}{2}=\frac{1}{4}$

$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$

Câu 3:

Coi số lớn là 5 phần thì số bé là 3 phần. Hiệu số phần bằng nhau là:
$5-3=2$ (phần)

Số lớn là: $36:2\times 5=90$

Số bé là: $36:2\times 3=54$

1)1/9 x 3^x = 2187:81=27

            3^x=27:1/9=243=3⁵

            =>x=5

\(\frac{1}{9}.3^4.3^x=3^7\)

\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

1) \(x^3+2x-3\)

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

\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+3\right)\)

2) \(x^3-6x+4\)

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

\(=x^2\left(x-2\right)+2x\left(x-2\right)-2\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+2x-2\right)\)

3) \(x^3-2x^2+1\)

\(=\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)\)

\(=x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-x-1\right)\)

4) \(x^3+5x^2-12\)

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

\(=x^2\left(x+2\right)+3x\left(x+2\right)-6\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+3x-6\right)\)

5) \(x^3-6x^2+9x\) (chắc đề như vậy)

\(=x\left(x^2-6x+9\right)\)

\(=x\left(x-3\right)^2\)

6) \(4x^3-9x^2+5x\)

\(=x\left(4x^2-9x+5\right)\)

\(=x\left[\left(4x^2-4x\right)-\left(5x-5\right)\right]\)

\(=x\left[4x\left(x-1\right)-5\left(x-1\right)\right]\)

\(=x\left(x-1\right)\left(4x-5\right)\)