3/15 chia cả mẫu cả tử cho 3 được 1/5+2/5=3/5 nhé

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

a) \(\dfrac{x^2+xy}{x^2-y^2}=\dfrac{x\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{x}{x-y}\)

b)\(\dfrac{x}{x-1}-\dfrac{x}{x+1}+\dfrac{2}{x^2-1}=\dfrac{x\left(x+1\right)-x\left(x-1\right)+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x-1}\)

b) x^2+xy/x^2-y^2= x(x+y) /(x+y) (x-y) =x/x-y

c) =2/x-1

Đây bạn nhé!

\(=\dfrac{x^2+2x-x^2+4x-4+6-5x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

a=52+53/71-180

2x ( x - 5 ) x . ( 3 - 2x ) = 26

2x\(^2\)- 10x . 3x - 2x\(^2\)= 26

2x\(^2\). ( 10x - 3x ) = 26

2x\(^2\). 7x              = 26

14x\(^3\)                   = 26

x\(^3\)                        = 26 : 14

x\(^3\)                       = \(\frac{13}{7}\)

→ X = 1.229.... \(\approx\)1,3

\(2x\left(x-5\right)\cdot x\left(3-2x\right)=26\)

\(\Leftrightarrow\left(2x^2-10x\right)\left(3x-6x\right)=26\)

\(\Leftrightarrow6x^3-30x^2-12x^3+60x^2=26\)

\(\Leftrightarrow-12x^3+30x^2=26\)

\(\Leftrightarrow2\left(-6x^3+15x^2\right)=26\)

\(\Leftrightarrow-6x^3+15x^2=13\)

\(\Leftrightarrow-6x^3+15x^2-13=0\)

...............mình chỉ làm được đến đây thôi!

Mơn nhiều nha :D

5) Ta có: \(\dfrac{x^3-x^2-2x-20}{x^2-4}-\dfrac{5}{x+2}+\dfrac{3}{x-2}\)

\(=\dfrac{x^3-x^2-2x-20}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^3-x^2-2x-20-5x+10+3x+6}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{\left(x-1\right)\left(x^2-4\right)}{\left(x^2-4\right)}\)

\(=x-1\)

6) Ta có: \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)

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

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

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

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

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

(6x+1)^2 +(6x-1)^2-2(1+6x)(6x-1)

Bài đó làm thế nào?

\(a,2x^2+6x=2x\left(x+3\right)\\ b,x^2+2xy+y^2-9z^2\\ =\left(x^2+2xy+y^2\right)-\left(3z\right)^2\\ =\left(x+y\right)^2-\left(3z\right)^2\\ =\left(x+y-3z\right)\left(x+y+3z\right)\\ b,x^3-2x^2+x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ d,x^2-2x-15=x^2-5x+3x-15\\ =x\left(x-5\right)+3\left(x-5\right)\\ =\left(x+3\right)\left(x-5\right)\)

a) 2x² + 6x

=2x (x+3)

b) x² + 2xy +y²- 9z²

(x² + 2xy +y2)- 9z²

(x + y)² - (3z)²

(x+y-3z) (x+y+3z) 

\(=2x^2-6x+x^2-1=x^2-6x-1\)

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

\(=2x^2-6x+x^2-1\)

\(=3x^2-6x+1\)

1335/157

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