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29 tháng 7 2021

\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)

\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)

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

a: \(=2x+x^3-5x^4\)

b: \(=\dfrac{8x^2+4x-7x-3}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x^2-3x-3}{\left(2x-1\right)\left(2x+1\right)}\)

AH
Akai Haruma
Giáo viên
12 tháng 3 2022

Lời giải:
a.

$5x-[2x+1-(2x-3)-(4x+1)]=5x-(2x+1-2x+3-4x-1)$

$=5x-(-4x+3)=5x+4x-3=9x-3$

b.

$(-3x^2+2x-1)+(4x^2-2x+3)$

$=-3x^2+2x-1+4x^2-2x+3=x^2+2$

14 tháng 4 2022

cô giải đúng qué

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

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AH
Akai Haruma
Giáo viên
25 tháng 10 2021

Lời giải:

a.

$2x(3x^2-4x+2)=2x.3x^2-2x.4x+2x.2$

$=6x^3-8x^2+4x$

b.

$2x(3x+5)-3(2x^2-2x+3)=2x.3x+2x.5-(3.2x^2-3.2x+3.3)$

$=6x^2+10x-6x^2+6x-9=16x-9$

\(3x\left(4x-3\right)-\left(2x-1\right)\left(6x+5\right)\\ =12x^2-9x-\left(12x^2+10x-6x-5\right)\)

\(=12x^2-9x-\left(12x^2+4x-5\right)\\ =5-13x\)

4 tháng 9 2021

undefined

Ta có:

\(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)\)

\(\Leftrightarrow B=\left(4x+2y-y\right)\left(2x+y\right)=\left(4x+y\right)\left(2x+y\right)=\left(4.\dfrac{1}{2}+\dfrac{-3}{5}\right)\left(2.\dfrac{1}{2}+\dfrac{-3}{5}\right)=\dfrac{14}{25}\)

21 tháng 8 2021

a) 2(x-1)2 - 4(x+3)2 + 2x(x-5)

= 2(x-2x +1)- 4(x2 + 6x +9) + 2x2 -10x

= 2x2 - 4x + 2 -4x2 - 24x - 36 + 2x2 - 10x

= (2x2 + 2x2 - 4x2) - (4x + 24x+10x) +(2-36)

= -38x-34

b) 2(2x+5)2  -3(4x+1)(1-4x)

= 2(4x2 + 20x + 25) + 3(4x+1)(4x-1)

= 8x2 +40x + 50 + 3(16x2 -1)

= 8x2 + 40x + 50 + 48x2 - 3

=56x2 +40x + 47

21 tháng 8 2021

a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)

\(=2\left(x^2-2x+1\right)-4\left(x^2+6x+9\right)+2x\left(x-5\right)\)

\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)

b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)

\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

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

 

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)