thực hiện phép tính (2x^2 + 4x) : 2x
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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.
$(-3x^2+2x-1)+(4x^2-2x+3)$
$=-3x^2+2x-1+4x^2-2x+3=x^2+2$
=3x(x^2-2)(3x^2+x-2)
=(3x^3-6x)(3x^2+x-2)
=9x^5+3x^4-6x^3-18x^3-6x^2+12x
=9x^5+3x^4-12x^3-6x^2+12x
2x(x^2-1)=2x^3-2x
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
\(\left(2x-1\right).\left(4x^2+2x+1\right)\)
\(=8x^3+4x+2x-4x^2-2x-1\)
\(=8x^3-4x^2+4x-1\)
a) 3x.(x² - 2)
= 3x.x² + 3x.(-2)
= 3x³ - 6x
b) (6x³ + 2x² - 4x) : 2x
= 6x³ : 2x + 2x² : 2x - 4x : 2x
= 3x² + x - 2
c) 2x(x² - 1)
= 2x.x² - 2x.1
= 2x³ - 2x
c: \(\dfrac{4x^3-8x^2+13x-5}{2x-1}=\dfrac{4x^3-2x^2-6x^2+3x+10x-5}{2x-1}\)
=2x^2-3x+5
\(\left(3x^{n+1}-2x^n\right).4x^2=12x^{n+3}-8x^{n+2}\)
a) 4x²(x² - 5x + 2)
= 4x².x² - 4x².5x + 4x².2
= 4x⁴ - 20x³ + 8x²
b) (2x² - 5x + 3) : (2x - 3)
= (2x² - 3x - 2x + 3) : (2x - 3)
= [(2x² - 3x) - (2x - 3)] : (2x - 3)
= [x(2x - 3) - (2x - 3)] : (2x - 3)
= (2x - 3)(x - 1) : (2x - 3)
= x - 1
\(\left(2x^2+4x\right):2x\)
\(=\left(2x.x+2x.2\right):2x\)
\(=\left(2x.x+2x.2\right):2x\)
\(=x+2\)