Thực hiện các phép tính sau: x - 2 x 3 + 8 . x 2 + x - 2 x 3 - 4 x 2 + 4 x
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Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{6.2}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}\\ =\dfrac{3\left(4+x\right)}{2x\left(x+4\right)}\\ =\dfrac{3}{2x}\)
________
\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\\ \left(\text{đ}k\text{x}\text{đ}:x\ne\pm2\right)\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{x^2-4}\\ =2\)
a: \(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: \(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-2\right)^2+x-14}{x^2-4}\)
\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{x^2-4}=\dfrac{2x^2-8}{x^2-4}=2\)
Ta có : 1x2x3x...x9-1x2x3x...x8-1x2x3x...x8^2
=1x2x3x...x8x(9-1-8)
=1x2x3x...x8x0
=0
Nhớ k cho mik nha !!!
\(=\dfrac{x^2+2x-3x+6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2\left(x-2\right)\left(x+2\right)}{x-3}\)
\(=\dfrac{2\left(x^2-x+6\right)}{x-3}\)
a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)
=(x-3)^2/(x-3)
=x-3
b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)
\(=\dfrac{2}{x+2}-\dfrac{4}{x-2}-\dfrac{5x+2}{x^2-4}\)
\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)
=-7(x+2)/(x-2)(x+2)
=-7/(x-2)
a)
160 : (4 × 8) = 160 : 32 = 5
160 : 4 : 8 = 40 : 8 = 5
Vậy 160 : (4 × 8) = 160 : 4 : 8.
96 : (3 × 8) = 96 : 24 = 4
96 : 3 : 8 = 32 : 8 = 4
Vậy 96 : (3 × 8) = 96 : 3 : 8.
105 : (5 × 7) = 105 : 35 = 3
105 : 5 : 7 = 21 : 7 = 3
Vậy 105 : (5 × 7) = 105 : 5 : 7.
Nhận xét: a : (b × c) = a : b : c
b)
270 : (9 × 6) = 270 : 9 : 6 = 30 : 6 = 5
420 : (7 × 3) = 420 : 7 : 3 = 60 : 3 = 20
144 : (2 × 8) = 144 : 2 : 8 = 72 : 8 = 9
Ta có: