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H
2
LP
21 tháng 12 2021
\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)
21 tháng 12 2021
\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)
TP
2
17 tháng 9 2021
\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)
17 tháng 9 2021
1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)
2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)
3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)
2 tháng 12 2023
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
28 tháng 7 2021
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
TT
1
NT
14 tháng 10 2021
1.(2x+3).(x-5)+2x(3-x)+x-10
=2x^2 -10x+3x-15+6x-2x^2+x-10
=2x-25
2.(-x-2)3+(2x-4).(x2+2x+4)-x2.(x-6)
=-x^3+6x^2-12x-8+2x^3+4x^2+8x-4x^2+8x-16-x^3+6x^2
23 tháng 12 2021
a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)
b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)
23 tháng 12 2021
\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)
2 tháng 12 2023
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:
LN
1
1 tháng 1 2022
\(=\dfrac{2x+6}{x\left(3x-1\right)}+\dfrac{x+3}{3x-1}\)
\(=\dfrac{2x+6+x^2+3x}{x\left(3x-1\right)}\)
\(=\dfrac{x^2+5x+6}{x\left(3x-1\right)}\)
NT
1
24 tháng 12 2021
\(=\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}+\dfrac{2-3x}{\left(x+1\right)^2}\\ =\dfrac{\left(3x+2\right)\left(x+1\right)^2-6\left(x^2-1\right)+\left(2-3x\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\\ =\dfrac{10x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(\left(x-6\right)\left(x+6\right)-2x\left(x+6\right)+\left(x+6\right)^2=x^2-36-2x^2-12x+x^2+12x+36=0\)
Ta có: \(\left(x-6\right)\left(x+6\right)-2x\left(x+6\right)+\left(x+6\right)^2\)
\(=x^2-36-2x^2-12x+x^2+12x+36\)
=0