A. 1 :1:1:1
B. 3:1:2:3
C. 2:3:1:1
D. 6:5:2:9
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MH
3 tháng 2 2021
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
M
9
H
18 tháng 4 2023
`a, 3/4 + 1/2 xx 7/2`
`= 3/4 + 7/4`
`=10/4`
`=5/2`
`b, 6/15 - 1/3 : 5/3`
`= 6/15 - 1/3 xx 3/5`
`= 6/15 - 3/15`
`= 3/15`
`=1/5`
`c, x-4/9 = 3/7 : 9/4`
`=> x-4/9= 3/7 xx 4/9`
`=> x-4/9= 12/63`
`=> x-4/9=4/21`
`=> x= 4/21 +4/9`
`=>x= 40/63`
`d, 7/9 xx 3/5 -1/2=1/5`
`->` sao lại bằng có `x` ko vậy ạ?
18 tháng 4 2023
`a,`
`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`
`b,`
`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`
`c,` Tìm x?
`x-4/9=3/7 \div 9/4`
`x-4/9=4/21`
`x=4/21+4/9`
`x=40/63`
`d, 7/9x \times 3/5-1/2=1/5`
`7/9x \times 3/5=1/5+1/2`
`7/9x \times 3/5=7/10`
`7/9x=7/10 \div 3/5`
`7/9x=7/6`
`x=7/6 \div 7/9=3/2`
B