\(\frac{1}{C\overset{1}{x}}-\frac{1}{C\overset{2}{x+1}}=\frac{7}{6C\overset{1}{x}+4}\)
JC
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Giải các bất phương trình sau
a) (x-4)2<x(x-8)
b) x+\(\dfrac{1}{2}\)\(\overset{>}{-}\)\(\dfrac{3-5x}{-3}\)
c) \(\dfrac{x-7}{-4}\)\(\overset{< }{-}\)\(\dfrac{4-2x}{-3}\)
a: =>x^2-8x+16<x^2-8x
=>16<0(loại)
b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)
=>x+1/2>=5/3x-1
=>-2/3x>=-3/2
=>x<=3/2:2/3=9/4
c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)
=>21-3x<=8x-16
=>-11x<=-37
=>x>=37/11
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Tính hợp lí:
a) \(\dfrac{2.8\overset{4}{ }.27\overset{2}{ }+4.6\overset{2}{ }}{2\overset{7}{ }.6\overset{7}{ }+2\overset{7}{ }.10.9\overset{4}{ }}\)- 2017
b) \(\dfrac{6\overset{6}{ }_{ }-6\overset{3}{ }.3\overset{3}{ }+3\overset{6}{ }}{-73}\): \(\dfrac{1}{3}\)
1) overset{lim}{xrightarrow1}dfrac{x^3-3x+2}{x^4-4x+3}
2)overset{lim}{xrightarrow2^-}dfrac{x^3+x^2-4x-4}{x^2-4x+4}
3) overset{lim}{xrightarrow2}dfrac{left(x^2-x-2right)^{20}}{left(x^3-12x+16right)^{10}}
4)overset{lim}{xrightarrow0^-}dfrac{left(1+xright)left(1+4xright)-1}{x^2}
5) overset{lim}{xrightarrow-1}dfrac{sqrt{x+2}-1}{sqrt{x+5}-2}
Đọc tiếp
1) \(\overset{lim}{x\rightarrow1}\)\(\dfrac{x^3-3x+2}{x^4-4x+3}\)\(\)
2)\(\overset{lim}{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}\)
3) \(\overset{lim}{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}\)
4)\(\overset{lim}{x\rightarrow0^-}\dfrac{\left(1+x\right)\left(1+4x\right)-1}{x^2}\)
5) \(\overset{lim}{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x+2}{x^4-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+2\right)\left(x-1\right)^2}{\left(x^2+2x+3\right)\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x+2}{x^2+2x+3}=\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}=\lim\limits_{x\rightarrow2^-}\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2^-}\dfrac{x^2+3x+2}{x-2}=-\infty\)
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}\left(x-2\right)^{20}}{\left(x+4\right)^{10}\left(x-2\right)^{20}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}}{\left(x+4\right)^{10}}=\dfrac{3^{10}}{2^{10}}\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{4x^2+5x}{x^2}=\lim\limits_{x\rightarrow0^-}\dfrac{4x+5}{x}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(\sqrt{x+5}+2\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2\)
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Rút gọn tổng: \(S=C\overset{0}{n}+C\overset{1}{n}+2.C\overset{2}{n}+...+nC\overset{n}{n}\) bằng:
A. \(n.2^n+1\)
B. \(2^n+1\)
C. \(n.2^{n-1}+1\)
D. \(n.2^{n+1}\)
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^nC_n^n\)
Đạo hàm 2 vế:
\(n\left(1+x\right)^{n-1}=C_n^1+2xC_n^2+...+n.x^{n-1}C_n^n\)
Thay \(x=1\)
\(\Rightarrow n.2^{n-1}=C_n^1+2C_n^2+...+nC_n^n\)
\(\Rightarrow n.2^{n-1}+1=C_n^0+C_n^1+2C_n^2+...+nC_n^n\)
\(\Rightarrow S=n.2^{n-1}+1\)
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Rút gọn: \(S=C\overset{1}{100}+2^2C\overset{2}{100}+3^2C\overset{3}{100}+...+100^2C\overset{100}{100}\)
Rút gọn tổng: \(S=-C\overset{1}{2019}+1.2C\overset{2}{2019}-2.3C\overset{3}{2019}+...+2017.2018C\overset{2018}{2019}-2018.2019C\overset{2019}{2019}\) bằng:
A. 1
B.. 2019
C. 0
D. -2019
Rút gọn tổng \(S=C\overset{1}{2019}-2C\overset{2}{2019}+...-2018C\overset{2018}{2019}+2019C\overset{2019}{2019}\) bằng:
A. 2019
B.1
C. -2019
D. 0
Rút gọn tổng: \(S=C\overset{1}{n}+1.2C\overset{2}{n}+2.3C\overset{3}{n}+...+\left(n-1\right)nC\overset{n}{n}\) bằng:
A. \(\left(n-1\right)n.2^{n-2}\)
B. \(n.2^{n-2}\)
C. \(\left(n-1\right)n.2^{n-1}+n\)
D. \(\left(n-1\right)n.2^{n-2}+n\)
Rút gọn tổng: \(S=C\overset{1}{n}+1.2C\overset{2}{n}+2.3C\overset{3}{n}+...+\left(n-1\right)nC\overset{n}{n}\) bằng:
A. \(\left(n-1\right)n.2^{n-2}\)
B. \(n.2^{n-2}\)
C. \(\left(n-1\right)n.2^{n-1}+n\)
D. \(\left(n-1\right)n.2^{n-2}+n\)
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^nC_n^n\)
Đạo hàm 2 vế:
\(n\left(1+x\right)^{n-1}=C_n^1+2xC_n^2+...+nx^{n-1}C_n^n\)
Tiếp tục đạo hàm 2 vế:
\(\left(n-1\right)n\left(1+x\right)^{n-2}=2C_n^2+2.3xC_n^3+...+\left(n-1\right)nx^{n-2}C_n^n\)
Thay \(x=1\)
\(\Rightarrow\left(n-1\right)n.2^{n-2}=1.2C_n^2+2.3C_n^3+...+\left(n-1\right)nC_n^n\)
\(\Rightarrow\left(n-1\right)n.2^{n-2}+n=C_n^1+1.2C_n^2+...+\left(n-1\right)n.C_n^n\)
\(\Rightarrow S=\left(n-1\right)n.2^{n-2}+n\)
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