Rút gọn:
\(E=x^{n-2}\left(x^2-1\right)-x\left(x^{n-1}-x^{n-3}\right)\)(với \(n\inℕ;x\ge3\))
Những câu hỏi liên quan
Rút gọn biểu thức Sleft(xright)dfrac{1}{x^2}+dfrac{2}{x^3}+dfrac{3}{x^4}+...+dfrac{n}{x^{n+1}} bằng:A. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^{n+1}left(x-1right)^2}B. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^{2n}left(x-1right)^2}C. Sdfrac{x^n-left(n+1right)x+n}{x^nleft(x-1right)^2}D. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^nleft(x-1right)^2}
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Rút gọn biểu thức \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) bằng:
A. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
B. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{2n}\left(x-1\right)^2}\)
C. \(S=\dfrac{x^n-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
D. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
Rút gọn biểu thức Sleft(xright)dfrac{1}{x^2}+dfrac{2}{x^3}+dfrac{3}{x^4}+...+dfrac{n}{x^{n+1}} bằng:A. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^{n+1}left(x-1right)^2}B. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^{2n}left(x-1right)^2}C. Sdfrac{x^n-left(n+1right)x+n}{x^nleft(x-1right)^2}D. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^nleft(x-1right)^2}
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Rút gọn biểu thức \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) bằng:
A. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
B. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{2n}\left(x-1\right)^2}\)
C. \(S=\dfrac{x^n-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
D. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)
\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)
\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)
\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)
\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
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Rút gọn biểu thức :
\(\left(x^n+1\right)\left(x^n-2\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2018\)
\(\left(x^n+1\right)\left(x^n-2\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2018=x^{2n}+x^n-2.x^n-2-x^{2n}+x^n+2018=2016.\)
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Rút gọn biểu thức sau
\(D=n^2\left(n+4\right)\left(n-4\right)+\left(1-n^2\right)\left(n^2+1\right)\)
\(E=\left(\frac{1}{2}x^m-y^n\right)×\left(y^n+\frac{1}{2}x^m\right)\)
Giúp mình với !!!
Cho đa thức: f( x ) = \(2\cdot\left(x^2\right)^n-5\cdot\left(x^n\right)^2+8\cdot x^{n-1}\cdot x^{1+n}-4\cdot x^{n^2+1}\cdot x^{2\cdot n-n^2-1}\left(n\inℕ\right)\)
a, Thu gọn đa thức f(x)
b, Tìm giá trị nhỏ nhất của f(x) + 2020
a) \(f\left(x\right)=2.\left(x^2\right)^n-5.\left(x^n\right)^2+8n^{n-1}.x^{1+n}-4.x^{n^2+1}.x^{2n-n^2-1}\)
\(=2x^{2n}-5x^{2n}+8x^{2x}-4x^{2n}\)
\(=x^{2n}\)
b) \(f\left(x\right)+2020=x^{2n}+2020\)
Vì \(n\in N\Rightarrow2n\in N\)và 2n là số chẵn
\(\Rightarrow x^{2n}\ge1\)
\(\Rightarrow x^{2n}+2020\ge2021\)
Dấu"="xảy ra \(\Leftrightarrow x^{2n}=1\)
\(\Leftrightarrow n=0\)
Vậy ...
( ko bít đúng ko -.- )
Rút gọn các phân thức sau:a, Afrac{x^3-7x-6}{x^2left(x-3right)^2+4xleft(x-3right)^2+4left(x-3right)^2} b, Bfrac{n!-left(n-1right)!}{left(n+1right)!}c, Cfrac{left(n+1right)!}{left(n+1right)!+left(n+2right)!} d, Dfrac{left(n+2right)!+left(n+3right)!}{left(n+2right)!-left(n+3right)!}e, Efrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^5+1}Ai biết câu nào thì trả lời câu đấy nha!!!
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Rút gọn các phân thức sau:
a, A=\(\frac{x^3-7x-6}{x^2\left(x-3\right)^2+4x\left(x-3\right)^2+4\left(x-3\right)^2}\) b, B=\(\frac{n!-\left(n-1\right)!}{\left(n+1\right)!}\)
c, C=\(\frac{\left(n+1\right)!}{\left(n+1\right)!+\left(n+2\right)!}\) d, D=\(\frac{\left(n+2\right)!+\left(n+3\right)!}{\left(n+2\right)!-\left(n+3\right)!}\)
e, E=\(\frac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^5+1}\)
Ai biết câu nào thì trả lời câu đấy nha!!!
rút gọn N
N =\(\left(\frac{2x^2+1}{x^2-1}-\frac{x}{x^2+x+1}\right).\left(\frac{x^3+1}{1+x}-x\right)\)
Rút gọn phân thức sau ( phân thức đều có nghĩa )
\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)
\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)
\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)
\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)
Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:
\(N=\dfrac{\left(y-1\right)\left(y+1\right)+1}{y}\)
\(=\dfrac{y^2-1+1}{y}\)
\(=\dfrac{y^2}{y}\)
\(=y\)
\(=x^2+7x+11\)
Vậy \(N=x^2+7x+11\).
\(\text{#}Toru\)
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Rút gọn biểuthức
\(N=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{x}\right)\)
mình nghĩ đề nên là \(N=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\) chớ như bạn rút gọn không hết
\(N=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\left(x>0,x\ne2\right)\)
\(=\dfrac{3\left(\sqrt{x}-2\right)-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\left(\dfrac{\sqrt{x}-6}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{2\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-8}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
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