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AH
Akai Haruma
Giáo viên
25 tháng 11 2018

Lời giải:

\(N=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a^2-2a+1)+15}=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a-1)^2+15}\)

Đặt \((a-1)^2=t\Rightarrow N=\frac{t^2-11t+30}{3t^2-18t+15}\)

\(=\frac{t^2-11t+30}{3(t^2-6t+5)}=\frac{(t-5)(t-6)}{3(t-1)(t-5)}\)

\(=\frac{t-6}{3(t-1)}=\frac{(a-1)^2-6}{3(a-1)^2-3}\)

25 tháng 11 2018

Đặt \(\left(a-1\right)^2=t\)

Ta có: \(\left(a-1\right)^4-11\left(a-1\right)^2+30\)

\(=t^2-11t+30\)

\(=t\left(t-5\right)-6\left(t-5\right)=\left(t-5\right)\left(t-6\right)\)

\(=\left[\left(a-1\right)^2-5\right]\left[\left(a-1\right)^2-6\right]\)

\(=\left(a^2-2a-4\right)\left(a^2-2a-5\right)\)

Đặt \(a^2-2a=k\)

Ta có: \(3\left(a-1\right)^4-18\left(a^2-2a\right)-3\)

\(=3\left(a^2-2a+1\right)^2-18\left(a^2-2a\right)-3\)

\(=3\left(k+1\right)^2-18k-3\)

\(=3k^2+6k+3-18k-3\)

\(=3k^2-12k=3k\left(k-4\right)\)

\(=3\left(a^2-2a\right)\left(a^2-2a-4\right)\)(Ở đây bạn ghi thêm điều kiện nhé)

Khi đó: \(N=\frac{\left(a^2-2a-4\right)\left(a^2-2a-5\right)}{3\left(a^2-2a\right)\left(a^2-2a-4\right)}=\frac{a^2-2a-5}{3\left(a^2-2a\right)}\)

11 tháng 11 2023

\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)

\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)

\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)

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 \(N=x^2+7x+11\).

\(\text{#}Toru\)

11 tháng 11 2023

À bạn ơi cho mình hỏi ngoài lề 1 chút được k ạ?

22 tháng 7 2023

a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)

b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

`c, (ab^2-a^2b)/(2a^2+a)`

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

9 tháng 8 2020

a. \(\left(a^2+a-1\right)\left(a^2-a+1\right)=a^4+a^2+1\)

b. \(\left(a+2\right)\left(a-2\right)\left(a^2+2a+4\right)\left(a^2-2x+4\right)=a^6-64\)

c. \(\left(2+3y\right)^2-\left(2x-3y\right)^2-12xy=4+12y-4x^2\)

d. \(\left(x+1\right)^3-\left(x-1\right)^3-\left(x^3-1\right)-\left(x-1\right)\left(x^2+x+1\right)=-2x^3+6x^2+4\)

9 tháng 8 2020

\(A=\left(a^2+\left(a-1\right)\right)\left(a^2-\left(a-1\right)\right)=a^4-\left(a-1\right)^2=a^4-\left(a^2-2a+1\right)=a^4-a^2+2a-1\)

\(B=\left(a+2\right)\left(a^2-2a+4\right)\left(a-2\right)\left(a^2+2a+4\right)=\left(a^3+8\right)\left(a^3-8\right)=a^6-64\)

\(C=9y^2+12y+4-\left(4x^2-12xy+9y^2\right)-12xy=12y+4-4x^2\)

\(D=x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-x+1=-x^3+6x^2-x+4\)

26 tháng 8 2017

Với mọi k thuộc N và k > 2 thì ta có :

\(1-\frac{1}{1+2+....+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=1-\frac{2}{k\left(k+1\right)}=\frac{k^2+k-2}{k\left(k+1\right)}=\frac{\left(k+2\right)\left(k-1\right)}{k\left(k+1\right)}\)

Áp dụng vào A ta được :

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)

\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)

29 tháng 8 2017

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

26 tháng 11 2017

Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak

Bài 2:

\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)

\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)

\(=\dfrac{1}{313}\)

26 tháng 11 2017

\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)

\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)

\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)

\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)

\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)

\(=\dfrac{7.2.10}{81-4}\)

\(=\dfrac{720}{77}\)