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26 tháng 3 2019

\(Đặt\) \(n^2+2n=h\) ta có:

\(h\left(h+2\right)+1=\)\(h^2+2h+1\)\(=\)\(\left(h+1\right)^2\)

\(\Rightarrow\left(x^2+2n\right)\left(x^2+2n+2\right)+1=\left(x^2+2n+1\right)^2\)

Đặt n^2 + 2n + 1= a, ta được:

(a - 1)(a + 1) +1= a^2  -  1 + 1= a^2=(n^2 + 2n +1)^2

=(n + 1)^4

19 tháng 12 2016

Bài 1:

\(x^5+x+1\)

\(=x^5-x^4+x^2+x^4-x^3+x+x^3-x^2+1\)

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

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

Bài 2:

\(\frac{2n^2-3n+1}{2n+1}=\frac{n\left(2n+1\right)-4n+1}{2n+1}=\frac{n\left(2n+1\right)}{2n+1}-\frac{4n+1}{2n+1}=n-\frac{4n+1}{2n+1}\in Z\)

\(\Rightarrow4n+1⋮2n+1\)

\(\Rightarrow\frac{4n+1}{2n+1}=\frac{2\left(2n+1\right)-1}{2n+1}=\frac{2\left(2n+1\right)}{2n+1}-\frac{1}{2n+1}=2-\frac{1}{2n+1}\in Z\)

\(\Rightarrow1⋮2n+1\)

\(\Rightarrow2n+1\inƯ\left(1\right)=\left\{1;-1\right\}\)

\(\Rightarrow2n\in\left\{0;-2\right\}\)

\(\Rightarrow n\in\left\{0;-1\right\}\)

 

9 tháng 2 2019

Bạn thử giải câu này xem

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9 tháng 2 2019

\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)

\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)

Đặt: \(x^2+2x=t\)

khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)

\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)

b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)

Khi đó:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)

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

\(A=1-\frac{1}{\left(n+1\right)^2}\)

6 tháng 10 2020

a) \(15x^ny^{2n}-3x^{n+1}\left(-y\right)^{2n}\)

\(=x^ny^{2n}\left(15-3x\right)\)

\(=3x^ny^{2n}\left(5-x\right)\)

b) \(4x^{2n}y^{n-1}+2\left(-x\right)^{2n+1}y^n\)

\(=4x^{2n}y^{n-1}-2x^{2n+1}y^n\)

\(=2x^{2n}y^{n-1}\left(2-xy\right)\)

4 tháng 10 2019

2. Ta có: P = 2x2 + y2 - 4x - 4y + 10

P = 2(x2 - 2x + 1) + (y2 - 4y + 4) + 4

P = 2(x - 1)2 + (y - 2)2 + 4 \(\ge\)\(\forall\)x;y

=> P luôn dương với mọi biến x;y

3 Ta có:

(2n + 1)(n2 - 3n - 1) - 2n3 + 1

= 2n3 - 6n2 - 2n + n2 - 3n - 1 - 2n3 + 1

= -5n2 - 5n = -5n(n + 1) \(⋮\)\(\forall\)\(\in\)Z

20 tháng 4 2020

1×2=2

5 tháng 11 2017

a) \(49-x^2+2xy-y^2\)

\(=49-\left(x^2-2xy+y^2\right)\)

\(=49-\left(x-y\right)^2\)

\(=\left(7-x+y\right)\left(7+x-y\right)\)

c) \(\frac{1}{36}a^2-\frac{1}{4}b^2\)

\(=\frac{1}{4}\left(\frac{1}{9}a^2-b^2\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}a-b\right)\left(\frac{1}{3}a+b\right)\)

26 tháng 10 2019

Ta có: 

\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)

\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)

\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)

\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)

...

\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)

\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)

=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)

\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)

2 tháng 10 2016

Phân tích thành nhân tử

\(=\left(my+nx\right)\left(ny+mx\right)\)

2 tháng 10 2016

mn(x+y2) +xy(m2 +n2)= mnx+mny+xym2 +xyn2

                                              =mx(nx + my) +ny( my +nx)

                                  =(mx+ny)(nx+my)