`a,` Tìm `n in N` * để `n^4 +4^n` là số nguyên tố

`b,` Tìm `x,y in Z: x^6 +3x^2 +1=y^4`

Cạnh hình vuông đó là:

`20:4=5(cm)`

Diện tích hình vuông đó là:

`5xx5=25(cm^2)`

Đáp số: `25cm^2`

`9-6x^2 +x^4`

`=(x^2)^2 -2.3x^2 +3^2`

`=(x^2 -3)^2`

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{a\left(a+1\right)}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{a}-\dfrac{1}{a+1}\\ =\left(1-\dfrac{1}{a+1}\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{a}-\dfrac{1}{a}\right)\\ =\left(\dfrac{a+1}{a+1}-\dfrac{1}{a+1}\right)-0-0-...-0\\ =\dfrac{a+1-1}{a+1}\\ =\dfrac{a}{a+1}\)

\(\dfrac{5932+6001\times5931}{5932\times6001-69}\\ =\dfrac{5932+6001\times5931}{\left(5931+1\right)\times6001-69}\\ =\dfrac{5932+6001\times5931}{5931\times6001+1\times6001-69}\\ =\dfrac{5932+6001\times5931}{5931\times6001+6001-69}\\ =\dfrac{5932+6001\times5931}{5931\times6001+5932}\\ =1\)

`ĐKXĐ: x >= 1`

`g(x) = x+2-4sqrt(x-1)`

`=x-1 - 4sqrt(x-1) +4 -1`

`=[(sqrt(x-1))^2 - 2.2.sqrt(x-1) +2^2]-1`

`=(sqrt(x-1) -2)^2 -1 >= -1`

Dấu "=" xảy ra `<=> sqrt(x-1) -2=0`

`<=>sqrt(x-1)=2`

`<=>x-1=4`

`<=>x=5`

Vậy `g(x)_(min) = -1 <=> x=5`

nó giống cái này thế nhỉ :V

https://hoidap247.com/cau-hoi/4784886

xem lại bài