\(a,x^8+14x^4+1=\left(x^8+14x^4+49\right)-48\)

\(=\left(x^4+7\right)^2-48\)

\(=\left(x^4+7+\sqrt{48}\right)\left(x^4+7-\sqrt{48}\right)\)

\(b,x^8+98x^4+1\)

\(=\left(x^8+98x^4+2401\right)-2400\)

\(=\left(x^4+49\right)^2-2400\)

\(=\left(x^4+49+20\sqrt{6}\right)\left(x^4+49-20\sqrt{6}\right)\)

Mình nghĩ vậy k bt đúng k :)

a) = a^10 - a + a^5 - a^2 + a^2 + a + 1

= a(a^9 - 1) + a^2(a^3 - 1) + (a^2 + a + 1)

= a.(a^3-1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)

= a.(a-1)(a^2 + a + 1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)

= (a^2 + a + 1)[(a.(a-1)(a^6 + a^3 + 1) + a^2 + 1]

b) x^5 - x^4 - 1 = x^5 - x^4 + x^3 - x^3 + x^2 - x - x^2 + x - 1

= x^3(x^2 - x + 1) - x(x^2 - x + 1) - (x^2 - x + 1)

= (x^2 - x + 1)(x^3 - x - 1)

a) \(a^{10}+a^5+1\)

\(=\left(a^{10}-a^9+a^7-a^6+a^5-a^3+a^2\right)\)

\(+\left(a^9-a^8+a^6-a^5+a^4-a^2+a\right)\)

\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(=a^2\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(+a\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

@hieu nguyen Em có nhân chéo hai vế và khai triển ra nhưng cũng không ra cái gì ạ. 

/nfvg6yguntbygynugytf6ht

\(\frac{3}{a^2+b^2}+\frac{2}{ab}=\frac{3}{a^2+b^2}+\frac{3}{2ab}+\frac{1}{2ab}\)

\(\ge\frac{12}{\left(a+b\right)^2}+\frac{1}{2ab}\ge12+\frac{2}{\left(a+b\right)^2}\ge12+2=14\)(đpcm)

Vậy..

Có lẽ đề là n nguyên dương:v

Với \(n=1\) thì \(\frac{1}{1+1}+\frac{1}{2\cdot1}=1>\frac{1}{2}\)

Giả sử bài toán đúng với \(n=k\) khi đó:\(A_k=\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+....+\frac{1}{2k}\)

Ta cần chứng minh bài toán đúng với \(n=k+1\) thật vậy:
\(A_{k+1}=\frac{1}{k+2}+\frac{1}{k+3}+\frac{1}{k+4}+....+\frac{1}{2k+2}\)

\(A_{k+1}=\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+.....+\frac{1}{2k}\right)+\left(\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\right)\)

\(A_{k+1}=A_k+\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)>\frac{1}{2}\) vì \(A_k>\frac{1}{2};\frac{1}{2k+1}-\frac{1}{2k+2}>0\) với mọi k nguyên dương.

Vậy bài toán được chứng minh. 

ĐKXĐ : \(x\ge0;y\ge1\)

\(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-1-6\sqrt{y-1}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{y-1}-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=10\end{cases}}}\)

n là tham số hay sao ah? 

Anh quên mất  \(n\ge0\)

\(\sqrt{16-6\sqrt{7}}-\sqrt{32+10\sqrt{7}}.\)

\(=\sqrt{9-6\sqrt{7}+7}-\sqrt{25+10\sqrt{7}+7}\)

\(=\sqrt{3^2-2.3.\sqrt{7}+\sqrt{7}^2}-\sqrt{5^2+2.5.\sqrt{7}+\sqrt{7^2}}\)

\(\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(5+\sqrt{7}\right)^2}\)

\(=3-\sqrt{7}-5-\sqrt{7}=-2-2\sqrt{7}\)

\(\sqrt{17-4}.\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{13}.\sqrt{5+4\sqrt{5}+4}\)

\(=\sqrt{13}\left(\sqrt{5}+2\right)\)

\(=\sqrt{65}+2\sqrt{13}\)