1) tìm sao cho $n^2+2n+12$ là số chính phương
2) cho $0^{o }<\alpha <90^o $ và $sin \alpha+cos \alpha=\frac{7}{5} $. Tính $tan\alpha $
3)chứng minh rằng với mọi số thực a,b ta luôn có: $a^2+b^2+b+\frac{5}{2}\geqslant ab +2a$

chỉ nói tìm n thôi, ko cho thêm đk, mà hình như là stn –  pnt2912003 07-02-18 09:36 PM
Câu 1 là STN hay ko vậy –  ๖ۣۜ➻❥Thảo(๖ۣۜST) 07-02-18 09:31 PM
3
$a^2+b^2+b+5/2\geq ab+2a\Leftrightarrow 2a^2+2b^2+2b+5-2ab-4a\geq 0\Leftrightarrow (a^2-2ab+b^2)+(a^2-4a+4)+(b^2+2b+1)\geq 0\Leftrightarrow (a-b)^2+(a-2)^2+(b+1)^2\geq 0$(luôn đúng với mọi a,b)=>đpcm
2
$0<\alpha <90\Rightarrow \cos \alpha \neq 0 $
$\sin\alpha +\cos \alpha =7/5\Rightarrow sin \alpha /cos \alpha  +1=7/(5cos\alpha )(chia  2 vế cho cos \alpha)$
$\Rightarrow (1 + tan\alpha)² = 49/25cos²\alpha = 49/25.(1 + tan²\alpha) $
 $\Rightarrow 25(1 + tan\alpha)² = 49(1 + tan²\alpha) $
 $\Rightarrow 25(1 + tan²\alpha + 50tan\alpha = 49(1 + tan²\alpha) $
$\Rightarrow 24(1 + tan²\alpha) - 50tan  \alpha = 0 $
$\Rightarrow 24tan²\alpha - 50tan\alpha + 24 = 0 $
có $Δ' = 25² - 24² = 7² $
$tan \alpha = 4/3 hoặc tan \alpha= 3/4 $
$mà 0 < \alpha < 90o => tan\alpha > 1 $
Vậy tan \alpha = 4/3.
1
$n^2+2n+12=(n+1)^2+11=a^2(a\in N)\Rightarrow a^2-(n+1)^2=11\Rightarrow (a-n-1)(a+n+1)=11$
hay a-n-1;a+n+1là ước của 11
cho (a-n-1;a+n+1)=(1;11);(-1;-11);(11;1);(-11;-1) rồi tìm a với n(chú ý điều kiện a ,n)
1,
$n^2+2n+12=(n+1)^2+11=a^2$
n là STN thì thay lần lượt n=1,2,3... thấy a nguyên là đk
2,
$VT=\sqrt{2} .sin(\alpha +45°)=7/5=>\alpha =...$

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