Quyển này là ...Những viên kim cương trong bất đẳng thức toán học của Trần Phương?

Áp dụng điểm rơi trong AM-GM đó bạn a^2 =1/16b^2 => a^4 =1/16 => a=1/2

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}\sin2x-\dfrac{\sqrt{2}}{2}\cos2x=\cos x\)

\(\Leftrightarrow\sin\left(2x-\dfrac{\pi}{4}\right)=\sin\left(\dfrac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=\dfrac{\pi}{2}-x+k2\pi\\2x-\dfrac{\pi}{4}=\pi-\dfrac{\pi}{2}+x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow2\alpha-\beta=\dfrac{2.3\pi}{4}-\dfrac{\pi}{4}=\dfrac{5\pi}{4}\)

\(\left(3+x+x^2\right)^8=\sum\limits^8_{k=0}C^k_8.\left(x^2+x\right)^k.3^{8-k}\)

\(\left(x^2+x\right)^k=\sum\limits^k_{t=0}C^t_k.x^{2t}.x^{k-t}\)

\(\Rightarrow\left(3+x+x^2\right)^8=\sum\limits^8_{k=0}C^k_8.C^t_k.x^{2t}.x^{k-t}.3^{8-k}\)

\(x^2\Leftrightarrow2t+k-t=2\Leftrightarrow t+k=2\)

\(k\ge t\Rightarrow\left(k;t\right)=\left(2;0\right);\left(1;1\right)\)

\(\left(k;t\right)=\left(2;0\right)\Rightarrow he-so:C^2_8.C^0_2.3^{8-2}=20412\)

\(\left(k;t\right)=\left(1;1\right)\Rightarrow he-so:C^1_8.C^1_1.3^{8-1}=17496\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-2m-\dfrac{1}{2}>-x^2+2x+\dfrac{1}{2}-m\\4x-2m-\dfrac{1}{2}< x^2-2x-\dfrac{1}{2}+m\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-\dfrac{1}{4}-m>0\\x^2-6x+3m>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}1+\dfrac{1}{4}+m< 0\\9-3m< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m< -\dfrac{5}{4}\\m>3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}m\le x\\x\le3\end{matrix}\right.\Rightarrow m\le3\Rightarrow\left[m;3\right]\) 

Vay \(m\le3\) thi ham so co tap xd la 1 doan tren truc so

P/s: Ve cai truc so ra la hieu

Ừ cậu xem giúp mình nhé!

1/ \(\lim\limits\dfrac{\dfrac{2^n}{7^n}-5.7.\left(\dfrac{7}{7}\right)^n}{\dfrac{2^n}{7^n}+\left(\dfrac{7}{7}\right)^n}=-35\)

2/ \(\lim\limits\dfrac{\dfrac{3^n}{7^n}-2.5.\left(\dfrac{5}{7}\right)^n}{\dfrac{2^n}{7^n}+\dfrac{7^n}{7^n}}=0\)

3/ \(\lim\limits\sqrt[3]{\dfrac{\dfrac{5}{n}-\dfrac{8n}{n}}{\dfrac{n}{n}+\dfrac{3}{n}}}=\sqrt[3]{-8}=-2\)

Đề bài có sai ko nhỉ, mình ngồi biến đổi thấy nó ko hợp lý chút nào

\(10\left[2\cos^2\left(\dfrac{\pi}{4}+x\right)-1\right]=10\cos\left(\dfrac{\pi}{2}+2x\right)\)

\(\sin\left(\dfrac{3\pi}{2}-4x\right)=\cos\left(\dfrac{\pi}{2}-\dfrac{3\pi}{2}+4x\right)=\cos\left(4x-\pi\right)\)

\(\Rightarrow pt:1+10\cos\left(4x-\pi\right)+10\cos\left(2x+\dfrac{\pi}{2}\right)+10=m\)

\(\Leftrightarrow10\left[\cos\left(4x-\pi\right)+\cos\left(2x+\dfrac{\pi}{2}\right)\right]+11=m\)

\(\cos\left(4x-\pi\right)=\cos2\left(2x-\dfrac{\pi}{2}\right)=2\cos^2\left(2x-\dfrac{\pi}{2}\right)-1\)

\(\Rightarrow10\left[2\cos^2\left(2x-\dfrac{\pi}{2}\right)+\cos\left(2x+\dfrac{\pi}{2}\right)-1\right]+11=m\)

\(\Leftrightarrow2\cos^2\left(2x-\dfrac{\pi}{2}\right)+\cos\left(2x+\dfrac{\pi}{2}\right)=\dfrac{1}{10}m-\dfrac{1}{10}\)

Nếu 2 cái trong ngoặc giống nhau thì mình sẽ biến đổi được, nhưng chả thuận lợi gì :(

Bạn dùng gõ công thức để mọi người hiểu đề bài nhé !