\(x^{12}-x^9+x^4-x+1>0\)\(\Leftrightarrow2x^{12}-2x^9+2x^4-2x+2>0\)

\(\Leftrightarrow\left(x^{12}-2x^9+x^6\right)+\left(x^{12}-x^6+\frac{1}{4}\right)+\left(2x^4-2x^2+\frac{1}{2}\right)+\)\(\left(2x^2-2x+\frac{1}{2}\right)+\frac{3}{4}>0\)

\(\Leftrightarrow\left(x^6-x^3\right)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{2}\right)^2+2\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

do đó ta có đpcm

\(D=x^{10}-x^9+x^4-x+1>0\)

\(D=x^9\left(x-1\right)+x\left(x^3-1\right)+1\)

Vậy ta xét : \(x\ge1\)\(\Rightarrow\)D Sẽ luôn dương (1)

Xét: \(x< 1\)

\(\Rightarrow\)\(D=x^{10}+x^4\left(1-x^5\right)+\left(1-x\right)\)

cái này mà môn Lí chứ k pk toán nha bn

Bạn kiểm tra lại đề nhé.

\(\sqrt{x-5}+\sqrt{y-2019}+\sqrt{z+2021}=\frac{1}{2}\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z-2\sqrt{x-5}-2\sqrt{y-2019}-2\sqrt{z+2021}=0\)

\(\Leftrightarrow x-5-2\sqrt{x-5}+1+y-2019-2\sqrt{y-2019}+z+2021-2\sqrt{x+2021}+1=0\)

\(\Leftrightarrow\left(\sqrt{x-5}-1\right)^2+\left(\sqrt{y-2019}-1\right)^2+\left(\sqrt{z+2021}-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-5}-1=0\\\sqrt{y-2019-1}=0\\\sqrt{z+2021-1=0}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2020\\z=-2020\end{cases}}\)

Vì A; B; C là 3 góc của một tam giác 

=> \(A+B+C=180^o\)

Ta có: \(\cos A+\cos B=2.\cos\frac{A+B}{2}.\cos\frac{A-B}{2}\le2.\cos\frac{180^o-C}{2}=2.\sin\frac{C}{2}\)

Tương tự: \(\cos A+\cos C\le2.\sin\frac{B}{2}\); \(\cos B+\cos C\le2.\sin\frac{A}{2}\)

=> \(9=5\cos A+6\cos B+7\cos C\)

\(=\left(3\cos A+3\cos C\right)+\left(2\cos A+2\cos B\right)+\left(4\cos B+4\cos C\right)\)

\(\le6.\sin\frac{B}{2}+4\sin\frac{C}{2}+8\sin\frac{A}{2}\)

Vì A; B; C là 3 góc của một tam giác  => \(A;B;C< 180^o\)=> \(\frac{A}{2};\frac{B}{2};\frac{C}{2}< 90^o\)

=> \(0< \sin\frac{B}{2};\sin\frac{C}{2};\sin\frac{A}{2}< 1\)

Đặt: \(\sin\frac{B}{2}=y;\sin\frac{C}{2}=z;\sin\frac{A}{2}=x\)

Đưa về bài toán: \(0< x;y;z< 1\); \(8x+6y+4z\ge9\)

Chứng minh: \(x^2+y^3+z^4\ge\frac{7}{16}\)

Ta có: \(\left(x^2+\frac{1}{4}\right)+\left(y^3+\frac{1}{8}+\frac{1}{8}\right)+\left(z^4+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\right)\)

\(\ge x+\frac{3}{4}y+\frac{4z}{8}\)( theo cauchy)

=> \(x^2+y^3+z^4\ge\frac{1}{8}\left(8x+6y+4z\right)-\frac{11}{16}\ge\frac{9}{8}-\frac{11}{16}=\frac{7}{16}\)

Dấu "=" xảy ra <=> x = y = z = 1/2 

<=> A = B = C = 60⁰

Câu 1) Sai đề thì phải, bạn xem lại đề nha. 

Câu 2) 

\(M=2\left(x^4+y^4\right)+\frac{1}{xy}\ge4\sqrt{x^4y^4}+\frac{1}{xy}=4x^2y^2+\frac{1}{xy}\)

\(=4x^2y^2+\frac{1}{1024xy}+\frac{1}{1024xy}+\frac{511}{512xy}\)

\(\ge3\sqrt[3]{4x^2y^2.\frac{1}{1024xy}.\frac{1}{1024xy}}+\frac{511}{512}.\left(\frac{2}{x+y}\right)^2\)

\(\ge\frac{3}{64}+\frac{511}{512}.\left(\frac{2}{\frac{1}{2}}\right)^2=\frac{1025}{64}\)

Suy ra \(M\ge\frac{1025}{64}\)

Dấu \(=\)xảy ra khi \(x=y=\frac{1}{4}\).

Vậy \(minM=\frac{1025}{64}\).