Mình cần gấp lắm rồi

Từ a⁴ + b⁴ \(\ge\)2a²b² cộng a² + b² vào 2 vế

\(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2\)

Tương tự\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)

Từ đó suy ra \(a^4+b^4\ge\frac{1}{8}\left(a+b\right)^2\)

Cái cuối là \(a^4+b^4\ge\frac{1}{8}\left(a+b\right)^4\)nha mình nhầm

\(S=2x+4y+6z\le2\sqrt{\left[x^2+\left(2y\right)^2+\left(3z\right)^2\right]\left(1^2+1^2+1^2\right)}=2\sqrt{3.3}=6\)

Dấu \(=\)khi \(\hept{\begin{cases}x^2+4y^2+9z^2=3\\\frac{x}{1}=\frac{2y}{1}=\frac{3z}{1}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=\frac{1}{3}\end{cases}}\).

\(4=x^2+y^2-xy=\frac{1}{2}\left(x^2+y^2\right)+\frac{1}{2}\left(x-y\right)^2\ge\frac{1}{2}\left(x^2+y^2\right)\)

\(\Leftrightarrow x^2+y^2\le8\)

Dấu \(=\)khi \(x=y=\pm2\).

\(4=x^2+y^2-xy=\frac{3}{2}\left(x^2+y^2\right)-\frac{1}{2}\left(x+y\right)^2\le\frac{3}{2}\left(x^2+y^2\right)\)

\(\Leftrightarrow x^2+y^2\ge\frac{8}{3}\)

Dấu \(=\)khi \(x=-y=\pm\frac{2}{\sqrt{3}}\).

\(x^3-3x^2-x+3\)

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

\(=x\left(x^2-1\right)-3\left(x^2-1\right)\)

\(=\left(x-3\right)\left(x^2-1\right)\)

𝑥^5+10^4

=x^5+10000

a) \(A\left(x\right)=2x^3-x^2-x+1\)

\(=\left(2x^3-4x^2\right)+\left(3x^2-6x\right)+\left(5x-10\right)+11\)

\(=\left(x-2\right).\left(2x^2+3x+5\right)+11\)

Vậy \(A\left(x\right):B\left(x\right)=2x^2+3x+5\) dư \(11\)

b) Để \(A\left(x\right)⋮B\left(x\right)\) thì \(11⋮B\left(x\right)\)

\(\Rightarrow x-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

\(\Rightarrow x\inơ\left\{13;3;2;-9\right\}\)

Tính (3x-2)(3x+2) bằng

Áp dụng hằng đẳng thức

(3x-2)(3x+2) = 9x²- 4

HT

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