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Không mất tính tổng quát, giả sử \(x=mid\left\{x;y;z\right\}\)
\(\Rightarrow\left(x-y\right)\left(x-z\right)\le0\)
\(\Rightarrow x^2+yz\le xy+xz\)
\(\Rightarrow zx^2+yz^2\le xyz+xz^2\)
\(\Rightarrow P\le x^3+y^3+z^3+8\left(xy^2+xz^2+xyz\right)\)
\(\Rightarrow P\le x^3+y^3+z^3+3yz\left(y+z\right)+8\left(xy^2+xz^2+2xyz\right)\)
\(\Rightarrow P\le x^3+\left(y+z\right)^3+8x\left(y+z\right)^2\)
\(\Rightarrow P\le x^3+\left(4-x\right)^3+8x\left(4-x\right)^2\)
\(\Rightarrow P\le8x^3-52x^2+80x+64\)
Tới đây, đơn giản nhất là khảo sát hàm \(f\left(x\right)=8x^3-52x^2+80x+64\) trên \(\left[0;4\right]\)
(Nếu ko khảo sát hàm, ta có thể tách như sau, tất nhiên là dựa trên điểm rơi có được từ việc khảo sát hàm):
\(\Rightarrow P\le\left(8x^3-52x^2+80x-36\right)+100\)
\(\Rightarrow P\le4\left(x-1\right)^2\left(2x-9\right)+100\)
Do \(0\le x\le4\Rightarrow2x-9< 0\Rightarrow P\le100\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;3;0\right)\) và 1 vài bộ hoán vị của chúng
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\(S_0=a_0+a_1+...+a_{16}=f\left(1\right)=1\)
Số hạng tổng quát trong khai triển:
\(\sum\limits^8_{k=0}C_8^k\left(x^2+2x\right)^k\left(-2\right)^{8-k}=\sum\limits^8_{k=0}C_8^k\left(-2\right)^{8-k}\sum\limits^k_{i=0}C_k^ix^{2i}\left(2x\right)^{k-i}\)
\(=\sum\limits^8_{k=0}\sum\limits^k_{i=0}C_8^kC_k^i\left(-2\right)^{8-k}2^{k-i}x^{i+k}\)
Số hạng không chứa x thỏa mãn: \(\left\{{}\begin{matrix}0\le i\le k\le8\\i+k=0\end{matrix}\right.\)
\(\Rightarrow i=k=0\Rightarrow a_0=C_8^0C_0^0\left(-2\right)^82^0=2^8\)
Số hạng chứa \(x^{16}\) thỏa mãn: \(\left\{{}\begin{matrix}0\le i\le k\le8\\i+k=16\end{matrix}\right.\)
\(\Rightarrow i=k=8\Rightarrow a_{16}=C_8^8C_8^8\left(-2\right)^0.2^0=1\)
\(\Rightarrow S=S_0-\left(a_0+a_{16}\right)=-2^8\)
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\(M+200=x^2+y^2+z^2+2xy-yz-xz\ge0\)
\(\Leftrightarrow x^2+x\left(2y-z\right)+y^2+z^2-yz\ge0\)
Can cm \(\left(2y-z\right)^2-4\left(y^2+z^2-yz\right)\le0\)
\(\Leftrightarrow3z^2\ge0\). TU dok ta co \(M+200\ge0\rightarrow M\ge-200\)
\("="\Leftrightarrow\left(x;y;z\right)=\left(10;-10;0\right)=\left(-10;10;0\right)\)
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a: \(log_22^{-13}=-13\)
b: \(lne^{\sqrt{2}}=\sqrt{2}\)
c: \(log_816-log_82=log_8\left(\dfrac{16}{2}\right)=log_88=1\)
c: \(log_26\cdot log_68=log_28=3\)
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Ta có \(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{1}{4}} \right)^2} - 1 = \frac{{ - 7}}{8}\)
Chọn B
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a: \(\left(\dfrac{1}{5}\right)^{-2}=25\)
b: \(4^{\dfrac{3}{2}}=8\)
c: \(\left(\dfrac{1}{8}\right)^{-\dfrac{2}{3}}=\left(\dfrac{1}{2}\right)^{3\cdot\dfrac{-2}{3}}=\left(\dfrac{1}{2}\right)^{-2}=4\)
d: \(\left(\dfrac{1}{16}\right)^{-0.75}=\left(\dfrac{1}{2}\right)^{4\cdot\left(-0.75\right)}=\left(\dfrac{1}{2}\right)^{-3}=8\)
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a) \(100+98+96+...+2-97-95-93-...-3\)
= \(100+98+\left(96-97\right)+\left(94-95\right)+...+\left(2-3\right)\)
= \(100+98-95\) = \(103\)
b) \(2-4-6+8+10-12-14+16+...-102+104\)
= \(\left(2-4\right)+\left(-6+8\right)+\left(10-12\right)+\left(-14+16\right)+...+\left(-102+104\right)\)
= \(-2+2-2+2-2+...+2\) = \(0\)
c) \(1+2-3-4+5+6-7-8+9+10-11-12+...-111-112+113+114\)
= \(\left(1+2\right)-\left(3+4\right)+\left(5+6\right)-\left(7+8\right)+...\left(113+114\right)\)
= \(3-7+11-15+19-23+...+219-223+227\)
= \(\left(3-7\right)+\left(11-15\right)+\left(19-23\right)+...+\left(219-223\right)+227\)
= \(-4-4-4-4-...-4+227\)
= \(54\left(-4\right)+227\) = \(-216+227\) = \(11\)
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\(\Leftrightarrow\left(sin^4x+cos^4x\right)^2-2\left(sinx.cosx\right)^4=\frac{17}{16}cos^22x\)
\(\Leftrightarrow\left[\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2\right]^2-2\left(\frac{1}{2}sin2x\right)^4=\frac{17}{16}cos^22x\)
\(\Leftrightarrow\left(1-\frac{1}{2}sin^22x\right)^2-\frac{1}{8}sin^42x=\frac{17}{16}cos^22x\)
\(\Leftrightarrow\frac{1}{8}sin^42x-sin^22x+1=\frac{17}{16}\left(1-sin^22x\right)\)
\(\Leftrightarrow2sin^42x+sin^22x-1=0\)
\(\Leftrightarrow\left(sin^22x+1\right)\left(2sin^22x-1\right)=0\)
\(\Leftrightarrow2sin^22x-1=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)