Đặt \(x+y=t,t\in\left[-2;2\right]\)

Biến đổi được \(P=-2t^3+6t\)

Xét \(f\left(t\right)=-2t^3+6t\) trên \(\left[-2;2\right]\)

Lập bảng biến thiên

Ta có \(P_{Max}=4\) khi t=1

          \(P_{Min}=-4\) khi t= -1

\(a^2=\left|z+\frac{1}{z}\right|^2=\left(z+\frac{1}{z}\right)\left(\overline{z}+\frac{1}{z}\right)=\left|z\right|^2+\frac{z^2+\overline{z}^2}{\left|z\right|^2}+\frac{1}{\left|z\right|^2}\)

                       \(=\frac{\left|z\right|^4+\left(z+\overline{z}\right)^2-2\left|z\right|^2+1}{\left|z\right|^2}\)

Do đó :

\(\left|z\right|^4-\left|z\right|^2\left(a^2+2\right)+1=-\left(z+\overline{z}\right)^2\le0\)

\(\Rightarrow\left|z\right|^2\in\left[\frac{a^2+2-\sqrt{a^4+4a^2}}{2};\frac{a^2+2+\sqrt{a^4+4a^2}}{2}\right]\)

\(\Rightarrow\left|z\right|\in\left[\frac{-a+\sqrt{a^4+4a^2}}{2};\frac{a+\sqrt{a^4+4a^2}}{2}\right]\)

max \(\left|z\right|=\frac{a+\sqrt{a^4+4a^2}}{2}\)

min \(\left|z\right|=;\frac{a+\sqrt{a^4+4a^2}}{2}\)

\(\Leftrightarrow z\in M,z=-\overline{z}\)

đây là đáp án

3.

\(y'=-3x^2-6x=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)

\(y\left(-1\right)=m-2\) ; \(y\left(1\right)=m-4\)

\(\Rightarrow y_{min}=y\left(1\right)=m-4\)

\(\Rightarrow m-4=0\Rightarrow m=4\)

4.

Hàm đã cho bậc nhất trên bậc nhất nên đơn điệu trên mọi khoảng xác định

\(\Rightarrow y_{min}+y_{max}=y\left(1\right)+y\left(2\right)=\frac{m+1}{2}+\frac{m+2}{3}=8\)

\(\Rightarrow m=\frac{41}{5}\)

Đáp án B

1.

\(y'=\frac{1}{\left(sinx+1\right)^2}.cosx>0\Rightarrow y\) đồng biến

\(m=y_{min}=y\left(0\right)=2\)

\(M=y_{max}=y\left(1\right)=\frac{5}{2}\)

\(\Rightarrow M^2+m^2=\frac{41}{4}\)

2.

Hàm xác định trên \(\left[-2;2\right]\)

\(y'=1-\frac{x}{\sqrt{4-x^2}}=0\Leftrightarrow x=\sqrt{2}\)

\(y\left(-2\right)=-2\) ; \(y\left(\sqrt{2}\right)=2\sqrt{2}\) ; \(y\left(2\right)=2\)

\(\Rightarrow N=-2;M=2\sqrt{2}\)

\(\Rightarrow M+2N=2\sqrt{2}-4\)

\(logu_1+\sqrt{2+logu_1-2logu_{10}}=2logu_{10}\)

\(\Leftrightarrow logu_1-2logu_{10}+\sqrt{2+logu_1-2logu_{10}}=0\)

\(\Leftrightarrow t^2-2+t=0\)(\(t=\sqrt{2+logu_1-2logu_{10}}\ge0\)) 

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=-2\end{cases}}\)

\(\Rightarrow2+logu_1-2logu_{10}=1\)

\(\Leftrightarrow2+logu_1-2log\left(2^9u_1\right)=1\)

\(\Leftrightarrow log\left(10u_1\right)=log\left(2^9u_1\right)^2\)

\(\Rightarrow10u_1=2^{18}u_1^2\)

\(\Leftrightarrow u_1=\frac{10}{2^{18}}\).

\(u_n=\frac{2^{n-1}.10}{2^{18}}>5^{100}\Leftrightarrow n>log_2\left(\frac{5^{100}.2^{19}}{10}\right)=-log_210+100log_25+19\)

Suy ra \(n\ge248\).