15.10-15 

=15.1-15

=0.1

Chiều rộng HCN là:

29.3=87(km)

Chiều dài HCN là:

87+29=116(km)

DT hình chữ nhật là:

87.116=10092(km²)

Đổi 10092km²=100920000000000cm²

^{Tick minh nha}

b)Giả sử \(\left\{{}\begin{matrix}x_1=\sqrt{t_1}\\x_2=-\sqrt{t_1}\\x_3=\sqrt{t_2}\\x_4=-\sqrt{t_2}\end{matrix}\right.\)

Ta có: \(\dfrac{x_1x_2x_3}{2x_4}=\dfrac{\sqrt{t_1}\left(-\sqrt{t_1}\right)\sqrt{t_2}}{-2\sqrt{t_2}}=\dfrac{t_1}{2}\)

Tương tự ta có: \(\dfrac{x_1x_2x_4}{2x_3}=\dfrac{t_1}{2};\dfrac{x_1x_3x_4}{2x_2}=\dfrac{t_2}{2};\dfrac{x_2x_3x_4}{2x_1}=\dfrac{t_2}{2}\)

\(\Rightarrow t_1+t_2=2013\Leftrightarrow m-1=2013\Leftrightarrow m=2014\left(TM\right)\)

a)Đặt \(t=x^2\) ta có: \(Pt\Leftrightarrow t^2+\left(1-m\right)t+2m-2=0\)

\(\Delta=\left(1-m\right)^2-4\left(2m-2\right)=1-2m+m^2-8m+8\\ =\left(m-1\right)\left(m-9\right)\)

Để phương trình có 4 nghiệm phân biệt thì \(\Delta>0\) tức là \(\left(m-1\right)\left(m-9\right)>0\Leftrightarrow\left[{}\begin{matrix}m< 1\\m>9\end{matrix}\right.\)

và \(t_1,t_2>0\)

Giả sử \(t_1>t_2>0\)

\(\Rightarrow m-1>\sqrt{\left(m-1\right)\left(m-9\right)}\Leftrightarrow m^2-2m+1>m^2-10m+9\\ \Leftrightarrow8m-8>0\Leftrightarrow m>1\)Vậy để phương trình có 4 nghiệm phân biệt thì \(m>9\)

\(\left\{{}\begin{matrix}x^2+y^2+xy+1=4y\left(1\right)\\y\left(x+y\right)=2x^2+7y+2\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow xy+y^2=2x^2+7y+2\left(3\right)\)

Thay \(\left(3\right)\) vào \(\left(1\right)\) ta có: \(\left(1\right)\Leftrightarrow x^2+2x^2+7y+2+1-4y=0\\ \Leftrightarrow x^2+y+1=0\\ \Leftrightarrow x^2+1=-y\)

Thay \(\left(4\right)\) vào \(\left(1\right)\): \(y^2+xy-5y=0\Leftrightarrow y\left(y+x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y=0\\y=5-x\end{matrix}\right.\)

Với y=0 thì \(x^2+1=0\) vô nghiệm

Với y=5-x thì \(x^2+1=x-5\Leftrightarrow x^2-x+6\) vô nghiệm

Vậy hpt vô nghiệm

uses crt;

var S: integer;

Begin

Clrscr;

Write('Nhap n: ');readln(n);

S:=1;

while S<= n do S:=S+2;

writeln('S = ', S-2);

End.

\(x_1=4+\sqrt{15};x_2=4-\sqrt{15}\)

\(U_{n+1}=x_1^{n+1}+x_2^{n+1}=x_1^nx_1+x_2^nx_2\\ =\left(x_1-x_2\right)x_1^n+x_2\left(x^n_1+x^n_2\right)\\ =\left(4+\sqrt{15}-4+\sqrt{15}\right)\left(4+\sqrt{15}\right)^n+\left(4-\sqrt{15}\right)U_n\\ =2\sqrt{15}\left(4+\sqrt{15}\right)^n+\left(4-\sqrt{15}\right)U_n\)

Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=2-a-b\\2ab-\left(2-a-b\right)^2=4\end{matrix}\right.\Leftrightarrow}}\left\{{}\begin{matrix}c=2-a-b\\2ab-4-a^2-b^2+4a+4a-2ab-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=2-a-b\\\left(a-2\right)^2+\left(b-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=-2\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\z=-\dfrac{1}{2}\end{matrix}\right.\)

\(\sqrt[3]{x+1}+\sqrt[3]{x+2}=\sqrt[3]{-\left(x+3\right)}\\ \Leftrightarrow\left(\sqrt[3]{x+1}\right)^3+\left(\sqrt[3]{x+2}\right)^3+3\sqrt[3]{\left(x+1\right)\left(x+2\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{x+2}\right)=-\left(x+3\right)\\ \Leftrightarrow x+1+x+2-3\sqrt[3]{x^2+3x+2}.\sqrt[3]{x+3}=-x-3\\ \Leftrightarrow3x+6=3\sqrt[3]{x^3+6x^2+11x+6}\\ \Leftrightarrow\left(x+2\right)^3=x^3+6x^2+11x+6\\ \Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2-11x-6=0\Leftrightarrow x=-2\)

\(x^2+y^2-4x-2=0\Leftrightarrow x^2+y^2=4x+2\)

\(-x^2+4x+2=y^2\ge0\Leftrightarrow2-\sqrt{6}\le x\le2+\sqrt{6}\\ \Rightarrow10-4\sqrt{6}\le4x+2\le10+4\sqrt{6}\\ \Leftrightarrow10-4\sqrt{6}\le x^2+y^2\le10+4\sqrt{6}\)