giá trị lớn nhất là:9999 + 9998 = 19997

giá trị nhỏ nhất là: (-9999) + (-9998)= -19997

nghĩ 3 năm mới ra:))

a) Có \(|x+3|\)\(\ge\)0

=>\(100-\)\(|x+3|\le100\)

=> GTLN của A là 100 khi x= -3

b) Tương tự GTNN của B là 100 khi x=2

uses crt;

var a:array[1..100]of integer;

n,i,k,max,min,t:integer;

begin

clrscr;

write('Nhap n='); readln(n);

for i:=1 to n do 

  begin

write('A[',i,']='); readln(a[i]);

end;

write('Nhap k='); readln(k);

max:=a[1];

for i:=1 to n do 

  if max<a[i] then max:=a[i];

writeln('Gia tri lon nhat la: ',max);

writeln('Vi tri la: ');

for i:=1 to n do 

  if max=a[i] then write(i:4);

writeln;

min:=a[1];

for i:=1 to n do

  if min>a[i] then min:=a[i];

writeln('Gia tri nho nhat la: ',min);

writeln('Vi tri la: ');

for i:=1 to n do

if min=a[i] then write(i:4);

writeln;

t:=0;

for i:=1 to n do 

  if i mod 2=0 then t:=t+sqr(a[i]);

writeln('Tong binh phuong cac gia tri o vi tri chan la: ',t);

readln;

end. 

\(T=x^2-xy+y^2\)

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

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

\(\ge\frac{3y^2}{4}\)

\(\ge0\)

Dấu "=" xảy ra khi x=y=0

\(y=-5\cdot\dfrac{1-cos2x}{2}+12sin2x+7\)

\(=-\dfrac{5}{2}+\dfrac{5}{2}\cdot cos2x+12\cdot sin2x+7\)

\(=12\cdot sin2x+\dfrac{5}{2}\cdot cos2x+\dfrac{9}{2}\)

\(=\dfrac{\sqrt{601}}{2}\cdot\left(\dfrac{12\cdot sin2x}{\dfrac{\sqrt{601}}{2}}+cos2x\cdot\dfrac{5}{2}\cdot\dfrac{2}{\sqrt{601}}\right)+\dfrac{9}{2}\)

\(=\dfrac{\sqrt{601}}{2}\cdot\left(sin2x\cdot cosa+cos2x\cdot sina\right)+\dfrac{9}{2}\)

\(=\dfrac{\sqrt{601}}{2}\cdot sin\left(2x+a\right)+\dfrac{9}{2}\)

\(-1< =sin\left(2x+a\right)< =1\)

=>\(\dfrac{-\sqrt{601}}{2}< =\dfrac{\sqrt{601}}{2}\cdot sin\left(2x+a\right)< =\dfrac{\sqrt{601}}{2}\)

=>\(\dfrac{-\sqrt{601}+9}{2}< =y< =\dfrac{\sqrt{601}+9}{2}\)

\(y_{min}\) khi sin(2x+a)=-1

=>\(2x+a=-\dfrac{pi}{2}+k2pi\)

=>\(2x=-\dfrac{pi}{2}-a+k2pi\)

=>\(x=-\dfrac{pi}{4}-\dfrac{a}{2}+kpi\)

\(y_{max}\) khi sin(2x+a)=1

=>\(2x+a=\dfrac{pi}{2}+k2pi\)

=>\(x=\dfrac{pi}{4}-\dfrac{a}{2}+kpi\)