3:

a:Các tia trên hình là Ax,Ay,Bx,By,Cx,Cy

=>Có 6 tia

b: AB<AC

=>B nằm giữa A và C

=>AB+BC=AC

=>BC=4cm

c: AI=3/2=1,5cm

CI=7-1,5=5,5cm

1:

a: BC=8-3=5cm

b: MN=MC+CN=1/2(CA+CB)

=1/2*AB=4cm

2: 

a: Có 2 tia là OA và OB

b: AB=OB+OA=11cm

c: AC=BC=11/2=5,5cm

Trả lời:

a, ( - x + 5 )² - 16 = ( - 2² ) . 5

=> ( - x + 5 )² - 16 = - 20

=> ( - x + 5 )² = - 20 + 16

=> ( - x + 5 )² = - 4 ( vô lí )

Vậy không tìm được x thỏa mãn đề bài.

b, 50 - ( 20 - x ) = - x - ( 45 - 85 )

=> 50 - 20 + x = - x - ( - 40 )

=> 30 + x = - x + 40

=> x + x = 40 - 30

=> 2x = 10

=> x = 10 : 2

=> x = 5

Vậy x = 5

Bài 2:

\(\sqrt{x^2-6x+9}=x-2\)

=>\(\sqrt{\left(x-3\right)^2}=x-2\)

=>|x-3|=x-2

=>\(\begin{cases}x-2\ge0\\ \left(x-3\right)^2=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge2\\ x^2-6x+9-x^2+4x-4=0\end{cases}\)

=>\(\begin{cases}x\ge2\\ -2x+5=0\end{cases}\Rightarrow x=\frac52\)

Bài 1:

a: \(-2\sqrt{12}+\frac15\cdot\sqrt{75}-\sqrt{27}\)

\(=-2\cdot2\sqrt3+\frac15\cdot5\sqrt3-3\sqrt3\)

\(=-4\sqrt3+\sqrt3-3\sqrt3=-6\sqrt3\)

b: \(\sqrt{52-16\sqrt3}+\sqrt{\left(4\sqrt3-7\right)^2}\)

\(=\sqrt{48-2\cdot4\sqrt3\cdot2+4}+\left|4\sqrt3-7\right|\)

\(=\sqrt{\left(4\sqrt3-2\right)^2}+7-4\sqrt3=4\sqrt3-2+7-4\sqrt3\)

=-2+7

=5

c; \(\frac{\sqrt{27}-3\sqrt2}{\sqrt3-\sqrt2}+\frac{2}{\sqrt5-2}-\frac{10}{\sqrt5}\)

\(=\frac{3\sqrt3-3\sqrt2}{\sqrt3-\sqrt2}+\frac{2\left(\sqrt5+2\right)}{\left(\sqrt5-2\right)\left(\sqrt5+2\right)}-2\sqrt5\)

\(=\frac{3\left(\sqrt3-\sqrt2\right)}{\sqrt3-\sqrt2}+2\left(\sqrt5+2\right)-2\sqrt5=3+2\sqrt5+4-2\sqrt5\)

=7

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\)

\(\sqrt{x^2-x-2}-\sqrt{x-2}=0\\ \Leftrightarrow\sqrt{x^2-x-2}=\sqrt{x-2}\\ \Leftrightarrow x^2-x-2=x-2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(a,ĐK:x\ge2\\ PT\Leftrightarrow x^2-x-2=x-2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=2\\ b,ĐK:\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\\ PT\Leftrightarrow\sqrt{x^2-1}=x^2-1\\ \Leftrightarrow x^2-1=\left(x^2-1\right)^2\\ \Leftrightarrow\left(x^2-1\right)\left(x^2-1-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\\x=\sqrt{2}\left(tm\right)\\x=-\sqrt{2}\left(tm\right)\end{matrix}\right.\)

\(c,ĐK:\left[{}\begin{matrix}x\le-2\\x\ge1\end{matrix}\right.\\ PT\Leftrightarrow\sqrt{x^2-x}=-\sqrt{x^2+x-2}\\ \Leftrightarrow x^2-x=x^2+x-2\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\left(tm\right)\)

Mỗi bó có số khóm hoa là:
\(3:6=\dfrac{1}{2}\)(khóm)
Đáp số: \(\dfrac{1}{2}\) khóm hoa

uses crt;

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

i,n,t:integer;

begin

clrscr;

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

for i:=1 to n do 

 begin

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

end;

t:=0;

for i:=1 to n do 

  t:=t+a[i];

writeln(t);

readln;

end.

=1540/220

=7

ANH GIẢI NHƯ THẾ NÀO