Var i, n,s: longint;

Begin

Write('Nhap so luong so le can tinh tong ');readln(n);

For i:=1 to n do

if i mod 2 <> 0 then s:=s+i;

write('Tong cua ',n,' so le dau tien la ',s);

readln;

end.

Ta có:

\(5\widehat{B}\) bù với \(\widehat{A}\Rightarrow5\widehat{B}+\widehat{A}=180^0\)

\(\Rightarrow\widehat{A}=180^0-5\widehat{B}\)              (1)

\(2\widehat{B}\) phụ với \(\widehat{A}\Rightarrow2\widehat{B}+\widehat{A}=90^0\)     (2)

Thay (1) vào (2), ta có:

\(2\widehat{B}+180^0-5\widehat{B}=90^0\)

\(3\widehat{B}=90^0\)

\(\widehat{B}=30^0\) thay vào (1) ta có:

\(\widehat{A}=180^0-5.30^0=30^0\)

Vậy \(\widehat{A}=\widehat{B}\)

Câu 2:

a) \(x-\dfrac{-5}{12}=\dfrac{-7}{12}\)

\(x=\dfrac{-7}{12}+\dfrac{-5}{12}\)

\(x=-1\)

Vậy \(x=-1\)

b) \(\dfrac{x}{20}=\dfrac{7}{10}+\dfrac{-13}{20}\)

\(\dfrac{x}{20}=\dfrac{14}{20}+\dfrac{-13}{20}\)

\(\dfrac{x}{20}=\dfrac{1}{20}\)

\(x=1\)

Vậy \(x=1\)

II. Tự luận

Câu 1:

a) \(\dfrac{3}{5}+\dfrac{-2}{5}=\dfrac{3+\left(-2\right)}{5}=\dfrac{1}{5}\)

b) \(\left(\dfrac{4}{5}+\dfrac{1}{2}\right).\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)

\(=\left(\dfrac{8}{10}+\dfrac{5}{10}\right).\dfrac{3-8}{13}\)

\(=\dfrac{13}{10}.\dfrac{-5}{13}\)

\(=\dfrac{-1}{2}\)

c) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\)

\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\)

\(=\dfrac{-5}{7}.1+1\)

\(=\dfrac{2}{7}\)

\(\dfrac{25.9-25.17}{-8.80-8.10}=\dfrac{25.\left(9-17\right)}{-8.\left(80+10\right)}=\dfrac{25.\left(-8\right)}{-8.90}=\dfrac{25}{90}=\dfrac{5}{18}\)

\(\dfrac{48.12-48.15}{-3.270-3.30}=\dfrac{48.\left(12-15\right)}{-3.\left(270+30\right)}=\dfrac{48.\left(-3\right)}{-3.300}=\dfrac{48}{300}=\dfrac{4}{25}\)

MC = 450

\(\dfrac{5}{18}=\dfrac{5.25}{18.25}=\dfrac{125}{450}\)

\(\dfrac{4}{25}=\dfrac{4.18}{25.18}=\dfrac{72}{450}\)

Giả sử ad âm

\(\Rightarrow\) a âm hoặc d âm

*) a âm

\(\Rightarrow\) -ac âm khi c âm

\(\Rightarrow\) -bc âm khi b âm

\(\Rightarrow\) -bd không âm (1)

*) d âm

\(\Rightarrow\) -bd âm khi b âm

\(\Rightarrow\) -bc âm khi c âm

\(\Rightarrow\) ac không âm (2)

Từ (1) và (2) \(\Rightarrow\) Các đơn thức ad, -bc, -ac, -bd không thể cùng âm

Ta có:

\(\dfrac{AB}{AM}=\dfrac{4}{2}=2\)

\(\dfrac{AC}{AN}=\dfrac{6}{3}=2\)

\(\Rightarrow\dfrac{AB}{AM}=\dfrac{AC}{AN}=2\)

Xét \(\Delta ABC\) và \(\Delta AMN\) có:

\(\dfrac{AB}{AM}=\dfrac{AC}{AN}=2\)

\(\widehat{A}\) chung

\(\Rightarrow\Delta ABC\) đồng dạng \(\Delta AMN\) (c-g-c)

Ta có:

|x| = \(\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3};x=-\dfrac{1}{3}\)

H và D ở đâu ra vậy em?

\(\dfrac{x-2}{2+x}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)    (1)

ĐKXĐ: \(x\ne2;x\ne-2\)

MTC = (x + 2)(x - 2)

\(\left(1\right)\Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-11\right)}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=2\left(x-11\right)\)

\(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\)

\(\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow\left(x^2-4x\right)-\left(5x-20\right)=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow x-4=0;x-5=0\)

*) \(x-4=0\)

\(\Leftrightarrow x=4\) (nhận)

*) \(x-5=0\)

\(\Leftrightarrow x=5\) (nhận)

Vậy \(S=\left\{4;5\right\}\)