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a: Trên tia Ox, ta có: OA<OB
nên A nằm giữa O và B
=>OA+AB=OB
=>AB+4=6
=>AB=2(cm)
b: C là trung điểm của OA
=>\(CO=CA=\dfrac{OA}{2}=2\left(cm\right)\)
Vì AO và AB là hai tia đối nhau
nên AC và AB là hai tia đối nhau
=>A nằm giữa hai điểm B và C
Ta có: A nằm giữa B và C
mà AB=AC(=2cm)
nên A là trung điểm của BC
d: Các góc đỉnh D trong hình vẽ là: \(\widehat{ODC};\widehat{ODA};\widehat{ODB};\widehat{CDA};\widehat{CDB};\widehat{ADB}\)
a: Trên tia Ox, ta có: OA<OB
nên A nằm giữa O và B
=>OA+AB=OB
=>AB+3=5
=>AB=2(cm)
b: Vì OC và OA là hai tia đối nhau
nên O nằm giữa C và A
Ta có: O nằm giữa C và A
mà OC=OA(=3cm)
nên O là trung điểm của AC
c: TH1: I nằm giữa O và B
=>OI+IB=OB
=>IB+4=5
=>IB=1(cm)
TH2: I nằm trên tia đối của tia OA
I nằm trên tia đối của tia OA
nên I nằm trên tia đối của tia OB
=>O nằm giữa I và B
=>IB=IO+OB=4+5=9(cm)
Lời giải:
Gọi số chia là $a$. Vì số chia luôn lớn hơn số dư nên $a>29$.
Theo bài ra thì: $65a+29< 1980$
$\Rightarrow 65a< 1951$
$\Rightarrow a< 30,02$
Mà $a>29$ nên $a=30$
Vậy số chia là $30$
Số số hạng trong dãy số 100;98;...;2 là:
\(\dfrac{100-2}{2}+1=\dfrac{98}{2}+1=50\left(số\right)\)
Tổng của dãy số 100;98;...;2 là:
\(\left(100+2\right)\cdot\dfrac{50}{2}=102\cdot25=2550\)
100+98+...+2+97-95-93
=2550+2-93
=2552-93
=2459
a: \(\dfrac{-3}{5}+\dfrac{28}{5}\left(\dfrac{13}{56}-\dfrac{5}{24}+\dfrac{1}{7}\right)\)
\(=\dfrac{-3}{5}+\dfrac{28}{5}\left(\dfrac{39}{168}-\dfrac{35}{168}+\dfrac{24}{168}\right)\)
\(=\dfrac{-3}{5}+\dfrac{28}{5}\cdot\dfrac{28}{168}=\dfrac{-3}{5}+\dfrac{28}{5}\cdot\dfrac{1}{6}=\dfrac{-3}{5}+\dfrac{14}{15}\)
\(=\dfrac{-9}{15}+\dfrac{14}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
b: \(\dfrac{5}{7}\cdot\dfrac{11}{18}+\dfrac{3}{7}\cdot\dfrac{5}{18}+\dfrac{4}{9}\)
\(=\dfrac{5}{18}\left(\dfrac{11}{7}+\dfrac{3}{7}\right)+\dfrac{4}{9}\)
\(=\dfrac{5}{18}\cdot2+\dfrac{4}{9}=\dfrac{5}{9}+\dfrac{4}{9}=\dfrac{9}{9}=1\)
c: \(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\dfrac{-5}{7}=10\cdot\dfrac{-7}{5}=-14\)
d: \(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
\(=\left(-\dfrac{3}{5}+\dfrac{4}{9}+\dfrac{-2}{5}+\dfrac{5}{9}\right)\cdot\dfrac{11}{7}\)
\(=\left(-\dfrac{5}{5}+\dfrac{9}{9}\right)\cdot\dfrac{11}{7}=\left(-1+1\right)\cdot\dfrac{11}{7}=0\)
e: \(\dfrac{-3}{4}\cdot5\dfrac{3}{13}-0,75\cdot\dfrac{36}{13}\)
\(=\dfrac{-3}{4}\left(5+\dfrac{3}{13}+\dfrac{36}{13}\right)\)
\(=\dfrac{-3}{4}\cdot8=-6\)
f: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{302\cdot305}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{302\cdot305}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{305}\right)=\dfrac{1}{3}\cdot\dfrac{12}{61}=\dfrac{4}{61}\)
g: \(6\dfrac{5}{12}:2\dfrac{3}{4}-11\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}-\dfrac{45}{4}\cdot\dfrac{2}{15}=\dfrac{77}{12}\cdot\dfrac{4}{11}-\dfrac{3}{2}\)
\(=\dfrac{7}{3}-\dfrac{3}{2}=\dfrac{14}{6}-\dfrac{9}{6}=\dfrac{5}{6}\)
h: \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right)\cdot2\dfrac{2}{3}\cdot0,25\)
\(=\left(0,6+0,415-0,015\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}\)
\(=1\cdot\dfrac{2}{3}=\dfrac{2}{3}\)
i: \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right)\cdot\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right)\cdot\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right)\cdot\dfrac{10}{11}=\dfrac{5}{2}-\dfrac{33}{20}\cdot\dfrac{10}{11}=\dfrac{5}{2}-\dfrac{3}{2}=\dfrac{2}{2}=1\)
\(C=\left(6-\dfrac{2}{3}+\dfrac{5}{16}\right)-\left(2+\dfrac{4}{3}-\dfrac{7}{8}\right)-\left(6-\dfrac{5}{8}+\dfrac{13}{12}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{5}{16}-2-\dfrac{4}{3}+\dfrac{7}{8}-6+\dfrac{5}{8}-\dfrac{13}{12}\)
\(=\left(6-2-6\right)+\left(-\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(\dfrac{5}{16}+\dfrac{7}{8}+\dfrac{5}{8}\right)-\dfrac{13}{12}\)
\(=-2-2+\left(\dfrac{5}{16}+\dfrac{12}{8}\right)-\dfrac{13}{12}\)
\(=-4-\dfrac{13}{12}+\dfrac{29}{16}=-\dfrac{157}{48}\)
\(C=\left(6-\dfrac{2}{3}-\dfrac{5}{10}\right)-\left(2+\dfrac{4}{3}-\dfrac{7}{8}\right)-\left(6-\dfrac{5}{8}+\dfrac{13}{12}\right)\)
\(=6-\dfrac{2}{3}-\dfrac{5}{10}-2-\dfrac{4}{3}+\dfrac{7}{8}-6+\dfrac{5}{8}-\dfrac{13}{12}\)
\(=\left(6-2-6\right)+\left(-\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(\dfrac{7}{8}+\dfrac{5}{8}\right)+\left(\dfrac{-5}{10}-\dfrac{13}{12}\right)\)
\(=\left(-2\right)+\left(-2\right)+\dfrac{3}{2}+\dfrac{-19}{12}\)
\(=\left(-4\right)+\dfrac{18}{12}+\dfrac{-19}{12}\)
\(=\left(-4\right)+\dfrac{-1}{12}\)
\(=\dfrac{-48}{12}+\dfrac{-1}{12}\)
\(=\dfrac{-49}{12}\)
\(B=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(-\dfrac{3}{4}-\dfrac{7}{4}+\dfrac{5}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)\)
\(=-3+\dfrac{-5}{4}+\dfrac{-7}{5}=-3-1,25-1,4=-5,65\)
giup em bai nay a kho qua
a) \(\dfrac{-3}{21}+\dfrac{-2}{7}+\dfrac{1}{3}=\dfrac{-3}{21}+\dfrac{-6}{21}+\dfrac{7}{21}=-\dfrac{2}{21}\)
b) \(\dfrac{-13}{15}+\dfrac{5}{-18}+\dfrac{4}{9}=\dfrac{-78}{90}+\dfrac{-25}{90}+\dfrac{40}{90}=\dfrac{63}{90}=\dfrac{7}{10}\)
c) \(\dfrac{-2}{5}-\left(\dfrac{-3}{11}\right)=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d) \(\left(-4\right)-\left(\dfrac{-4}{5}\right)-\dfrac{2}{3}=\left(-4\right)+\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{-60}{15}+\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{-58}{15}\)
e) \(\dfrac{3}{5}+\left(\dfrac{-4}{3}\right)-\dfrac{-3}{4}=\dfrac{3}{5}+\left(\dfrac{-4}{3}\right)+\dfrac{3}{4}=\dfrac{36}{60}+\dfrac{-80}{60}+\dfrac{45}{60}=\dfrac{1}{60}\)
g) \(\dfrac{5}{8}-\left(-\dfrac{2}{5}\right)-\dfrac{3}{10}=\dfrac{5}{8}+\dfrac{2}{5}-\dfrac{3}{10}=\dfrac{25}{40}+\dfrac{16}{40}-\dfrac{12}{40}=\dfrac{29}{40}\)
h) \(\dfrac{3}{4}-\left(-\dfrac{5}{3}\right)+\left(\dfrac{1}{12}+\dfrac{2}{9}\right)=\dfrac{3}{4}+\dfrac{5}{3}+\left(\dfrac{3}{36}+\dfrac{8}{36}\right)=\dfrac{27}{36}+\dfrac{60}{36}+\dfrac{11}{36}=\dfrac{98}{36}=\dfrac{49}{18}\)
a) \(\dfrac{-3}{21}+\dfrac{-2}{7}+\dfrac{1}{3}=\dfrac{-3}{21}-\dfrac{6}{21}+\dfrac{7}{21}\\ =\dfrac{-3-6+7}{21}=-\dfrac{2}{21}\)
b) \(\dfrac{-13}{15}+\dfrac{5}{-18}+\dfrac{4}{9}=\dfrac{-78}{90}-\dfrac{25}{90}+\dfrac{40}{90}\\ =\dfrac{-78-25+40}{90}=\dfrac{-63}{90}=-\dfrac{7}{10}\)
c) \(\dfrac{-2}{5}-\left(\dfrac{-3}{11}\right)=-\dfrac{2}{5}+\dfrac{3}{11}\\ =-\dfrac{22}{55}+\dfrac{15}{55}=-\dfrac{7}{55}\)