Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
$-\frac38\times\frac16+\frac{3}{-8}\times\frac56+\frac{-10}{16}$
$=-\frac38\times(\frac16+\frac56)-\frac58$
$=-\frac38\times1-\frac58$
$=-\frac38-\frac58=-\frac88=-1$
\(\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{3}{-8}\cdot\dfrac{5}{6}+\dfrac{-10}{16}\\ =\dfrac{-3}{8}\cdot\dfrac{1}{6}+\dfrac{-3}{8}\cdot\dfrac{5}{6}+\dfrac{-5}{8}\\ =\dfrac{-3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}\\ =-\dfrac{3}{8}\cdot1+\dfrac{-5}{8}\\ =\dfrac{-3-5}{8}\\ =-1\)
B(6) = {0; 6; 12; 18;…}
B(12) = {0; 12; 24;….}
B(42) = {0; 42; 84;…}
BC(6; 12; 42) = {0; 84; 168,…
Đặt \(A=2-4+8-16+...+512-1024+2048\)
\(2A=4-8+16-32+...+1024-2048+4096\)
\(A+2A=\left(2-4+8-16+...+512-1024+2048\right)+\left(4-8+16-32+...+1024-2048+4096\right)\)
\(3A=2+4096=4098\)
\(\Rightarrow A=\dfrac{4098}{3}=1366\)
\(A=2-4+8-16+...+512-1024+2048\)
=>\(2A=4-8+16-32+...+1024-2048+4096\)
=>\(2A+A=4-8+16-32+...+1024-2048+4096+2-4+8-16+...+512-1024+2048\)
=>\(3A=4096+2=4098\)
=>A=4098/3=1366
\(\left(\dfrac{5}{7}+\dfrac{-7}{15}\right)-\left(-\dfrac{8}{15}+\dfrac{5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-7}{15}+\dfrac{8}{15}-\dfrac{5}{7}\)
\(=\dfrac{8}{15}-\dfrac{7}{15}=\dfrac{1}{15}\)
a: \(48\cdot64+96\cdot18\)
\(=48\cdot64+48\cdot36\)
\(=48\left(64+36\right)=48\cdot100=4800\)
b: \(\left(42\cdot6565-4242\cdot65\right)\cdot\left(2001\cdot2002\cdot...\cdot2016\right)\)
\(=42\cdot65\left(101-101\right)\left(2001\cdot2002\cdot...\cdot2016\right)\)
=0
c: \(D=\left(2+4+6+...+100\right)\left(36\cdot333-108\cdot111\right)\)
\(=\left(36\cdot3\cdot111-36\cdot3\cdot111\right)\left(2+4+6+...+100\right)\)
\(=0\)
d: Tích này có 1000 thừa số nên n=1000
=>n-1000=0
=>\(G=0\cdot\left(1000-1\right)\cdot\left(1000-2\right)\cdot...\cdot\left(1000-999\right)=0\)
Ta có:
\(S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\\ \Rightarrow\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}+\dfrac{3}{14}< S< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}\\ \Rightarrow\dfrac{3+3+3+3+3}{14}< S< \dfrac{3+3+3+3+3}{10}\\ \Rightarrow\dfrac{15}{14}< S< \dfrac{3}{2}\\ \Rightarrow\dfrac{14}{14}< S< \dfrac{3}{2}\\ \Rightarrow1< S< \dfrac{3}{2}\)
=> S không phải số tự nhiên vì giữa 1 và `3/2` không có số tự nhiên
\(-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\\ =-\dfrac{3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\\ =\left(-\dfrac{3}{7}-\dfrac{4}{7}\right)+\left(\dfrac{3}{8}-\dfrac{3}{8}\right)\\ =\dfrac{-7}{7}+0\\ =-1\)
Ta có: \(\widehat{HBF}+\widehat{ABF}=180^{\circ}\) (hai góc kề bù)
\(\Rightarrow\widehat{ABF}=180^{\circ}-\widehat{HBF}=180^{\circ}-130^{\circ}=50^{\circ}\)
Khi đó: \(\widehat{CAB}=50^{\circ};\widehat{ABF}=50^{\circ}\)
\(\Rightarrow\widehat{CAB}=\widehat{ABF}\)
Mà hai góc này nằm ở vị trí so le trong
Do đó: \(CD//EF\)
Ta có: \(\widehat{FBH}+\widehat{ABF}=180^o\) (kề bù)
\(\Rightarrow\widehat{ABF}=180^o-\widehat{FBH}=180^o-130^o=50^o\)
\(\Rightarrow\widehat{CAE}=\widehat{ABF}=50^o\)
Mà 2 góc này ở vị trí so le trong
=> CD//EF