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a ) \(\frac{15}{2}-\left(-\frac{7}{3}\right)-\frac{7}{3}-10\)
= \(\frac{15}{2}+\frac{7}{3}-\frac{7}{3}-10\)
=\(\frac{15}{2}-10\)
= \(\frac{-5}{2}\)
b ) \(\frac{65}{2}+\left(-\frac{65}{3}\right)-\frac{130}{3}\)
= \(\frac{65}{6}-\frac{130}{3}\)
=\(\frac{-65}{2}\)
c ) \(\left(\frac{-16}{3}\right)+\left(\frac{-2}{3}\right)\)
= -6
d ) \(\frac{1}{7}+\frac{9}{-8}+\left(-\frac{3}{14}\right)\)
= \(\frac{-55}{56}+\left(-\frac{3}{14}\right)\)
= \(\frac{-67}{56}\)
e ) - 1,8 +\(\frac{5}{-6}\)
= \(\frac{-79}{30}\)
a ) \(\frac{15}{2}-\left(-\frac{7}{3}\right)-\frac{7}{3}-10\)
= \(\frac{15}{2}+\frac{7}{3}-\frac{7}{3}-10\)
=\(\frac{15}{2}-10\)
= \(\frac{-5}{2}\)
b ) \(\frac{65}{2}+\left(-\frac{65}{3}\right)-\frac{130}{3}\)
= \(\frac{65}{6}-\frac{130}{3}\)
=\(\frac{-65}{2}\)
c ) \(\left(\frac{-16}{3}\right)+\left(\frac{-2}{3}\right)\)
= -6
d ) \(\frac{1}{7}+\frac{9}{-8}+\left(-\frac{3}{14}\right)\)
= \(\frac{-55}{56}+\left(-\frac{3}{14}\right)\)
= \(\frac{-67}{56}\)
e ) - 1,8 +\(\frac{5}{-6}\)
= \(\frac{-79}{30}\)
bạn ghi sai đề phải ko?
Δ ABCD= hình thang ABCD
góc BCx=góc BCD
A B C D
ta có: AB//CD
=>góc A+ góc D=180*=góc C+ góc B
=> góc A=180*-65*=115*
ta có:
góc B+ góc C=180*
=> góc B=180*-130*=50*
vậy góc A=115 độ; góc B=50 độ
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-1\)
Ta đặt: \(P=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
=> \(2P=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
=> \(2P-P=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
=> \(P=1-\frac{1}{64}\)
Mà \(S=P-1\)
=> \(S=1-\frac{1}{64}-1=-\frac{1}{64}\)
Vậy \(S=-\frac{1}{64}\)
=-65/2