1 bỏ ngoặc rồi tính
\(\dfrac{3}{2}+\left(\dfrac{1}{5}-\dfrac{3}{4}\right)\)
\(1\dfrac{1}{3}+\left(\dfrac{2}{3}-\dfrac{3}{4}\right)-\left(0,8+1\dfrac{1}{5}\right)\)
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a) Ta có: ∠A + ∠B + ∠C = 180 độ
Mà ∠A=∠B=2∠C
⇒ 5∠C = 180 độ
⇒ ∠C = 180 độ : 5 = 36 độ
⇒ ∠A=∠B=2∠C=2.36 độ = 72 độ
Vậy ∠A=∠B=72 độ, ∠C=36 độ
b) Δ ABC là Δ nhọn
\(\dfrac{3}{8}:\left(\dfrac{7}{22}-\dfrac{2}{11}\right)+\dfrac{3}{8}:\left(\dfrac{2}{5}-\dfrac{1}{10}\right)\)
\(=\dfrac{3}{8}:\left(\dfrac{7}{22}-\dfrac{4}{22}\right)+\dfrac{3}{8}:\left(\dfrac{4}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{3}{8}:\left(\dfrac{7-4}{22}\right)+\dfrac{3}{8}:\left(\dfrac{4-1}{10}\right)\)
\(=\dfrac{3}{8}:\dfrac{3}{22}+\dfrac{3}{8}:\dfrac{3}{10}\)
\(=\dfrac{3}{8}.\dfrac{22}{3}+\dfrac{3}{8}.\dfrac{10}{3}\)
\(=\dfrac{3}{8}.\left(\dfrac{22}{3}+\dfrac{10}{3}\right)\)
\(=\dfrac{3}{8}.\left(\dfrac{22+10}{3}\right)\)
\(=\dfrac{3}{8}.\dfrac{32}{3}\)
\(=\dfrac{3.32}{8.3}\)
`=32/8`
`=4`
M = (100 – 1).(100 – 22). (100 – 32)…(100 – 502)
M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92) .(100 – 102) .(100 – 112) …(100 – 502)
M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92). (100 – 100) .(100 – 112) …(100 – 502)
M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92) .0.(100 – 112) …(100 – 502)
M = 0
Vậy M = 0
Lời giải:
\(\frac{3}{5}-[\frac{-1}{15}+\frac{16}{15}.(\frac{1}{4})^2]=\frac{3}{5}-[\frac{-1}{15}+\frac{16}{15}.\frac{1}{16}]\)
\(=\frac{3}{5}-(\frac{-1}{15}+\frac{1}{15})=\frac{3}{5}-0=\frac{3}{5}\)
=\(\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{3}{4}\)
=\(\dfrac{30}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)
=\(\dfrac{19}{20}\)
=\(\dfrac{4}{3}+\dfrac{2}{3}-\dfrac{3}{4}-\dfrac{4}{5}-\dfrac{6}{5}\)
=(\(\dfrac{4}{3}+\dfrac{2}{3}\)) -\(\dfrac{4}{3}-\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\)
=2\(-\dfrac{4}{3}-\dfrac{-2}{5}\)
=\(\dfrac{30}{15}-\dfrac{20}{15}-\dfrac{-6}{15}\)
=\(\dfrac{4}{15}\)