a) \(2\dfrac{1}{2}\) - x + \(\dfrac{4}{5}\) = \(\dfrac{2}{3}\) - (\(-\dfrac{4}{7}\) )
b) \(-\dfrac{4}{7}\) - x = \(\dfrac{3}{5}\) - 2x
c) (\(\dfrac{3}{8}\) - \(\dfrac{1}{5}\) ) + (\(\dfrac{5}{8}\) - x) = \(\dfrac{1}{5}\)
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Đồng vị: \(\widehat{C_2}=\widehat{A_2};\widehat{E_2}=\widehat{A_2};\widehat{E_2}=\widehat{C_2}\)
So le trong: \(\widehat{C_4}=\widehat{A_2};\widehat{C_3}=\widehat{A_1};\widehat{A_3}=\widehat{E_1}\)
Trong cùng phía: \(\widehat{A_3}+\widehat{E_2}=180^o;\widehat{A_4}+\widehat{E_1}=180^o;\widehat{C_3}+\widehat{A_1}=180^o\)
\(\widehat{B_1}\) = \(\widehat{B_2}\) = 1000 (hai góc đối đỉnh)
\(\widehat{C_2}\) = \(\widehat{B_1}\) = 1000 (hai góc đồng vị)
\(\widehat{C_3}\) + \(\widehat{C_2}\) = 1800 ⇒ \(\widehat{C_3}\) = 1800 - 1000 = 800
\(\widehat{D_1}\) = \(\widehat{A_1}\) = 600 (so le trong)
\(\widehat{DAH}\) = 900 - 600 = 300
a) \(-\dfrac{3}{5}-x=-0,75\)
\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)
b) \(1\dfrac{4}{5}=-0,15-x\)
\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)
a) \(-\dfrac{3}{5}-x=-0,75\)
\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)
\(x=\dfrac{27}{20}\)
________
b) \(1\dfrac{4}{5}=-0,15-x\)
\(=>-0,15-x=\dfrac{9}{5}\)
\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)
\(x=\dfrac{-39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)
\(x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)
\(x=\dfrac{6}{15}=\dfrac{2}{5}\)
\(a,0,\left(123\right)+0,\left(876\right)=\dfrac{123}{999}+\dfrac{876}{999}=\dfrac{999}{999}=1\left(đpcm\right)\\ b,0,\left(123\right).3+0,\left(630\right)=\dfrac{123}{999}.3+\dfrac{630}{999}=\dfrac{369}{999}+\dfrac{630}{999}=\dfrac{999}{999}=1\left(đpcm\right)\)
Đổ vào bể số nước là:
30 x 120 = 3600 (L nước)
= 3,6m3
Chiều rộng của bể nước là:
3,6 : 0,6 : 3 = 2 (m)
Đáp số: 2m
\(3,\left(5\right)=3+0,\left(5\right)=3+5.0,\left(1\right)=3+5.\dfrac{1}{9}=3+\dfrac{5}{9}=\dfrac{32}{9}\)
\(\left(-39,82\right):\left(-7,24\right)\)
\(=39,82:7,24\)
\(=5,5\)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)