Bài 2 ; 2 ; 3 ; 4 phần tự luận bạn tham khảo ở Câu hỏi của Phùng minh long - Toán lớp 7 - Học trực tuyến OLM nhé !

Bài 1 :

\(1)\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)

\(\Rightarrow\left(\frac{11}{24}+\frac{13}{24}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)+\frac{1}{2}\)

\(\Rightarrow1-1+\frac{1}{2}=\frac{1}{2}\)

\(2)\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac{2}{3}\)

\(=\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1-\frac{15}{17}+\frac{14}{21}\)

\(=\left(\frac{15}{34}+\frac{19}{34}-1\right)+\left(\frac{7}{21}+\frac{14}{21}\right)-\frac{15}{17}\)

\(=1-\frac{15}{17}=\frac{2}{17}\)

\(3)1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5=2,5\)

\(4)\left(-12\right):\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

\(=\left(-12\right):\left(\frac{-1}{12}\right)^2\)

\(=\left(-12\right).12^2=-12^3\)

\(5)\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)

\(=\frac{-891}{25}:4=\frac{-891}{100}\)

\(6)\left(-6,5\right).5,7+5,7.\left(-3,5\right)\)

\(=\left[\left(-6,5\right)+\left(-3,5\right)\right].5,7\)

\(=-10.5,7=57\)

\(7)-3,75.\left(-7,2\right)+2,8.3,75\)

\(=3,75.7,2+2,8.3,75\)

\(=3,75.\left(7,2+2,8\right)\)

\(=3,75.10=37,5\)

\(8)23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)

\(=23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)

\(=\left(23\frac{1}{4}-13\frac{1}{4}\right).1,4\)

\(=10.1,4=14\)

\(9)16\frac{2}{7}:\left(\frac{-3}{5}\right)+28\frac{2}{7}:\frac{3}{5}\)

\(=-16\frac{2}{7}:\frac{3}{5}+28\frac{2}{7}:\frac{3}{5}\)

\(=-16\frac{2}{7}.\frac{5}{3}+28\frac{2}{7}.\frac{5}{3}\)

\(=\left(-16\frac{2}{7}+28\frac{2}{7}\right).\frac{5}{3}\)

\(=12.\frac{5}{3}=20\)

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đây nè bạn

Ta có Ax // zc

Suy ra góc xAc=C1=40(So le trong)

Lại có By // zc

Suy ra góc cBy=C2=60(so le trong)

Vì góc ACB=C1+C2

Mà C1=40,C2=60

Suy ra góc ACB=40+60=100

Vậy góc ACB=100

Good luck!

xem boif

1x10=10 k mk nhe

2323333344+33443343312:2= 2323333344+16721671656  = 19045005000

Bài 2 :

\(\left(2x+1\right)^2=\frac{16}{25}\)

\(\Rightarrow\left(2x+1\right)^2=\left(\frac{4}{5}\right)^2=\left(-\frac{4}{5}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{4}{5}\\2x+1=-\frac{4}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{-1}{5}\\2x=-\frac{9}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{10}\\x=-\frac{9}{10}\end{cases}}\)

b) \(0,5-\left|\frac{5}{4}-2x\right|=\frac{1}{3}\)

\(\Rightarrow\frac{1}{2}-\left|\frac{5}{4}-2x\right|=\frac{1}{3}\)

\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}\\2x=\frac{5}{4}-\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{13}{12}\\2x=\frac{17}{12}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{24}\\x=\frac{17}{24}\end{cases}}\)

c) Ta có :

\(\frac{x}{-3}=\frac{y}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x}{-3}=\frac{y}{7}=\left(\frac{x}{-3}\right)^2=\left(\frac{y}{7}\right)^2=\frac{x^2}{\left(-3\right)^2}=\frac{y^2}{7^2}=\frac{x^2-y^2}{\left(-3\right)^2-7^2}=\frac{-160}{40}=4\)

\(\Rightarrow\hept{\begin{cases}x=-3.4=-12\\y=7.4=28\end{cases}}\)

Bài  5 :

Ta có : a + c = 2b và 2bd = c . ( b + d ) ( b và d khác 0 )

=> ( a + c ) . d = c . ( b + d )

=> ad + cd = cb + cd

=> ad = cb

\(\Rightarrow\frac{ad}{bd}=\frac{cb}{bd}\Rightarrow\frac{a}{b}=\frac{c}{b}\)

bài nào vậy bn

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