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c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
a)11 3/13-(2 4/7+5 3/13)=11+3/13-(2+4/7+5+3/13)=11+3/13-2-4/7-5-3/13=(11-2-5)+(3/13-4/7-3/13)=4+(0-4/7)=4+-4/7=28/7-4/7=28-4/7=24/7=3 3/7(phải tính ta hỗn số nha bn)
SORRY MIK CHỈ LM` 1 CÁI CÒN LẠI ĐỂ BN ĐÓ
THỰC RA MIK BIK HẾT R` NHƯNG ĐỂ BN TỰ MIK LM` ĐÓ NHA
NẾU KO LM` ĐC THÌ NHỜ MIK NHẮN TIN HOẶC GỌI QUA 01288449416 NHA R` MIK LÊN GIẢI CHO.
\(\text{}\text{}11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)=11\frac{3}{13}-2\frac{4}{7}-5\frac{3}{13}=11\frac{3}{13}-5\frac{3}{13}-2\frac{4}{7}=6-2\frac{4}{7}=5\frac{7}{7}-2\frac{4}{7}=3\frac{3}{7}\)
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
a) \(\frac{1}{3}-\frac{-1}{6}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
b) \(2\frac{1}{3}+4\frac{1}{5}=\frac{7}{3}+\frac{21}{5}=\frac{98}{15}\)
c) \(\frac{4}{9}-\frac{13}{3}-\frac{4}{9}-\frac{10}{3}=\left(\frac{4}{9}-\frac{4}{9}\right)-\left(\frac{13}{3}+\frac{10}{3}\right)\)
\(=0-\frac{23}{3}=\frac{-23}{3}\)
d) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{4.78}{104}=\frac{312}{104}=3\)
~ Chúc em học tốt ~
b, \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)
=\(\frac{2^2.78}{104}\)=\(\frac{312}{104}\)=3