\(\frac{2^3.3}{2^23^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha 

cho mik hỏi bạn tùng lâm là bạn hướng dẫn mik cách giải bài này đc ko tại vì mik vẫn còn mấy bài giống vậy

\(\frac{1212}{3131}=\frac{1212:101}{3131:101}=\frac{12}{31}\)

12/31 nhé

12/31 nhé

Ảnh hơi mờ 

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)

(-5)/9

-14 nha bạn

1/2 . P = 1/2.6 + 1/6.10 + 1/10.14 + ... + 1/198.202

4.1/2. P= 4/2.6 + 4/6.10 + 4/10.14 + ... + 4/198.202

2P=1/2-1/6+1/6-1/10+1/10-1/14+...+1/198-1/202

2P=1/2-1/202=50/101

P=50/101:2=50/101.1/2=25/101

\(\frac{-5^3\cdot40\cdot4^3}{135\cdot\left(-2\right)^{14}\left(-100\right)^0}=\frac{-125\cdot2^3\cdot5\cdot\left(2^2\right)^3}{5\cdot27\cdot2^{14}\cdot1}=\frac{-125\cdot2^6}{27\cdot2^{11}}=\frac{-125}{27\cdot2^5}=\frac{-125}{864}\)

\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}\)\(=\frac{\left(-5\right)^3.5.2^3.2^6}{3^3.5.2^{14}.1}\)\(=\frac{-125}{864}\)

a) e  chỉ  cần nhân chúng lại với nhau =  cách tách từng cái ra

b)đặt 4/2.5+4/5.8+4/8.11+......+4/62.65 là S

\(.S=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\right)\)

\(S=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\right)\)

\(S=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{65}\right)\)

\(S=\frac{4}{3}\left(\frac{65}{130}-\frac{2}{130}\right)\)

\(S=\frac{4}{3}\left(\frac{63}{130}\right)\)

\(S=\frac{42}{65}\)

Bài 2 : \(\frac{15+a}{29+a}=\frac{3}{5}\)\(\Leftrightarrow\left(15+a\right)5=\left(29+a\right)3\Leftrightarrow75+5a=87+3a\Leftrightarrow5a-3a=87-75\Rightarrow2a=12\Rightarrow a=6\)

vậy a =6

cái nài từ 2 năm tr rùi h bạn có cần nxko ạ:)))

\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

Ta có: \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{20}{41}\)

Lại có: \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{31}\)

\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)

\(B=\dfrac{10}{31}\)

Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên...