123+234=357

12345+9302747=9315092

456:12=38

\(\dfrac{1}{3}.\)\(\dfrac{3}{5}.\)\(\dfrac{5}{12}+\dfrac{2}{6}.\)\(\dfrac{6}{10}.\)\(\dfrac{10}{24}+\dfrac{3}{9}.\)\(\dfrac{9}{15}.\)\(\dfrac{15}{36}\)

\(=\)\(\dfrac{1}{3}.\)\(\dfrac{3}{5}.\)\(\dfrac{5}{12}+\dfrac{1}{3}.\)\(\dfrac{3}{5}.\)\(\dfrac{5}{12}+\dfrac{1}{3}.\)\(\dfrac{3}{5}.\)\(\dfrac{5}{12}\)

\(=\)(\(\dfrac{1}{3}.\)\(\dfrac{3}{5}.\)\(\dfrac{5}{12}\))\(.3\)

\(=\dfrac{1}{12}.3\)

\(=\dfrac{3}{36}\)

\(\)

y × 6/11 + y × 5/11 =2025

y × (6/11+5/11)      =2025

y ×1                           =2025

y                                 =2025

y × 6/11 + y × 5/11 =2025

y × (6/11+5/11)      =2025

y ×1                           =2025

y                                 =2025

\(\dfrac{1}{3}\times\dfrac{3}{5}\times\dfrac{5}{12}+\dfrac{2}{6}\times\dfrac{6}{10}\times\dfrac{10}{24}+\dfrac{3}{9}\times\dfrac{9}{15}\\ =\dfrac{1}{12}+\dfrac{2}{24}+\dfrac{3}{15}\\ =\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{5}\\=\dfrac{1}{6}+\dfrac{1}{5}\\ =\dfrac{5}{30}+\dfrac{6}{30}\\ =\dfrac{11}{30} \)

\(\left(1+\dfrac{1}{2}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{4}\right)\times...\times\left(1+\dfrac{1}{2020}\right)\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{2021}{2020}\\ =\dfrac{2021}{2}\)

a: Sau buổi sáng thì số mét vải còn lại chiếm:

\(1-\dfrac{3}{4}=\dfrac{1}{4}\)(tấm vải)

15m vải còn lại chiếm: \(\dfrac{1}{4}\times\left(1-\dfrac{1}{5}\right)=\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)(tấm vải)

Độ dài tấm vải là \(15:\dfrac{1}{5}=75\left(m\right)\)

b: Buổi sáng bán được: \(75\times\dfrac{3}{4}=56,25\left(m\right)\)

Buổi chiều bán được:

75-56,25-15=3,75(m)

Ta có:

\(\dfrac{43}{41}-1=\dfrac{2}{41};\dfrac{53}{51}-1=\dfrac{2}{51}\)

Vì \(\dfrac{2}{41}>\dfrac{2}{51}\) nên \(\dfrac{43}{41}>\dfrac{53}{51}\)

lộn tìm x xin lỗi nha

780-(x-45):2=120

=>(x-45):2=780-120=660

=>\(x-45=660\times2=1320\)

=>x=1320+45=1365

\(F=\dfrac{\dfrac{2010}{1}+\dfrac{2009}{2}+...+\dfrac{1}{2010}+2010}{1+\dfrac{1}{2}+...+\dfrac{1}{2010}}\)

\(=\dfrac{\left(\dfrac{2010}{1}+1\right)+\left(\dfrac{2009}{2}+1\right)+...+\left(\dfrac{1}{2010}+1\right)}{1+\dfrac{1}{2}+...+\dfrac{1}{2010}}\)

\(=\dfrac{\dfrac{2011}{1}+\dfrac{2011}{2}+...+\dfrac{2011}{2010}}{1+\dfrac{1}{2}+...+\dfrac{1}{2010}}=2011\)

Help me có ai làm nhanh hum

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