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1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)
\(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)=\(\frac{27.1}{33.27}\)=\(\frac{1}{33}\)
mik dang ban moi giai duoc mot bai ha, sorry
\(\frac{27\cdot18+27\cdot103-27\cdot120}{15\cdot33+12\cdot33}=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)\(=\frac{27\cdot1}{33\cdot27}=\frac{1}{33}\)
\(\Leftrightarrow x-\frac{1}{3}=\left(+-\right)\frac{5}{6}\)
nếu \(x-\frac{1}{3}=\frac{5}{6}\) nếu \(x-\frac{1}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}+\frac{1}{3}\) \(\Leftrightarrow x=-\frac{5}{6}+\frac{1}{3}\)
\(\Leftrightarrow x=\frac{7}{6}\) \(\Leftrightarrow x=-\frac{1}{2}\)
Vậy ....
nhớ k mk nha bạn , mk nhanh nhất
thanks
\(\frac{27.18+27.103-120.27}{15.33+33.12}\)
\(=\)\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)
\(=\)\(\frac{27.1}{33.27}\)
\(=\)\(\frac{1}{33}\)
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
1.a)\(\dfrac{5}{14}+\left(\dfrac{2}{201}+\dfrac{9}{14}\right)\)
\(=\dfrac{5}{14}+\dfrac{1837}{2814}\)
\(=\dfrac{203}{201}\)
b) \(\dfrac{22.18+27.103-120.27}{15.33+33.12}\)
\(=\dfrac{22.18+27.\left(103-120\right)}{33.\left(15+12\right)}\)
\(=\dfrac{22.18+27.\left(-17\right)}{33.27}\)
\(=\dfrac{22.18}{33.27}+\dfrac{27.\left(-17\right)}{33.27}\)
\(=\dfrac{2.11.3.3.2}{11.3.3.3.3}-\dfrac{27.17}{33.27}\)
\(=\dfrac{4}{9}-\dfrac{17}{27}\)
\(=\dfrac{-5}{27}\)
Mình nghĩ là đề của bạn sai nên mình sửa lại nhan!( Vì \(\dfrac{2}{4}=\dfrac{1}{2}\). Một nửa của chiều dài lại chưa bằng 50% chiều rộng. Vô lí)
Câu 2:Một miếng đất hình chữ nhật có chiều rộng 70m.Biết 40% chiều dài bằng \(\dfrac{2}{4}\) chiều rộng. Tính chu vi và diện tích miếng đất
Gọi độ dài chiều dài là a, độ dài chiều rộng là b \(\left(a,b\in N^{ }\right)\)
Theo bài ra ta có:\(a.40\%=\dfrac{2}{4}b\)
\(\Leftrightarrow a.40\%=\dfrac{2}{4}.70\)
\(\Leftrightarrow a.40\%=35\)
\(\Leftrightarrow a=35:40\%\)
\(\Leftrightarrow a=87,5\)
Chu vi miếng đất là:\(P=\left(a+b\right)2=\left(87,5+70\right)2=315\)(cm)
Diện tích miếng đất là:
\(S=a.b=87,5.70=6125\)(cm2)
a) \(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{3}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3}{12}+\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3+1+4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\dfrac{2}{3}-\dfrac{2011}{2012}=\dfrac{298}{719}\cdot\dfrac{3}{2}-\dfrac{2011}{2012}=\dfrac{149.3}{719.1}-\dfrac{2011}{2012}=\dfrac{447}{719}-\dfrac{2011}{2012}=\dfrac{889364}{1446628}-\dfrac{1445909}{1446628}=\dfrac{889364-1445909}{1446628}=-\dfrac{556545}{1446628}.\)b)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)