N=\(\frac{27.18+27.103-120.27}{15.33+33.12}\)

N=\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)

N=\(\frac{21.1}{33.27}\)

N=\(\frac{1}{33}\)

Vậy N=\(\frac{1}{33}\)

N=1/33

\(\frac{27.18+27.103-120.27}{15.33+33.12}=\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}=\frac{27.1}{33.27}=\frac{1.1}{33.1}=\frac{1}{33}\)

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\(=\frac{1}{33}\)

= 1/33

ai k mk thì mk k lại

\(\frac{27\cdot18+27\cdot103-27\cdot120}{15\cdot33+12\cdot33}=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)\(=\frac{27\cdot1}{33\cdot27}=\frac{1}{33}\)

\(\Leftrightarrow x-\frac{1}{3}=\left(+-\right)\frac{5}{6}\)

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

\(\frac{27.18+27.103-120.27}{15.33+33.12}\)

\(=\)\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)

\(=\)\(\frac{27.1}{33.27}\)

\(=\)\(\frac{1}{33}\)

\(|x-\frac{1}{3}|=\frac{5}{6}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x-\frac{1}{3}=\frac{5}{6}\\x-\frac{1}{3}=-\frac{5}{6}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{5}{6}+\frac{1}{3}\\x=-\frac{5}{6}+\frac{1}{3}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{7}{6}\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x=\frac{7}{6};-\frac{1}{2}\)

1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)

1) \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)\) \(=\dfrac{-5}{2}:\dfrac{1}{4}=-10\)

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)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)

\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)=\(\frac{27.1}{33.27}\)=\(\frac{1}{33}\)

mik dang ban moi giai duoc mot bai ha, sorry

k biet

\(\frac{27\cdot18\cdot27\cdot103-102\cdot27}{15\cdot33+33\cdot12}\)

=\(\frac{27\cdot\left(18+103-102\right)}{33\cdot\left(15+12\right)}\)

=\(\frac{27\cdot19}{33\cdot27}\)

=\(\frac{19}{33}\)

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) Ta có: \(\widehat{xOt}=\widehat{xOy}+\widehat{yOt}\)

\(\Leftrightarrow180^0=120^0+\widehat{yOt}\)

\(\Leftrightarrow\widehat{yOt}=180^0-120^0=60^0\)

Vậy \(\widehat{yOt}=60^0\)

b) Ta có:\(\widehat{MOt}=\widehat{MOy}+\widehat{yOt}\)

Mà: \(\widehat{MOy}=\widehat{xOM}=\dfrac{\widehat{xOy}}{2}=\dfrac{120^0}{2}=60^0\)

Nên: \(\widehat{MOt}=60^0+60^0=120^0\)

Vậy \(\widehat{MOt}=120^0\)

c) Ta có: \(\widehat{MON}=\widehat{MOt}-\widehat{tON}\)

Mà: \(\widehat{tON}=\widehat{NOy}=\dfrac{\widehat{yOt}}{2}=\dfrac{60^0}{2}=30^0\)

Nên:\(\widehat{MON}=120^0-30^0=90^0\)

Vậy \(\widehat{MON}=90^0\)