Mk gợi ý nhưng chưa chắc đúng đâu nhé vì mk ít khi làm dạng này

`12=4xx3`

`15=3xx5`

`18=2xx9`

Vậy số này chia hết cho `4;3;5;2;9`

\(\frac{7}{x}=\frac{y}{27}=-\frac{42}{54}\)

\(\Leftrightarrow\frac{7}{x}=\frac{y}{27}=-\frac{7}{9}\)

Có \(\frac{7}{x}=-\frac{7}{9}\)

\(\Leftrightarrow x=-9\)

Lại có \(\frac{y}{27}=-\frac{7}{9}\)

\(\Leftrightarrow x=-21\)

X= -9, Y= -21   CHƯA CHẮC ĐÂY ĐẤY

Bài 3: 

Ta có: a//b

nên \(x+y=180\)

mà \(2x-3y=0\)

nên \(\left\{{}\begin{matrix}x+y=180\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y=180\\2x-3y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=180\\x+y=180\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=36\\x=144\end{matrix}\right.\)

ok bn noi di

bài gì???

BÀi 2:

\(\left|x-\frac{2019}{2020}\right|\ge0\forall x\)

=>\(-\frac{2020}{2019}\left|x-\frac{2019}{2020}\right|\le0\forall x\)

=>\(-\frac{2020}{2019}\left|x-\frac{2019}{2020}\right|+\frac{2019}{2020}\le\frac{2019}{2020}\forall x\)

Dấu '=' xảy ra khi \(x-\frac{2019}{2020}=0\)

=>\(x=\frac{2019}{2020}\)

Bài 1:

a: \(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{402-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac35+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(=\frac{5\left(31-\frac27-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac27-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac15+\frac{1}{13}-\frac{3}{10}\right)}{\frac15+\frac{1}{13}-\frac{3}{10}}=\frac{5}{13}+3=\frac{44}{13}\)

b: \(B=-\frac54+\frac35-1\frac{3}{14}:\left|-\frac{34}{21}\right|+\frac{-5}{17}:\left|-\frac{1}{34}\right|+\frac53\cdot\left(\frac12-\frac25\right)\)

\(=-\frac54+\frac35-\frac{17}{14}\cdot\frac{21}{34}+\frac{-5}{17}\cdot34+\frac53\cdot\frac{1}{10}\)

\(=-\frac54+\frac35-\frac{3}{2\cdot2}+\left(-10\right)+\frac{5}{30}=-\frac54+\frac35-\frac34+\left(-10\right)+\frac16\)

\(=-\frac{75}{60}+\frac{36}{60}-\frac{45}{60}+\frac{\left(-600\right)}{60}+\frac{10}{60}=\frac{-674}{60}=-\frac{337}{30}\)

c: \(C=1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\cdots-\frac{1}{95\cdot100}\)

\(=1-\frac15\left(\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+\cdots+\frac{5}{95\cdot100}\right)\)

\(=1-\frac15\left(\frac15-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\cdots+\frac{1}{95}-\frac{1}{100}\right)\)

\(=1-\frac15\left(\frac15-\frac{1}{100}\right)=1-\frac15\cdot\frac{19}{100}=1-\frac{19}{500}=\frac{481}{500}\)

d: \(D=\frac17+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\cdots+\frac{1}{31\cdot37}\)

\(=\frac16\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\cdots+\frac{6}{31\cdot37}\right)=\frac16\left(1-\frac17+\frac17-\frac{1}{13}+\cdots+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac16\left(1-\frac{1}{37}\right)=\frac16\cdot\frac{36}{37}=\frac{6}{37}\)

a) \(2^4+8\left[\left(-2\right)^2:\dfrac{1}{2}\right]^0-2^{-2}.4+\left(-2\right)^2\)

\(=2^4+8.1-\dfrac{1}{4}.4+4\)

\(=16+8-1+4\)

\(=24-1+4\)

\(=23+4\)

\(=27\)

Bài 4: 
\(8^{12}\)\(-2^{33}\)-\(2^{30}\) chia hết cho 55
= \(2^{36}\)-\(2^{33}\)-\(2^{30}\)
= \(2^{30}.\left(2^6-2^3-1\right)\)
= \(2^{30}\). 55 chia hết cho 55 (đpcm)

365-(120+80-365-350)

Số học sinh giỏi chiếm sô % so với cả lớp là

15 : 50 = 0,3 = 30%

Số cuốn tập có là

15 x 10= 150 (cuốn)

Số tiền thưởng là

9000 x 150 = 1350000 (đồng)

Bài 1:

a)Vì \(m\perp CD\)

        \(n\perp CD\)

nên \(m//n\)

b)Vì \(m//n\) nên

\(\widehat{CFE}+\widehat{FED}=180^o\) (trong cùng phía)

\(110^o+\widehat{FED}=180^o\)

\(\widehat{FED}=70^o\)

Bài 2:

Vì \(AB\perp AD\)

        \(AB\perp CB\)

nên \(AD//BC\)

Vì \(AD//BC\) nên

\(\widehat{D_1}+\widehat{C_1}=180^o\) (trong cùng phía)

\(115^o+\widehat{C_1}=180^o\)

\(\widehat{C_1}=65^o\)

Tớ ko nhìn rõ nhé Xin lỗi!