Mỗi loại có số viên là: 

`100 : 4 = 25` (viên) 

Phải lấy ra ít nhất số viên bi để chắc chắn có đủ 3 màu là: 

`25 xx 3 = 75` (viên)

Đáp số: `75` viên

51 viên là đủ rồi em

aaa + mmm + nnn

= a x 111 + m x 111 + n x 111

= 111 x (a + m + n) 

`overline{aaa} + overline{mmm} + overline{nnn} `

`= 111 a + 111m + 111n`

`= 111(a+m+n)`

Đề là \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge9\) với đúng chứ em?

thầy trả lời giúp em với

\(1)-\dfrac{3}{7}+\dfrac{5}{13}-\dfrac{4}{7}+\dfrac{8}{13}\\ =\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\\ =\dfrac{-7}{7}+\dfrac{13}{13}\\ =-1+1\\ =0\\ 2)-\dfrac{5}{14}-\dfrac{2}{14}+\dfrac{1}{8}+\dfrac{1}{8}\\ =\left(\dfrac{-5}{14}-\dfrac{2}{14}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\\ =\dfrac{-7}{14}+\dfrac{2}{8}\\ =\dfrac{-1}{2}+\dfrac{1}{4}\\ =\dfrac{-1}{4}\\ 3)\dfrac{-5}{22}-1+\dfrac{3}{2}-\dfrac{6}{22}\\ =\left(\dfrac{-5}{22}-\dfrac{6}{22}\right)+\left(\dfrac{3}{2}-1\right)\\ =\dfrac{-11}{22}+\dfrac{1}{2}\\ =\dfrac{-1}{2}+\dfrac{1}{2}\\ =0\)

\(4,\dfrac{7}{16}+\dfrac{5}{9}+\dfrac{-3}{16}+\dfrac{-2}{9}\\ =\left(\dfrac{7}{16}-\dfrac{3}{16}\right)+\left(\dfrac{5}{9}-\dfrac{2}{9}\right)\\ =\dfrac{4}{16}+\dfrac{3}{9}\\ =\dfrac{1}{4}+\dfrac{1}{3}\\ =\dfrac{7}{12}\\ 5,\dfrac{2}{5}-\dfrac{3}{11}-\dfrac{7}{35}+\dfrac{14}{11}-\dfrac{1}{5}\\ =\left(-\dfrac{3}{11}+\dfrac{14}{11}\right)+\left(\dfrac{2}{5}-\dfrac{1}{5}\right)-\dfrac{7}{35}\\ =\dfrac{11}{11}+\dfrac{1}{5}-\dfrac{7}{35}\\ =1+\dfrac{1}{5}-\dfrac{1}{5}\\ =1\\ 6,\dfrac{3}{4}-\dfrac{3}{17}+\dfrac{-5}{6}+\dfrac{20}{17}-\dfrac{1}{4}\\ =\left(\dfrac{3}{4}-\dfrac{1}{4}\right)+\left(\dfrac{-3}{17}+\dfrac{20}{17}\right)+\dfrac{-5}{6}\\ =\dfrac{1}{2}+1+\dfrac{-5}{6}\\ =\dfrac{3}{2}-\dfrac{5}{6}\\ =\dfrac{9}{6}-\dfrac{5}{6}\\ =\dfrac{4}{6}=\dfrac{2}{3}\)

\(7-\left(x-1\right)=15+3\left(x+1\right)\\ 7-x+1=15+3x+3\\ 8-x=18+3x\\ 3x+x=8-18\\ 4x=-10\\ x=-\dfrac{10}{4}\\ x=\dfrac{-5}{2}\)

Vậy: ... 

 A = (14⁸)²⁰²⁰ + 10 

A =  (14⁸)^5.404 + 10

A = (14⁵)^8.404 + 10

A = 537824³²³² + 10

537824 \(\equiv\) 1 (mod 11)

537824³²³² \(\equiv\) 1³²³² (mod 11) \(\equiv\) 1 (mod 11)

10 \(\equiv\) 10 (mod 11)

⇒ 537824³²³² + 10  \(\equiv\) 1 + 10 (mod 11)

⇒537824³²³² + 10 \(\equiv\) 11 (mod 11) \(\equiv\) 0 (mod 11)

⇒ A = (14⁸)²⁰²⁰ + 10 \(⋮\) 11 (đpcm)

\(14\equiv3\left(mod11\right)\Rightarrow\left(14^8\right)^{2020}\equiv\left(3^8\right)^{2020}\left(mod11\right)\)

\(\left(3^8\right)^{2020}=3^{8.404.5}=\left(3^5\right)^{3232}=\left(243\right)^{3232}\)

\(243\equiv1\left(mod11\right)\Rightarrow243^{3232}\equiv1\left(mod11\right)\)

\(\Rightarrow\left(14^8\right)^{2020}\equiv1\left(mod11\right)\)

\(\Rightarrow\left(14^8\right)^{2020}+10⋮11\)

\(C=\dfrac{5}{2\cdot7}+\dfrac{16}{7\cdot9}-\dfrac{2}{9\cdot11}-\dfrac{29}{1\cdot11}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{29}{11}\)

\(=\dfrac{1}{2}-\dfrac{28}{11}=\dfrac{11-56}{22}=\dfrac{-45}{22}< \dfrac{1}{3}\)

67

a: 25-(5-x)=-7

=>5-x=25+7=32

=>x=5-32=-27

b: \(\dfrac{1}{2}+\dfrac{2}{3}:\left(x-1\right)=\dfrac{3}{4}\)

=>\(\dfrac{2}{3}:\left(x-1\right)=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)

=>\(x-1=\dfrac{2}{3}:\dfrac{1}{4}=\dfrac{2}{3}\cdot4=\dfrac{8}{3}\)

=>\(x=\dfrac{8}{3}+1=\dfrac{11}{3}\)

d: \(\left(2x+1\right)\left(x-\dfrac{1}{7}\right)=0\)

=>\(\left[{}\begin{matrix}2x+1=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{7}\end{matrix}\right.\)

c: \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

=>\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}=\dfrac{7}{21}-\dfrac{9}{21}=-\dfrac{2}{21}\)

=>\(x=-\dfrac{2}{21}:\dfrac{4}{7}=-\dfrac{2}{21}\cdot\dfrac{7}{4}=\dfrac{-1}{6}\)

Tham khảo có chọn lọc ạ:

"Hổ mang bò lên núi" hiểu theo 2 nghĩa:

1) Con hổ đưa (mang) con bò đi lên núi.

2) Con hổ mang (rắn) di chuyển (bò) lên núi.

_________________

+) Con hổ mang con bò lên núi.

+) Con rắn hổ mang bò lên núi.

\(#NqHahh\)

c.on bạn nhiều nhaaa✿

\(\left|x-y+1\right|>=0\forall x,y\)

=>\(-2\left|x-y+1\right|< =0\forall x,y\)

\(\left|y-2\right|>=0\forall y\)

=>\(-3\left|y-2\right|< =0\forall y\)

Do đó: \(-2\left|x-y+1\right|-3\left|y-2\right|< =0\forall x,y\)

=>\(C=-2\left|x-y+1\right|-3\left|y-2\right|-4< =-4\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1=2-1=1\end{matrix}\right.\)

Sao mình không nhìn thấy đề bài bạn nhỉ?