/ là gì????

dấu  / ....  /  là dấu giá trị tuyệt đối nha

\(A=3x^3y^4+4xy^3-8y^3+2021-3y^4x^3\)

\(\Rightarrow A=\left(3x^3y^4-3y^4x^3\right)+4xy^3-8y^3+2021\)

\(\Rightarrow A=4xy^3-8y^3+2021\)

Thay x = 2; y = -3 ta có:

\(A=4\cdot2\cdot\left(-3\right)^3-8\cdot\left(-3\right)^3+2021\)

\(\Rightarrow A=-216-\left(-216\right)+2021\)

\(\Rightarrow A=2021\)

~~ Chúc bạn học tốt ~~

d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)

\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)

\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)

\(\ge\left|6-2x+2x+5\right|=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).

e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)

\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)

Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).

f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)

\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)

\(=\left|2x-3\right|+\left|2x-7\right|\)

\(\ge\left|2x-3+7-2x\right|=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{2}=\frac{y}{4}=\frac{x^4+y^4}{16+256}=\frac{16}{272}=\frac{1}{17}\)

\(\Rightarrow x=\frac{2}{17};y=\frac{4}{17}\)

Ta có \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

=> b + c = 2a ; c + a = 2b ; a  + b = 2c

Khi đó P = \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)

      \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{1}{2}-\frac{2}{27}\)

\(=\frac{23}{54}\)

Chi do you me => Chi có thik tôi ko

                HOK TỐT ~

tui thiếu mất từ sorry

a, \(A=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}=\frac{3}{5}\)

b, \(2^{17}+2^{14}=2^{17}\left(2^3+1\right)=9.2^{17}⋮9\)( đpcm )

a) A=\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)

A=\(\frac{\frac{3}{10}-\frac{3}{10}+\frac{36}{132}+\frac{33}{132}}{\frac{5}{10}-\frac{5}{10}+\frac{60}{132}+\frac{55}{132}}\)

A=\(\frac{69}{132}\):\(\frac{115}{132}\)

A=\(\frac{69}{115}\)

b) 2¹⁷+2¹⁴ chia hết cho 9

Ta có :\(2^{17}\)+\(2^{14}\)

         =>\(2^3\)+\(2^{14}\)+\(2^{14}\)

         =>\(2^{14}\).(8+1)

         =>\(2^{14}\).9

Vì 9 chia hết cho 9 nên \(2^{14}\).9 chia hết cho 9 

Vậy \(2^{17}\)+\(2^{14}\) chia hết cho 9