địt con mẹ mày lên

\(\dfrac{1}{3}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)=\(\dfrac{3}{15}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)

=\(\dfrac{13}{15}\)

\(B=\frac{1}{3}.b+\frac{2}{9}.b– b: \frac{9}{4}= \frac{1}{3}.b+\frac{2}{9}.b – b. \frac{4}{9}\)

\(=b(\frac{1}{3}+\frac{2}{9}-\frac{4}{9})=b. (\frac{3}{9}+\frac{2}{9}-\frac{4}{9})= b. \frac{1}{9} \)

Thay \(b=\frac{9}{10}\) vào B, ta được

B= \(b=\frac{9}{10}. \frac{1}{9}= \frac{1}{10}\)

a) A= (x-3)^2

thay x=203 vào 3, ta có 

A=(203-3)^2=200^2=40000

bạn  làm chi tiết ra cho mình đc ko ạ

Không mất tính tổng quát ta giả sử \(a\ge b\ge c\)

Vì \(a^2+b^2+c^2=1\Rightarrow lal,lbl,lcl\le1\)

\(\Rightarrow\hept{\begin{cases}a^2\ge a^3\\b^2\ge b^3\\c^2\ge c^3\end{cases}}\Rightarrow a^2+b^2+c^2\ge a^3+b^3+c^3=1\)

Dấu = xảy ra khi \(\hept{\begin{cases}a^2=a^3\\b^2=b^3\\c^2=c^3\end{cases}}\)

Mà theo giả thuyết thì \(\hept{\begin{cases}a\ge b\ge c\\a^2+b^2+c^2=1\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=c=0\end{cases}}}\)

Vậy C = 1

Tương tự với các trường hợp giả sử về a,b,c khác ta luôn có giá trị C = 1

Giả sử\(a\ge b\ge c\)(ko mất tính tổng quát) .Ta có :\(\hept{\begin{cases}a^2+b^2+c^2=1\\a^2;b^2;c^2\ge0\end{cases}\Rightarrow a^2;b^2;c^2\le1\Rightarrow|a|;|b|;|c|\le1\Rightarrow\hept{\begin{cases}a^2\ge a^3\\b^2\ge b^3\\c^2\ge c^3\end{cases}\Rightarrow}a^2+b^2+c^2\ge a^3+b^3+c^3=1}\)

\(\Rightarrow\hept{\begin{cases}a^2=a^3\\b^2=b^3\\c^2=c^3\end{cases}\Rightarrow\hept{\begin{cases}a,b,c\in\left\{0;1\right\}\\a^2+b^2+c^2=1\\a\ge b\ge c\end{cases}}\Rightarrow a=1;b=c=0\Rightarrow a^2+b^9+c^{1945}=1}\)

a) \(\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\)

=> \(\frac{12}{13}x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)

=> \(x=2:\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)

b) \(x:\frac{13}{3}=-2,5\)

=> \(x:\frac{13}{3}=-\frac{5}{2}\)

=> \(x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)

c) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

=> \(\frac{4x-3}{12}=-\frac{10}{12}\)

=> 4x - 3 = -10

=> 4x = -10 + 3 = -7

=> x = -7/4

Bài 2 :

\(A=a\cdot\frac{1}{3}+a\cdot\frac{1}{4}-a\cdot\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\cdot\frac{5}{12}\)

Thay a = -3/5 vào biểu thức ta có : \(A=\left(-\frac{3}{5}\right)\cdot\frac{5}{12}=\frac{-3}{12}=\frac{-1}{4}\)

\(B=b\cdot\frac{5}{6}+b\cdot\frac{3}{4}-b\cdot\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\cdot\frac{13}{12}\)

Thay b = 12/13 vào ta được kết quả là 1

a ) \(\frac{25}{9}-\frac{12}{13}\cdot x=\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}\cdot x=\frac{25}{9}-\frac{7}{9}=\frac{18}{9}=2\)

\(\Rightarrow x=2\div\frac{12}{13}=2\cdot\frac{13}{12}=\frac{13}{6}\)

Vậy ...

b ) \(x\div\frac{13}{3}=-\frac{5}{2}\)

\(\Rightarrow x\div\frac{13}{3}=-\frac{5}{2}\)

\(\Rightarrow x=\left(-\frac{5}{2}\right)\cdot\frac{13}{3}=-\frac{65}{6}\)

Vậy ..

c ) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

\(\Rightarrow\frac{4x-3}{12}=-\frac{10}{12}\)

\(\Rightarrow4x-3=-10\)

\(\Rightarrow4x=-10+3=-7\)

\(\Rightarrow x=-\frac{7}{4}\)

Vậy ....

a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)

\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)

\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)

\(M=1-\frac{1}{49}\)

\(M=\frac{48}{49}\)

b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)

=  \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)

\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)

\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=2\times\frac{9}{22}\)

\(=\frac{9}{11}\)

Mình trả lời câu a nha  M= 4/1*5+4/5*9+4/9*13+...+4/45*49   M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49    M=1-1/49=48/49

a) \(\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{4}{9}\times2=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{23}{9}\)

b) \(\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{2}{5}\times\dfrac{3}{2}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)

Câu 4 

\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times4\times3\times5\times4\times5}{5\times2\times4\times4\times5\times5}=\dfrac{3\times3}{5\times2}=\dfrac{9}{10}\)

Câu 3:

\(a.\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{15}{9}+\dfrac{8}{9}=\dfrac{23}{9}\)

\(b.\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)

Câu 4:

\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times3\times1}{2\times1\times5}=\dfrac{9}{10}\)