22 giờ - 10 giờ 35 phút = 12 giờ 35 phút

22 giờ - 10 giờ 35 phút = 11 giờ 25 phút

\(-\dfrac{5}{7}\cdot\dfrac{2}{11}+\left(-\dfrac{5}{7}\right)\cdot\dfrac{4}{11}+\dfrac{5}{7}\cdot\left(-\dfrac{5}{11}\right)\\ =-\dfrac{5}{7}\cdot\dfrac{2}{11}+\left(-\dfrac{5}{7}\right)\cdot\dfrac{4}{11}+\left(-\dfrac{5}{7}\right)\cdot\dfrac{5}{11}\\ =\left(-\dfrac{5}{7}\right)\cdot\left(\dfrac{2}{11}+\dfrac{4}{11}+\dfrac{5}{11}\right)\\ =\left(-\dfrac{5}{7}\right)\cdot1=-\dfrac{5}{7}\)

\(\dfrac{5}{7}\cdot\dfrac{6}{13}-\left(-\dfrac{7}{13}\cdot\dfrac{5}{7}\right)+\dfrac{1}{7}:\dfrac{1}{2}\\ =\dfrac{5}{7}\cdot\dfrac{6}{13}-\left(-\dfrac{7}{13}\cdot\dfrac{5}{7}\right)+\dfrac{1}{7}\cdot2\\ =\dfrac{5}{7}\cdot\left[\dfrac{6}{13}-\left(-\dfrac{7}{13}\right)\right]+\dfrac{2}{7}\\ =\dfrac{5}{7}\cdot1+\dfrac{2}{7}=1\)

\(\left(2\dfrac{2}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\\ =\left(\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right)\cdot\left(-\dfrac{17}{3}\right)\\ =\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\cdot\dfrac{-17}{3}\\ =\left(\dfrac{32}{15}\cdot\dfrac{3}{32}\right)\cdot\left(\dfrac{9}{17}\cdot\dfrac{-17}{3}\right)\\ =\dfrac{1}{5}\cdot\left(-3\right)=-\dfrac{3}{5}\)

a) 

\(\dfrac{-5}{11}\cdot\dfrac{7}{15}\cdot\dfrac{11}{-5}\cdot\left(-30\right)\\ =\left(\dfrac{-5}{11}\cdot\dfrac{11}{-5}\right)\cdot\left(\dfrac{7}{15}\cdot-30\right)\\ =1\cdot-14\\ =-14\)

b) 

\(-\dfrac{1}{3}\cdot\left(-\dfrac{15}{19}\right)\cdot\dfrac{38}{45}\\ =-\dfrac{1}{3}\cdot\left(-\dfrac{15}{29}\cdot\dfrac{38}{45}\right)\\ =-\dfrac{1}{3}\cdot\left(-\dfrac{15}{29}\cdot\dfrac{2\cdot19}{3\cdot15}\right)\\ =-\dfrac{1}{3}\cdot-\dfrac{2}{3}\\ =\dfrac{2}{9}\)

\(g)\dfrac{x}{xy+y^2}-\dfrac{y}{x^2+xy}\\ =\dfrac{x}{y\left(x+y\right)}-\dfrac{y}{x\left(x+y\right)}\\ =\dfrac{x^2}{xy\left(x+y\right)}-\dfrac{y^2}{xy\left(x+y\right)}\\ =\dfrac{x^2-y^2}{xy\left(x+y\right)}\\ =\dfrac{\left(x+y\right)\left(x-y\right)}{xy\left(x+y\right)}\\ =\dfrac{x-y}{xy}\) 

h) 

\(\dfrac{x^2+4}{x^2-4}-\dfrac{x}{x+2}-\dfrac{x}{2-x}\\ =\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}-\dfrac{x}{x+2}+\dfrac{x}{x-2}\\ =\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2+4-x^2+2x+x^2+2x}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2+4x+4}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x+2}{x-2}\) 

i) 

\(\dfrac{5}{6x-6}+\dfrac{9}{14x-14}+\dfrac{6}{7x-7}\\ =\dfrac{5}{6\left(x-1\right)}+\dfrac{9}{14\left(x-1\right)}+\dfrac{6}{7\left(x-1\right)}\\ =\dfrac{7\cdot5}{42\left(x-1\right)}+\dfrac{3\cdot9}{42\left(x-1\right)}+\dfrac{6\cdot6}{42\left(x-1\right)}\\ =\dfrac{35+27+36}{42\left(x-1\right)}\\ =\dfrac{98}{42\left(x-1\right)}\\ =\dfrac{7}{3\left(x-1\right)}\)

\(9x-3=14\)
\(9x=14+3=17\)
\(9x=17\)
\(x=17\div9=\dfrac{17}{9}\)
\(x=\dfrac{17}{9}\)
~nhớ tk và hok tốt~
@Đinh Sơn Tùng

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\\ =\left[\left(x+6\right)\left(x+4\right)\right]\left[x\left(x+10\right)\right]+128\\ =\left(x^2+6x+4x+24\right)\left(x^2+10x\right)+128\\ =\left(x^2+10x+24\right)\left(x^2+10x\right)+128\\ =\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\\ =\left(x^2+10x\right)^2+8\left(x^2+10x\right)+16\left(x^2+10x\right)+128\\ =\left(x^2+10x\right)\left[\left(x^2+10x\right)+8\right]+16\left[\left(x^2+10x\right)+8\right]\\ =\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(9\left(x-1\right)^2-\left(x-1\right)^4=0\\ \Rightarrow\left(x-1\right)^2\left[9-\left(x-1\right)^2\right]\\ \Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\9-\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\9=\left(x-1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\3^2=x-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x-1=3\\x-1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-2\end{matrix}\right.\) 

________________________

\(\left(2x-3\right)^2+2x=3\\ \Rightarrow\left(2x-3\right)^2+\left(2x-3\right)=0\\ \Rightarrow\left(2x-3\right)\left(2x-3+1\right)=0\\ \Rightarrow\left(2x-3\right)\left(2x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\2x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\2x=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{2}{2}=1\end{matrix}\right.\)

đừng đăng linh tinh nhé bạn (Phạm Minh Khuê)
 

uh, mình sẽ làm !

Chào các bạn đã kết bạn với tớ, các bạn có khỏe mạnh không, giờ tớ đang học nè, còn các cậu đang làm gì.