\(b,\dfrac{85}{16}x-3,77x+0,66=13\)

\(\dfrac{85}{16}x-3,77x=12,34\)

\(1,5425x=12,34\)

\(x=12,34:1,5425=8,0064\approx8\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{10}=\dfrac{y-x}{10-9}=\dfrac{120}{1}=120\) 

\(\Rightarrow x=120.9=1080\)

\(y=120.10=1200\)

\(C-A-B=\left(-x^2+3xy+2y^2\right)-\left(4x^2-5xy+3y^2\right)-\left(3x^2+2xy+y^2\right)\)

\(=-x^2+3xy+2y^2-4x^2+5xy-3y^2-3x^2-2xy-y^2\)

\(=\left(-x^2-4x^2-3x^2\right)+\left(3xy+5xy-2xy\right)+\left(2y^2-3y^2-y^2\right)\)

\(=-8x^2+6xy-2y^2\)

Các câu con lại bạn tự làm nha 

\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\)

\(=\dfrac{-5}{7}.1+1\)

\(=\dfrac{-5}{7}+1=\dfrac{2}{7}\)

\(a,\dfrac{3}{2}=\dfrac{9}{6};\dfrac{3}{9}=\dfrac{2}{6};\dfrac{9}{3}=\dfrac{6}{2};\dfrac{2}{3}=\dfrac{6}{9}\)

\(b,\dfrac{-2,5}{5}=\dfrac{-1,5}{3};\dfrac{-2,5}{-1,5}=\dfrac{5}{3};\dfrac{5}{-2,5}=\dfrac{3}{-1,5};\dfrac{-1,5}{-2,5}=\dfrac{3}{5}\)

\(c,\dfrac{12}{-9}=\dfrac{1}{\dfrac{-3}{4}};\dfrac{-9}{12}=\dfrac{-\dfrac{3}{4}}{1};\dfrac{12}{1}=\dfrac{-9}{\dfrac{-3}{4}};\dfrac{1}{12}=\dfrac{\dfrac{-3}{4}}{-9}\)

\(x^2.25=6.24\)

\(x^2.25=144\)

\(x^2=144:25\)

\(x^2=5,67\)

\(x=\sqrt{5,67}=2,4;-2,4\)

\(a,2+4+6+...+2x=9900\)

\(\Leftrightarrow\dfrac{2+2x}{2}:\left(\dfrac{2-2x}{2}+1\right)=9900\)

\(\Leftrightarrow\left(x+1\right).x=9900\)

\(\Leftrightarrow x^2+x=9900\)

\(\Leftrightarrow x=99\)

\(b,2+4+6+...+2x=2550\)

\(\Leftrightarrow\dfrac{2+2x}{2}:\left(\dfrac{2-2x}{2}+1\right)=2550\)

\(\Leftrightarrow\left(x+1\right).x=2550\)

\(\Leftrightarrow x^2+x=2550\)

\(\Leftrightarrow x=50\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(\Rightarrow n_{H_2SO_4}=n_{H_2}=1,5\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{1,5}{0,5}=3\left(M\right)\)

\(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)

\(\Rightarrow n_{O_2}=\dfrac{1}{2}n_{H_2}=0,75\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,75.22,4=16,8\left(l\right)\)

Một giờ người thứ hai đuổi được là

\(45-30=15\)(km/giờ)

1 giờ người thứ nhất đi doạn đường cách A là

\(30.1=30\left(km\right)\)

Thời gian để người  thứ hai đuổi kịp là

\(30:15=2\left(h\right)\)

2 người gặp nhau lúc :

\(6+2=8\left(h\right)\)

Nơi gặp nhau cách A là

\(30.2=60\left(km\right)\)

\(=y\left(1-x^2-2xy-y^2\right)\)

\(=y\left[1-\left(x+y\right)^2\right]\)

\(=y\left(1-x-y\right)\left(1+y+x\right)\)