\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 

0,2                0,2

\(m_{Cu}=0,2.64=12,8\left(g\right)\)

ĐK : \(a\ne b\ne c\ne0\)

Đặt \(x=\dfrac{a-b}{c};y=\dfrac{b-c}{a};z=\dfrac{c-a}{b}\)  (x;y;z \(\ne0\) )

Ta có : \(N=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+1\)

\(=4+\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)

Thấy : \(\dfrac{y+z}{x}=\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right).\dfrac{c}{a-b}=\dfrac{b-a}{ab}.\dfrac{c}{a-b}=\dfrac{-c}{ab}\)

Tương tự ; ta có : \(\dfrac{x+z}{y}=\dfrac{-a}{bc};\dfrac{x+y}{z}=-\dfrac{b}{ac}\)

Suy ra : \(N=4-\left(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\right)=4-\dfrac{a+b+c}{abc}=4\)  ( do a + b + c = 0 ) 

Vậy ... 

Bn tra là có lời giải nhé ; lg của người ta rất hay 

Giả sử (C) tâm I ; BK R 

\(I\in d':2x+y=0\)  \(\Rightarrow I\left(t;-2t\right)\) 

 \(\Rightarrow R^2=IA^2=\left(t-4\right)^2+\left(-2t-2\right)^2\)  \(=5t^2+20\)

Ta có : \(IA=\dfrac{\left|t-7.\left(-2t\right)+10\right|}{\sqrt{1+7^2}}\)  \(\Rightarrow IA^2=\dfrac{\left(15t+10\right)^2}{50}=\dfrac{\left(3t+2\right)^2}{2}\)

Suy ra : \(5t^2+20=\dfrac{\left(3t+2\right)^2}{2}\)  \(\Leftrightarrow10t^2+40=9t^2+12t+4\)

\(\Leftrightarrow t^2-12t+36=0\) \(\Leftrightarrow t=6\)

Suy ra : \(I\left(6;-12\right)\) ; \(R^2=200\)

PT (C) : \(\left(x-6\right)^2+\left(y+12\right)^2=200\)

có ai học Nap ko ? :? 

Bn 12 à ? :)) 

ok 

\(\dfrac{45}{19}-\left(\dfrac{1}{2}+\left(\dfrac{1}{3}+\left(\dfrac{1}{4}\right)^{-1}\right)^{-1}\right)^{-1}\)

\(=\dfrac{45}{19}-\left[\dfrac{1}{2}+\left(\dfrac{1}{3}+4\right)^{-1}\right]^{-1}\) 

\(=\dfrac{45}{19}-\left[\dfrac{1}{2}+\left(\dfrac{13}{3}\right)^{-1}\right]^{-1}\)

\(=\dfrac{45}{19}-\left[\dfrac{1}{2}+\dfrac{3}{13}\right]^{-1}\)

\(=\dfrac{45}{19}-\left(\dfrac{19}{26}\right)^{-1}\)

\(=\dfrac{45}{19}-\dfrac{26}{19}=\dfrac{19}{19}=1\)

Từ GT ta lấy tích phân 2 vế cận từ 0 đến 1 ; sẽ được : 

\(\int\limits^1_0f\left(x+1\right)dx+\int\limits^1_03f\left(3x+2\right)dx-\int\limits^1_04f\left(4x+1\right)dx-\int\limits^1_0f\left(2^x\right)dx=\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}\left(1\right)\)

\(\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}=\int\limits^1_03\left(\sqrt{x+2}-\sqrt{x+1}\right)dx\)  = 

\(2\left[\left(x+2\right)\sqrt{x+2}-\left(x+1\right)\sqrt{x+1}\right]\dfrac{1}{0}\)  = \(2+6\sqrt{3}-8\sqrt{2}\left(2\right)\)

Dễ thấy : \(\int\limits^1_0f\left(x+1\right)dx=\int\limits^2_1f\left(t\right)dt=\int\limits^2_1f\left(x\right)dx\)

\(\int\limits^1_03f\left(3x+2\right)dx=\int\limits^5_2f\left(t\right)dt=\int\limits^5_2f\left(x\right)dx\)  (3)

\(\int\limits^1_04f\left(4x+1\right)=\int\limits^5_1f\left(t\right)dt=\int\limits^5_1f\left(x\right)dx\left(4\right)\)

\(\int\limits^1_0f\left(2^x\right)dx=\int\limits^2_1\dfrac{f\left(t\right)dt}{tln2}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(t\right)dt}{t}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}\)  (5)

Thay (2) ; (3) ; (4) ; (5) vào (1) ta được : 

\(\int\limits^2_1f\left(x\right)dx+\int\limits^5_2f\left(x\right)dx-\int\limits^5_1f\left(x\right)dx-\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=2+6\sqrt{3}-8\sqrt{2}\)

\(\Leftrightarrow\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=\left(2+6\sqrt{3}-8\sqrt{2}\right)ln2\)