Áp dụng tc dtsbn:

\(\dfrac{2y+z-x}{x}=\dfrac{2z-y+x}{y}=\dfrac{2x+y-z}{z}=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}2y+z-x=2x\\2z-y+x=2y\\2x+y-z=2z\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2y+z=3x\\2z+x=3y\\2x+y=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-2y=z\\3y-2z=x\\3z-2x=y\end{matrix}\right.;\left\{{}\begin{matrix}3x-z=2y\\3y-x=2z\\3z-y=2z\end{matrix}\right.\\ \Rightarrow P=\dfrac{xyz}{2x\cdot2y\cdot2z}=\dfrac{1}{8}\)

Chọn D

\(\dfrac{2y+z-x}{x}=\dfrac{2z-y+x}{y}=\dfrac{2x+y-z}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\left\{{}\begin{matrix}2y+z-x=2x\\2z-y+x=2y\\2x+y-z=2z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2y+z=3x\\2z+x=3y\\2x+y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x-2y=z\\3y-2z=x\\3z-2x=y\end{matrix}\right.\) và \(\left\{{}\begin{matrix}3x-z=2y\\3y-x=2z\\3z-y=2x\end{matrix}\right.\)

Thay vào A:

\(A=\dfrac{z.x.y}{2y.2z.2x}=\dfrac{1}{8}\)

Vì x>0, y>0, z>0 ⇒ x+y+z>0

Áp dụng t/c dtsbn ta có:

\(\dfrac{2y+z-x}{x}=\dfrac{2z-y+x}{y}=\dfrac{2x+y-z}{z}=\dfrac{2y+z-x+2z-y+x+2x+y-z}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\dfrac{2y+z-x}{x}=2\Rightarrow2y+z-x=2x\Rightarrow2y+z=3x\\ \dfrac{2z-y+x}{y}=2\Rightarrow2z-y+x=2y\Rightarrow2z+x=3y\\ \dfrac{2x+y-z}{z}=2\Rightarrow2x+y-z=2z\Rightarrow2x+y=3z\)

\(A=\dfrac{\left(3x-2y\right)\left(3y-2z\right)\left(3z-2x\right)}{\left(3x-z\right)\left(3y-x\right)\left(3z-y\right)}\)

\(\Rightarrow A=\dfrac{\left(2y+z-2y\right)\left(2z+y-2z\right)\left(2x+y-2x\right)}{\left(2y+z-z\right)\left(2z+x-x\right)\left(2x+y-y\right)}\)

\(\Rightarrow A=\dfrac{xyz}{2x.2y.2z}\)

\(\Rightarrow A=\dfrac{1}{8}\)

Chiều từ cực dương đến cực âm

\(I=I_1=I_2=0,5A\\ U_2=U-U_1=3,5V\) 

Đèn sẽ không sáng do hiệu điện thế trong mạch nhỏ hơn hiệu điện thế của 2 đèn 

a) 

b)vì các đèn mắc nối tiếp nên:

\(I=I_1=I_2=0,5A\)

vì các đèn mắc nối tiếp nên:

\(U=U_1+U_2\)

\(=>U_2=U-U_1=8-4,5=3,5\left(V\right)\)

c)nếu với một bóng đèn nhất định(có HĐT định mức), thì hiệu điện thế đặt vào hai đầu bóng đèn sẽ giảm dần ( tức là độ sáng của hai đèn yếu đi)

Giải:

\(A=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\) 

\(A=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\) 

\(A=\dfrac{1}{6}.\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+\dfrac{6}{19.25}+\dfrac{6}{25.31}+\dfrac{6}{31.37}\right)\) 

\(A=\dfrac{1}{6}.\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\) 

\(A=\dfrac{1}{6}.\left(1-\dfrac{1}{37}\right)\) 

\(A=\dfrac{1}{6}.\dfrac{36}{37}\) 

\(A=\dfrac{6}{37}\)

\(6A=\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{31.37}\)

= \(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{31}-\dfrac{1}{37}=\dfrac{36}{37}\)

<=> A = \(\dfrac{6}{37}\)

Ta có: \(A=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+\dfrac{6}{19\cdot25}+\dfrac{6}{25\cdot31}+\dfrac{6}{31\cdot37}\right)\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

x^{3 _} x^{2 _} 4x - 4 = 0

x mũ 2(x+1)- 4(x+1)=0

(x mũ 2 - 4) (x+1)=0

(x+2) (x-2) (x+1)  =0

suy ra (x+2)=0

            (x-2)=0

            (x+1)=0

vậy      x=-2

            x=2

            x= -1

good luck!

Sửa đề : \(x^3-x^2-4x+4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow x=\pm2;1\)