A=F.s=2500.4000=10000000 Nm

hoàng hữu bình bn có thể làm rõ hơn đc ko 

\(-4x^2+x+30=0\)

Ta có: \(\Delta=1^2+4.4.30=481,\sqrt{\Delta}=\sqrt{481}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+\sqrt{481}}{-8}=\frac{1-\sqrt{481}}{8}\\x=\frac{-1-\sqrt{481}}{-8}=\frac{1+\sqrt{481}}{8}\end{cases}}\)

a) \(2\left(x-2\right)+x-2=3\left(x-2\right)\)

\(\Leftrightarrow\left(2+1\right)\left(x-2\right)=3\left(x-2\right)\)

Vì phương trình trên luôn đúng với mọi x nên có vô số nghiệm

B) \(4\left(1-x\right)+3x=1-x\)

\(4-4x+3x=1-x\Leftrightarrow4-x=1-x\)(vô nghiệm)

3(x - 1) - x = 2(x - 3)

=> 3x - 3 - x = 2x - 6

=> 2x - 3 = 2x - 6

=> -3 = -6 (vô lí)

=> x thuộc tập hợp rỗng

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

\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

\(ĐKXĐ:x\ne\pm1\)

a) \(A=\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)

\(\Leftrightarrow A=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)

\(\Leftrightarrow A=x-1+x+1-3\)

\(\Leftrightarrow A=2x-3\)

b) Thay x = 3 vào A, ta được :

\(A=2.3-3=3\)

Thay x = 0 vào A, ta được :

\(A=2.0-3=-3\)

c) Để A = 2

\(\Leftrightarrow2x-3=2\)

\(\Leftrightarrow2x=5\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy để \(A=2\Leftrightarrow x=\frac{5}{2}\)

b,\(A=\frac{4}{3x-6}-\frac{x}{x^2-4}\)

\(A=\frac{4}{3\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x+8}{3\left(x-2\right)\left(x+2\right)}-\frac{3x}{3\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{x-8}{3\left(x-2\right)\left(x+2\right)}\)

c, Thay x = 1 vào A ta đc

\(\frac{1-8}{3\left(1-2\right)\left(1+2\right)}=\frac{7}{9}\)

a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x-6\ne0\\x^2-4\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne6\\x^2\ne4\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne2\\x\ne\pm2\end{cases}\Leftrightarrow}x\ne\pm2}\)

Vậy A xác định khi \(x\ne\pm2\)

b) \(A=\frac{4}{3x-6}-\frac{x}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow A=\frac{4}{3\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{4\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}-\frac{3x}{3\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{4x+8}{3\left(x+2\right)\left(x-2\right)}-\frac{3x}{3\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{4x+8-3x}{3\left(x-2\right)\left(x+2\right)}=\frac{x+8}{3\left(x-2\right)\left(x+2\right)}\)

Vậy \(A=\frac{x+8}{3\left(x-2\right)\left(x+2\right)}\left(x\ne\pm2\right)\)

c) Thay x=1 (tmđk) vào A ta có: \(A=\frac{1+8}{3\left(1-2\right)\left(1+2\right)}=\frac{9}{-9}=-1\)

Vậy \(A=-1\)khi x=1