展开 1*(x+λy+z=λ)

解答

展开

1 \cdot (x + λ y + z = λ)

x + λ y + z = λ

求解步骤

1 \cdot (x + λ y + z = λ)

= 1 \cdot (x + λ y + = λ z)

打开括号

1 \cdot x + 1 \cdot λ y + 1 \cdot z = λ

乘开

1 \cdot x + 1 \cdot λ y + 1 \cdot z : x + λ y + z

x + λ y + z = λ

e x p an d - \frac{4}{( x + 1 ) ( x + 2 ) ( x - 1 )}

e x p an d \frac{2 - x}{x ^{3} ( x - 3 )}

e x p an d \frac{2}{( 4 x + 3 ) ( 4 x - 1 )}

e x p an d \frac{x}{x ^{2} + 1} + \frac{1}{3 x} - \frac{1}{x ^{2}}

e x p an d \frac{( n + 3 ) ( n + 2 )}{( n + 1 ) ( n + 4 )}