`(0,6-1 2/5):(-1/2)+1`

`=(3/5-7/5)xx(-2)+1`

`=[-5]/5xx(-2)+1`

`=2+1=3`

chụp lại bài nhìn cho rõ hơn .

Tớ sửa đề đa thức Q nhé.

\(P+Q=\left(xyz-xy^2-xz^2\right)+\left(z^3+y^3\right)\)

Theo đề cho, ta có: \(x-y=z\Rightarrow x=z+y\)

Thay \(x=z+y\), ta được:

\(P+Q=\left(z+y\right).yz-\left(z+y\right).y^2-\left(z+y\right).z^2+z^3+y^3\)

\(=yz^2+y^2z-y^2z-y^3-z^3-yz^2+z^3+y^3\)

\(=\left(yz^2-yz^2\right)+\left(y^2z-y^2z\right)+\left(-y^3+y^3\right)+\left(-z^3+z^3\right)\)

\(=0\)

\(\Rightarrowđpcm\)

\(\dfrac{14}{15}.\left(\dfrac{1}{6}-\dfrac{5}{21}\right):\dfrac{11}{2}.\dfrac{121}{12}+\dfrac{5}{6}-\dfrac{1}{3}\)

\(=\dfrac{14}{15}.\left(\dfrac{7}{42}-\dfrac{10}{42}\right).\dfrac{2}{10}.\dfrac{121}{12}+\dfrac{5}{6}-\dfrac{2}{6}\)

\(=\dfrac{14}{15}.-\dfrac{1}{14}.\dfrac{6}{11}+\dfrac{1}{2}\)

\(=-\dfrac{1}{15}.\dfrac{6}{11}+\dfrac{1}{2}\)

\(=-\dfrac{22}{5}+\dfrac{1}{2}\)

\(=\dfrac{51}{110}\)

\(M_{\left(x\right)}=-\left(x^2+2\right)\)

\(x^2\ge0\forall x\Rightarrow x^2+2\ge2\forall x\Rightarrow-\left(x^2+2\right)< 2\forall x\)

\(\Rightarrow-\left(x^2+2\right)< 0\forall x\Rightarrow M_{\left(x\right)}\) không có nghiệm

a) A(x) = x³- 3x² - 2x - 8

=> A(x) = (x³- 3x²) +(- 2x - 8)

b) A(x) = x³- 3x² - 2x - 8

=> A(x) = (x³- 3x²) - (2x + 8)

ui em mới lớp 5 chuk hiểu phân giác là j