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\(B=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)....\left(1-\frac{2}{99}\right)\)
\(B=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{97}{99}\)
\(B=\frac{3\cdot5\cdot7\cdot...\cdot97}{5\cdot7\cdot9\cdot...\cdot99}=\frac{3}{99}=\frac{1}{33}\)
Vậy B = \(\frac{1}{33}\)
\(\left[1-\frac{2}{5}\right]\left[1-\frac{2}{7}\right]\left[1-\frac{2}{9}\right]...\left[1-\frac{2}{99}\right]\)
\(=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{97}{99}\)
\(=\frac{3\cdot5\cdot7\cdot...\cdot97}{5\cdot7\cdot9\cdot...\cdot99}=\frac{3}{99}=\frac{1}{33}\)
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`a,`
\(\text {P(x)+Q(x)=}\)\(\left(x^3-\dfrac{1}{2}x^3+8x-1\right)+\left(-x^3+\dfrac{3}{2}x^2-10x+4\right)\)
`= x^3-1/2x^3+8x-1-x^3+3/2x^2-10x+4`
`= (x^3-x^3-1/2x^3)+3/2x^2+(8x-10x)+(-1+4)`
`= -1/2x^3+3/2x^2-2x+3`
`b,`
\(\text {P(x)-Q(x)=}\)\(\left(x^3-\dfrac{1}{2}x^3+8x-1\right)-\left(-x^3+\dfrac{3}{2}x^2-10x+4\right)\)
`= x^3-1/2x^3+8x-1+x^3-3/2x^2+10x-4`
`= (x^3+x^3-1/2x^3)-3/2x^2+(8x+10x)+(-1-4)`
`= 3/2x^3-3/2x^2+18x-5`
`c,`
`\text {M(x)-P(x)=Q(x)}`
`->` `\text {M(x)=P(x)+Q(x)}`
`M(x)=-1/2x^3+3/2x^2-2x+3=0`
\(\text{Vì }\dfrac{3}{2}x^2\ge0\text{ }\)\(\forall\)\(\text{x}\)
`->`\(-\dfrac{1}{2}x^3+\dfrac{3}{2}x^2-2x+3>0\text{ }\forall\text{ }x\)
`-> \text {Đa thức vô nghiệm.}`
`@`\(\text {Dnammv}\)
1-2+3-4+..........+99-100
=(1-2)+(3-4)+...+(99-100)
=(-1)+(-1)+...+(-1)
=(-1)*50
=-50
Đặt \(A=1-2+3-4+...+99-100\)
\(\Rightarrow A=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\) ( 50 cặp số )
\(\Rightarrow A=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\) ( 50 số -1 )
\(\Rightarrow A=\left(-1\right).50\)
\(\Rightarrow A=-50\)
Vậy \(A=-50\)