Tính C=(x2+x+1)(x3+1)(x6+1)(x12+1)(x24+1) x=2
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2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X1: HCl X2: H2S X3: FeCl2
X4: CuS X5: H2SO4 X6: O2
X7: S X8: H2O X9: Cl2
X10: FeCl3 X11:I2 X12: MnO2
Đáp án D
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn D
X1: HCl X2: H2S
X3: FeCl2 X4: CuS
X5: H2SO4 X6: O2
X7: S X8: H2O
X9: Cl2 X10: FeCl3
X11:I2 X12: MnO2
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án D
X1: HCl
X2: H2S
X3: FeCl2
X4: CuS
X5: H2SO4
X6: O2
X7: S
X8: H2O
X9: Cl2
X10: FeCl3
X11:I2
X12: MnO2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Khi m=4 thì phương trình trở thành \(x^2-4x+3=0\)
=>(x-3)*(x-1)=0
=>x=3 hoặc x=1
b: \(x_1+x_2=m\)
\(x_1x_2=m-1\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=m^2-2\left(m-1\right)=m^2-2m+2\)
\(x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2\)
\(=\left(m^2-2m+2\right)^2-2\cdot\left(m-1\right)^2\)
\(=m^4+4m^2+4-4m^3+4m^2-8m-2m^2+4m-2\)
\(=m^4-4m^3+2m^2-4m+2\)
Thay \(x=2\)vào \(C\)ta được :
\(C=\left(2^2+2+1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(\Rightarrow C=\left(2-1\right)\left(2^2+2+1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(\Rightarrow C=\left(2^3-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(\Rightarrow C=\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(\Rightarrow C=\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(\Rightarrow C=\left(2^{24}-1\right)\left(2^{24}+1\right)\)
\(\Rightarrow C=2^{48}-1\)
Vậy \(C=2^{48}-1\)