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![](https://rs.olm.vn/images/avt/0.png?1311)
Sai thì thôi nha
Từ M ta có:
\(M=2^n-2.2^n+8+2.2^n-16-16.2^n\)
\(M=2^n.\left(-2+2+2-16\right)+8-16\)
M=\(2^n.\left(-14\right)-8\)
Vậy thu gọn M ta được....
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)
\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)
\(=2^{n-1}+2^{n+4}-2^{n-1}-2^{n+4}\)
\(=0\)
b) \(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)
\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}-2^{n+1}\right)-3^{2n+2}+2^{2n+2}\)
\(=3^{2n+2}-2^{2n+2}-3^{2n+2}+2^{2n+2}\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2^{n-1}+2^{n+4}-2^3\cdot2^{n-4}-2^4\cdot2^n\)
\(A=2^{n-1}+2^{n+4}-2^{n-1}-2^{n+4}\)
\(A=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)
\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)
\(=2.2^{n-1}+2.2^{n+4}=2^n+2^{n+5}\)
b,
\(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)
\(=\left(3^{n+1}\right)^2-\left(2.2^n\right)^2-\left(3^{n+1}\right)^2+\left(2^{n-2+3}\right)^2\)
\(=-2^{n+1}+2^{n+1}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ thấy 5=4+1=x+1
Thay vào C,ta có:
\(C=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-1=x-1=4-1=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P = (x-1)(2x+3)
=> P=2x2+3x-2x-3
=> P=2x2+x-3
=> P=\(2x^2+x+\dfrac{1}{8}-\dfrac{25}{8}\)
=> P=2\(\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{25}{8}\)
=> P=\(2\left(x+\dfrac{1}{4}\right)^2-\dfrac{25}{8}\)
=> min P =\(\dfrac{-25}{8}\) khi \(x+\dfrac{1}{4}=0\Rightarrow x=-\dfrac{1}{4}\)