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A= 216 và B=(2+1)(22+1)(24+1)(28+1)
Xet B=(2+1)(22+1)(24+1)(28+1)
=(2-1)(2+1)(22+1)(24+1)(28+1)
=(22-1)(22+1)(24+1)(28+1)
=(24-1)(24+1)(28+1)
=(28-1)(28+1)
=216-1
So sanh A=216 va B=216-1 ta co A>B
Ta có 1999*2001 = (2000-1)*(2000+1)
= 2000^2 - 1^2
Biết 2000^2 = 2000^2
=> 2000^2 - 1^2 < 2000^2
<=> 1999*2001 < 2000^2
Ax(2-1)=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)=(2^4-1)(2^4+1)(2^8+1)(2^16+1)=(2^8-1)(2^8+1)(2^16+1)=(2^16-1)(2^16+1)=2^32-1
Vậy A=B
Áp dụng hằng đẵng thức A^2-B^2 đó bạn
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
a ) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)
Do \(2016^2>2016^2-1\)
\(\Rightarrow B>A\)
b ) \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}=D\)
Vậy \(C< D\)
so sánh :a)A=2015.2017 va B=20162
Ta có: A = 2015.2017 = (2016-1)(2016+1)
= 20162-1<20162
=> A < B
Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1=A-1\)
Vậy B < A