tính tích
1) P = \(\left(a+b\right)\left(a^2+b^2\right)...\left(a^{2^{1997}}+b^{2^{1997}}\right)\)
2) P= \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2^n}+1\right)\)
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\(=\dfrac{\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)...\left(a^{2^{1997}}+b^{2^{1997}}\right)}{a-b}\)
\(=\dfrac{\left(a^2-b^2\right)\left(a^2+b^2\right)\cdot...\cdot\left(a^{2^{1997}}+b^{2^{1997}}\right)}{a-b}\)
\(=\dfrac{a^{2^{1998}}-b^{2^{1998}}}{a-b}\)
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
A=[1+(-2)+(3)+4]+[5+(-6)+(-7)]+.....+[1997+(-1998)+(-1999)+2000] A=0+0+0+...+0=0
Nhiều quá làm 1 bài tiêu biểu thôi nhé:
a/ \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+a\right)\left(b+c\right)\left(a+b\right)\left(c+a\right)\left(b+c\right)}=1\)
Ta có
\(D=\frac{2^{2x+1}}{2^{2x}-2}+\frac{2^{2\left(1-x\right)+1}}{2^{2\left(1-x\right)}-2}=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{2^{1-2x}-1}\)
Mà \(2^{1-2x}=\frac{1}{2^{2x-1}}\)(do 1-2x+2x-1=0)
=>\(D=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{\frac{1}{2^{2x-1}}-1}=\frac{2^{2x}-2^{2\left(1-x\right)}.2^{2x-1}}{2^{2x-1}-1}=\frac{2^{2x}-2^1}{2^{2x-1}-1}=\frac{2\left(2^{2x-1}-1\right)}{2^{2x-1}-1}=2\)
Áp dụng D ta được
\(P\left(\frac{1}{1998}\right)+P\left(\frac{1997}{1998}\right)=2\)
\(P\left(\frac{2}{1998}\right)+P\left(\frac{1996}{1998}\right)=2\)
..............................................................
Do \(x\ne\frac{1}{2}\)nên không có \(P\left(\frac{999}{1998}\right)\)
\(P\left(\frac{998}{1998}\right)+P\left(\frac{1000}{1998}\right)=2\)
=> \(A=1997+2+2+....+2\)(998 số 2)
=> \(A=1997+2.998=3993\)
Vậy A=3993
Tíc mình rồi mình giải cho