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a,
Ta có: \(a\left(b+1\right)b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow ab=\left(a+1\right)\left(b+1\right):\left(a+1\right)\left(b+1\right)=1\)
=>đpcm
b,
Ta có: \(2\left(a+1\right)\left(a+b\right)=\left(a+b\right)\left(a+b+2\right)\)
\(\Rightarrow2a+2=a+b+2\)
\(\Rightarrow a-b=0\)
\(\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2+b^2=2\) (đpcm)
Nhận xét nào sau đây là sai?
A:Sự oxi hóa chậm là quá trình oxi hóa có kèm theo tỏa nhiệt nhưng không phát sáng
B:Oxi là chất oxi hóa trong các phản ứng hóa học.
C:Sự cháy là sự oxi hóa có kèm theo tỏa nhiệt và không phát sáng.
D:Sự oxi hóa là quá trình tác dụng của một chất với oxi.
# HOK TỐT #
a,\(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{a+b}{2}\right)^2-1^2=\left(\frac{a+b}{2}-1\right)\left(\frac{a+b}{2}+1\right)\)
b,\(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(x-y\right)^2\left(9-4\right)=5\left(x-y\right)^2\)
c,\(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)\)
\(=\left(-p-4q\right)3p\)
d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{p^2}{6}\right)^2=\left(5pm-\frac{p^2}{6}\right)\left(5pm+\frac{p^2}{6}\right)\)
a, \(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{1}{2}a+\frac{1}{2}b\right)^2-1=\left(\frac{a}{2}+\frac{b}{2}-1\right)\left(\frac{a}{2}+\frac{b}{2}+1\right)\)
b, \(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(3x-3y\right)^2-\left(2x-2y\right)^2\)
\(=\left(3x-3y-2x+2y\right)\left(3x-3y+2x-2y\right)=5\left(x-y\right)^2\)
c, \(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q\right)^2-\left(2p+2q\right)^2\)
\(=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)^2=9p^2\left(-p-4q\right)\)
d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{1}{6}p^2\right)^2=\left(5pm^2-\frac{1}{6}p^2\right)\left(5pm^2+\frac{1}{6}p^2\right)\)
\(=p^2\left(5m^2-\frac{1}{6}p\right)\left(5m^2+\frac{1}{6}p\right)\)
\(a.\)
Ta sẽ biến đổi biểu thức \(B\) quy về dạng có thể dùng được hằng đẳng thức \(\left(x-y\right)\left(x+y\right)=x^2-y^2\), khi đó:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
Vì \(2^{16}>2^{26}-1\) nên \(2^{16}>\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
Vậy, \(A>B\)
Tương tự với câu \(b\) kết hợp với phương pháp tách hạng tử, khi đó xuất hiện hằng đẳng thức mới và dễ dàng đơn giản hóa biểu thức \(A\). Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)=\frac{1}{2}\left(3^{128}-1\right)\)
Mặt khác, do \(\frac{1}{2}<1\) nên \(\frac{1}{2}\left(3^{128}-1\right)<3^{128}-1\)
Vậy, \(B>A\)
a) <=> \(A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\ldots\left(3^{16}+1\right)\)
\(A=3^{32}-1\)
b) \(B=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\ldots\left(4^{64}+1\right)\)
\(B=4^{128}-1\)
c) \(\Leftrightarrow C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\ldots\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(C=5^{256}-1+5^{256}-1=2\cdot5^{256}-2\)
a: \(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
b: \(B=15\left(4^2+1\right)\left(4^4+1\right)\cdot\ldots\cdot\left(4^{64}+1\right)\)
\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\cdot\ldots\cdot\left(4^{64}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\cdot\left(4^8+1\right)\cdot\left(4^{16}+1\right)\cdot\left(4^{32}+1\right)\cdot\left(4^{64}+1\right)\)
\(=\left(4^8-1\right)\cdot\left(4^8+1\right)\cdot\left(4^{16}+1\right)\cdot\left(4^{32}+1\right)\cdot\left(4^{64}+1\right)\)
\(=\left(4^{16}-1\right)\cdot\left(4^{16}+1\right)\cdot\left(4^{32}+1\right)\cdot\left(4^{64}+1\right)\)
\(=\left(4^{32}-1\right)\cdot\left(4^{32}+1\right)\cdot\left(4^{64}+1\right)\)
\(=\left(4^{64}-1\right)\cdot\left(4^{64}+1\right)=4^{128}-1\)
c: \(C=24\left(5^2+1\right)\left(5^4+1\right)\cdot\ldots\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\cdot\ldots\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\cdot\ldots\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^{64}-1\right)\left(5^{64}+1\right)\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^{128}-1\right)\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=5^{256}-1+5^{256}-1=2\cdot5^{256}-2\)