(1010)2 + (1101)2 = ...
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a: \(1CA_{16}=458_{10}=\text{111001010}_2\)
b: \(1101_2=13_{10}=D_{16}\)
c: \(998_{10}=3E6_{16}\)
d: \(67_{10}=\text{1000011}_2\)
e: \(1111001_{10}=10F3D9_{16}\)
a) 1CA(16)=458(10)=111001010(2)
b)1101(2)=13 (10)=D (16)
c)998(10)=3E6(16)
d)67(10)=1000011(2)
e)1111001(10)=10F3D9(16)
\(bx^2=ay^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow\left(\dfrac{x^2}{a}\right)^{1010}=\left(\dfrac{y^2}{b}\right)^{1010}\\ \Leftrightarrow\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{a^{1010}}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{b^{1010}}=\dfrac{x^{2020}+y^{2020}}{a^{1010}+b^{1010}}\left(3\right)\)
Đặt \(\dfrac{x^2}{a}=\dfrac{y^2}{b}=k\Leftrightarrow x^2=ak;y^2=bk\)
\(x^2+y^2=1\Leftrightarrow ak+bk=1\Leftrightarrow k\left(a+b\right)=1\Leftrightarrow a+b=\dfrac{1}{k}\)
\(\Leftrightarrow\dfrac{2}{\left(a+b\right)^{1010}}=\dfrac{2}{\left(\dfrac{1}{k}\right)^{1010}}=2:\dfrac{1}{k^{1010}}=k^{1010}\left(1\right)\)
Mà \(\dfrac{x^{2020}}{a^{1010}}=\dfrac{\left(x^2\right)^{1010}}{a^{1010}}=\dfrac{a^{1010}k^{1010}}{a^{1010}}=k^{1010}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta được đpcm
a: \(1CA_{16}=458_{10}=\text{111001010}_2\)
b: \(1101_2=13_{10}=D_{16}\)
c: \(998_{10}=3E6_{16}\)
d: \(67_{10}=\text{1000011}_2\)
e: \(1111001_{10}=10F3D9_{16}\)
a) 1CA(16)=458(10)=111001010(2)
b)1101(2)=13 (10)=D (16)
c)998(10)=3E6(16)
d)67(10)=1000011(2)
e)1111001(10)=10F3D9(16)
a: \(1CA_{16}=458_{10}=\text{111001010}_2\)
b: \(1101_2=13_{10}=D_{16}\)
c: \(998_{10}=3E6_{16}\)
d: \(67_{10}=\text{1000011}_2\)
e: \(1111001_{10}=10F3D9_{16}\)
a) 1CA(16)=458(10)=111001010(2)
b)1101(2)=13 (10)=D (16)
c)998(10)=3E6(16)
d)67(10)=1000011(2)
e)1111001(10)=10F3D9(16)
a: \(1CA_{16}=458_{10}=\text{111001010}_2\)
b: \(1101_2=13_{10}=D_{16}\)
c: \(998_{10}=3E6_{16}\)
d: \(67_{10}=\text{1000011}_2\)
e: \(1111001_{10}=10F3D9_{16}\)
a) 1CA(16)=458(10)=111001010(2)
b)1101(2)=13 (10)=D (16)
c)998(10)=3E6(16)
d)67(10)=1000011(2)
e)1111001(10)=10F3D9(16)