(y+1/2)+(y+1/4)+(y+1/8)+(y+1/16)+...+(y+1/1024)=1
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Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
\(\left(y-1\right)\left(y+1\right)\left(y^2+1\right)\left(y^4+1\right)\left(y^8+1\right)=2^{16}-1\)
\(y^{16}+y^8+y^{12}+y^4-y^{12}-y^4-y^8-1=65535\)
\(y^{16}-1=65535\)
\(y^{16}=65536\)
\(y=\pm\sqrt[16]{65536}=\pm2\)
y x 4 + 1/2 + 1/4 + 1/8 +1/16 = 1
y x 4 + 15/16 = 1
y x 4 = 1/16
y = 1/16 : 4
y = 1/64
\(\left(y+\frac{1}{2}\right)+\left(y+\frac{1}{4}\right)+\left(y+\frac{1}{8}\right)+\left(y+\frac{1}{16}\right)=1\)
\(\Rightarrow4y+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4y+\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4y+\frac{15}{16}=1\)
\(\Leftrightarrow4y=\frac{1}{16}\)
\(\Leftrightarrow y=\frac{1}{64}\)
a: =>4y+15/16=1
=>4y=1/16
hay y=1/64
b: =>10y+1023/1024=1
=>10y=1/1024
hay y=1/10240
3 x y+7/16 = 2
3 x y = 2 - 7/16
3 x y = 25/16
y = 25/16 : 3
y = 25/48
1: =(x+y-3x)(x+y+3x)
=(-2x+y)(4x+y)
2: =(3x-1-4)(3x-1+4)
=(3x+3)(3x-5)
=3(x+1)(3x-5)
3: =(2x)^2-(x^2+1)^2
=-[(x^2+1)^2-(2x)^2]
=-(x^2+1-2x)(x^2+1+2x)
=-(x-1)^2(x+1)^2
4: =(2x+1+x-1)(2x+1-x+1)
=3x(x+2)
5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]
=(2x^2+2)*4x
=8x(x^2+1)
6: =(5x-5y)^2-(4x+4y)^2
=(5x-5y-4x-4y)(5x-5y+4x+4y)
=(x-9y)(9x-y)
7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)
=(x^2+2xy+y^2)(x^2-y^2)
=(x+y)^3*(x-y)
8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)
=[(x-2y)^2-4][(x+2y)^2-36]
=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)
= (y+y+y+y+...+y)+(1/2+1/4+1/8+1/16+...+1/1024)=1
y x 512 + 1023/1024 =1
y x 512 =1-1023/1024
y x 512 =1/1024
y =1/1024:512
y=1/524,288
mk trình bày hơi lệch xíu