a,
ta có:
(x2+7x+3)2=x4+14x3+55x2+42x+9
(8x+4)(x2+5x+2)=8x3+44x2+36x+8
=>x4+14x3+55x2+42x+9=8x3+44x2+36x+8
<=>x4+6x3+11x2+6x+1=0
xét x=0 ko phải no của pt
xét x khác 0
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+6\left(x+\frac{1}{x}\right)+11=0\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+6\left(x+\frac{1}{x}\right)+9=0\Leftrightarrow\left(x+\frac{1}{x}+3\right)^2=0\Rightarrow x=\frac{-3+\sqrt{5}}{2};\frac{-3-\sqrt{5}}{2}\)
d,
xét n=1=> mệnh đề luôn đúng
giả sử mệnh đề đúng với n=k
ta sẽ cm nó đúng với n=k+1
với n=k+1
=>(n+1)(n+2)..(n+n)=2n(n+1)(n+2)...(2n-1)
=2(k+1)(k+2).....2k chia hết cho 2k+1
=>(n+1)(n+2)(n+3)...(n+n) chia hết cho 2n
c,
ta có:
\(\left(1+x\right)\left(1+\frac{y}{x}\right)=1+x+y+\frac{y}{x}\ge1+y+2\sqrt{y}=\left(\sqrt{y}+1\right)^2\)
\(\Rightarrow\left(1+x\right)\left(1+\frac{y}{x}\right)\left(1+\frac{9}{\sqrt{y}}\right)^2\ge\left[\left(\sqrt{y}+1\right)\left(1+\frac{9}{\sqrt{y}}\right)\right]^2\)
\(=\left(\sqrt{y}+\frac{9}{\sqrt{y}}+10\right)^2\ge\left(6+10\right)^2=256\left(Q.E.D\right)\)
dấu = xảy ra khi y=9;x=3
b,
x7+xy6=y14+y8
<=>(x7-y14)+(xy6-y8)=0
<=>(x-y2)(x+y2)+y6(x-y2)=0
<=>(x-y2)(x+y2+y6)=0
xét x=y2
\(\Rightarrow\sqrt{4x+5}+\sqrt{y^2+8}=\sqrt{4y^2+5}+\sqrt{y^2-1}\)
\(\Rightarrow\sqrt{4y^2+5}+\sqrt{y^2+8}=6\)
\(\Rightarrow\left(\sqrt{4y^2+5}-3\right)+\left(\sqrt{y^2+8}-3\right)=0\)
\(\Rightarrow\frac{4y^2-4}{\sqrt{4y^2+5}+3}+\frac{y^2-1}{\sqrt{y^2+8}+3}=0\)
\(\Rightarrow\left(y^2-1\right)\left(\frac{4}{\sqrt{4y^2+5}+3}+\frac{1}{\sqrt{y^2+8}+3}\right)=0\)
\(\frac{4}{\sqrt{4y^2+5}+3}+\frac{1}{\sqrt{y^2+8}+3}>0\Rightarrow y^2=1\Rightarrow\left(x;y\right)=\left(1;1\right);\left(1;-1\right)\)
xét x+y2+y6=0
<=>x=-y2-y6
lại có:
x7+xy6=y14+y8
<=>x(x6+y6)=y14+y8
<=>-(y2+y6)(x6+y6)=y14+y8
mà \(-\left(y^2+y^6\right)\left(x^6+y^6\right)\le0\le y^{14}+y^8\)
<=>y=0=>x=0(ko thỏa mãn)
vậy nghiệm của pt:(x;y)=(1;-1);(1;1)
