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1.
Ta có: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}\)
<=> \(pq\left(x+y\right)=xy\)
Đặt: \(x=ta;y=tb\) với (a; b)=1
Ta có: \(pq.\left(a+b\right)=tab\)
<=> \(pq=\frac{t}{a+b}.ab\left(1\right)\)
vì (a; b) =1 => a, b, a+b đôi một nguyên tố cùng nhau. (2)
(1); (2) => \(t⋮a+b\)
=> \(pq⋮ab\Rightarrow pq⋮a\)vì p; q là hai số nguyên tố nên \(a\in\left\{1;p;q;pq\right\}\)
TH1: a=1 => \(pq⋮b\Rightarrow b\in\left\{1;p;q;pq\right\}\)
+) Khả năng 1: b=1
(1) => \(t=2pq\)=> \(x=y=2pq\)( thỏa mãn)
+) Khả năng 2: b=p
(1) => \(pq=\frac{t}{1+p}.p\Leftrightarrow t=\left(1+p\right)q=q+pq\)
=> \(x=at=q+pq;\)
\(y=at=pq+p^2q\)(tm)
+) Khả năng 3: b=q
tương tự như trên
(1) => \(t=p\left(1+q\right)=p+pq\)
=> \(x=at=p+pq\)
\(y=bt=q\left(p+pq\right)=pq+pq^2\)
+) Khả năng 4: \(b=pq\)
(1) =>\(t=1+pq\)
=> \(x=1+pq;y=pq\left(1+pq\right)=1+p^2q^2\)
TH2: \(a=p\)
=> \(q⋮b\Rightarrow\orbr{\begin{cases}b=1\\b=q\end{cases}}\)
+) KN1: \(b=1\)
Em làm tiếp nhé! Khá là dài
2. \(x^4+4=p.y^4\)
+) Với x chẵn
Đặt x=2m ( m thuộc Z)
=> \(16m^2+4=py^4\)
=> \(py^4⋮4\Rightarrow y^4⋮4\Rightarrow y^2⋮2\Rightarrow y⋮2\)=> Đặt y=2n ;n thuộc Z
Khi đó ta có:
\(16m^2+4=p.16n^2\Leftrightarrow4m^2+1=p.4n^2⋮4\)=> \(1⋮4\)( vô lí)
=> X chẵn loại
+) Với x lẻ
pt <=> \(x^4+4=py^4\)
<=> \(\left(x^2+2x+2\right)\left(x^2-2x+2\right)=py^4\)(i)
Gọi \(\left(x^2+2x+2;x^2-2x+2\right)=d\)(1)
=> \(x^2+2x+2⋮d\)
\(x^2-2x+2⋮d\)
=.> \(\left(x^2+2x+2\right)-\left(x^2-2x+2\right)=4x⋮d\)
Vì x lẻ => d lẻ
=> \(x⋮d\)
=> \(2⋮d\Rightarrow d=1\)
Do đó: \(\left(2x^2+2x+2;2x^2-2x+2\right)=1\)(ii)
Từ (i) và (ii) có thể đặt: với \(ab=y^2\)sao cho:
\(x^2+2x+2=pa^2;\)
\(x^2-2x+2=b^2\)<=> \(\left(x-1\right)^2+1=b^2\)\(\Leftrightarrow\left(x-1-b\right)\left(x-1+b\right)=-1\)
<=> x=b=1 hoặc x=1; b=-1
Với x=1 => a^2.p=5 => p=5
Đặt \(a=\sqrt{2x-3}\) ; \(b=\sqrt{y-2}\) ; \(c=\sqrt{3z-1}\) (\(a,b,c>0\))
Ta có : \(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c=14\)
\(\Leftrightarrow\left(\sqrt{2x-3}+\frac{1}{\sqrt{2x-3}}-2\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{3z-1}+\frac{16}{\sqrt{3z-1}}-8\right)=0\)
\(\Leftrightarrow\left[\frac{\left(2x-3\right)-2\sqrt{2x-3}+1}{\sqrt{2x-3}}\right]+\left[\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}\right]+\left[\frac{\left(3z-1\right)-8\sqrt{3z-1}+16}{\sqrt{3z-1}}\right]=0\)
\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)^2}{\sqrt{2x-3}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{3z-1}-4\right)^2}{\sqrt{3z-1}}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2x-3}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{3z-1}-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}}\)(TMĐK)
Vậy : \(\left(x;y;z\right)=\left(2;6;\frac{17}{3}\right)\)
Với mọi số thực ta luôn có:
`(x-y)^2>=0`
`<=>x^2-2xy+y^2>=0`
`<=>x^2+y^2>=2xy`
`<=>(x+y)^2>=4xy`
`<=>(x+y)^2>=16`
`<=>x+y>=4(đpcm)`
\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)
\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))
=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)
<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)
<=>\(6x+6y+26-5x-5y-30\)≥\(0\)
<=> \(x+y-4\)≥\(0\)
Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)
Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)
<=>\(x+y\) ≥ 2\(\sqrt{xy}\)
=>2\(\sqrt{xy}-4\)≥\(0\)
<=> \(4-4\)≥0
<=>0≥0 ( Luôn đúng )
Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)
lp 8 mà khó thế -,-
Có \(4=a^4+b^4+c^4+1\ge4\sqrt[4]{\left(abc\right)^4}=4abc\)\(\Leftrightarrow\)\(-abc\ge-1\)
\(\Rightarrow\)\(\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca}=\frac{a+b+c}{4-abc}\le\frac{a+b+c}{4-1}=\frac{a+b+c}{3}\)
Lại có \(3=a^4+b^4+c^4\ge\frac{\left(a^2+b^2+c^2\right)^2}{3}\ge\frac{\frac{\left(a+b+c\right)^4}{9}}{3}=\frac{\left(a+b+c\right)^4}{27}\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^4\le81\)\(\Leftrightarrow\)\(a+b+c\le3\)
\(\Rightarrow\)\(\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca}\le\frac{a+b+c}{3}\le\frac{3}{3}=1\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
Ta có: \(4n^4+1=\left(4n^4+4n^2+1\right)-4n^2=\left(2n^2+2n+1\right)\left(2n^2-2n+1\right)\)
\(\frac{4n}{4n^4+1}=\frac{\left(2n^2+2n+1\right)-\left(2n^2-2n+1\right)}{\left(2n^2-2n+1\right)\left(2n^2+2n+1\right)}=\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\)
Thay vào ta có:
\(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^2+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow\frac{2n^2+2n}{2n^2+2n+1}=\frac{220}{221}\Rightarrow n=10\)