Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{bc\left(a+1\right)}=\dfrac{a}{abc+bc}=\dfrac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\\\dfrac{b}{ca\left(b+1\right)}=\dfrac{b}{abc+ca}=\dfrac{b}{\left(ab+ca\right)+\left(bc+ca\right)}\\\dfrac{c}{ab\left(c+1\right)}=\dfrac{c}{abc+ab}=\dfrac{c}{\left(ab+bc\right)+\left(ca+ab\right)}\end{matrix}\right.\)(ab+bc+ca=abc)

Áp dụng bđt Svácxơ, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

\(P\le\dfrac{a}{4}\left[\dfrac{1}{bc+ca}+\dfrac{1}{ab+bc}\right]+\dfrac{b}{4}\left[\dfrac{1}{ab+ca}+\dfrac{1}{bc+ca}\right]+\dfrac{c}{4}\left[\dfrac{1}{ab+bc}+\dfrac{1}{ca+ab}\right]\)

\(=\dfrac{a}{4\left(a+b\right)c}+\dfrac{a}{4b\left(c+a\right)}+\dfrac{b}{4\left(b+c\right)a}+\dfrac{b}{4\left(a+b\right)c}+\dfrac{c}{4\left(c+a\right)b}+\dfrac{c}{4\left(b+c\right)a}\)

 \(=\dfrac{a+b}{4\left(a+b\right)c}+\dfrac{b+c}{4\left(b+c\right)a}+\dfrac{c+a}{4\left(c+a\right)b}\)

\(=\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}=\dfrac{ab+bc+ca}{4abc}=\dfrac{abc}{4abc}=\dfrac{1}{4}\)(ab+bc+ca=abc)

\(MaxP=\dfrac{1}{4}\Leftrightarrow a=b=c=3\)

Dạ, em sẽ cố

a) \(A+B=\left(x^2-3xy-y^2+1\right)+\left(2x^2+y^2-7xy-5\right)\)

\(=3x^2-10xy-4\)

b) \(B-A=\left(2x^2+y^2-7xy-5\right)-\left(x^2-3xy-y^2+1\right)\)

\(=x^2-4xy+2y^2-6\)

c) Tại x=2; \(y=-\dfrac{1}{2}\)

C\(=\left(2^2\right)-4.2.\dfrac{-1}{2}+2\left(\dfrac{-1}{2}\right)^2-6\)

\(=4+4+\dfrac{1}{2}-6=2+\dfrac{1}{2}=\dfrac{5}{2}\)

Đk:x \(\ge0\)

+) x không là số chính phương

=> \(\sqrt{x}\) là số vô tỉ (loại)

+) x là số chính phương

\(A=3+\dfrac{\sqrt{x}-5}{2\sqrt{x}+1}\)

Để A nhận giá trị nguyên dương

\(\Rightarrow\left(\sqrt{x}-5\right)⋮\left(2\sqrt{x}+1\right)\)

\(\Leftrightarrow\left(2\sqrt{x}-10\right)⋮\left(2\sqrt{x}+1\right)\)

\(\Leftrightarrow-11⋮\left(2\sqrt{x}+1\right)\)

\(\Rightarrow\left(2\sqrt{x}+1\right)\inƯ\left(11\right)=\left\{1;11\right\}\left(2\sqrt{x}+1>0\right)\)

Thay vào => x=25

\(\left(3x-5y\right)^2+\left(xy-135\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(3x-5y\right)^2\ge0\forall x,y\\\left(xy-135\right)^2\ge0\forall x,y\end{matrix}\right.\)

=>  \(\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\xy=135\end{matrix}\right.\)

\(\Rightarrow\dfrac{5}{3}y.y=135\)\(\Rightarrow y^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}y=9\Rightarrow x=15\\y=-9\Rightarrow x=-15\end{matrix}\right.\)

Vì MN là tiếp tuyến với đường tròn O tại A

=>\(AB\perp MN\)

a) Xét tứ giác HANB có: \(\widehat{A}=\widehat{H}=90^o\) mà 2 góc cùng nhìn cạnh BN

=> tứ giác HANB nội tiếp

=> \(\widehat{ANO}=\widehat{HBO}\)

Xét tam giác NAB và tam giác OAN có:

\(\widehat{MAB}=\widehat{NAO}\)

\(\widehat{ANO}=\widehat{MBA}\)

=> \(\Delta NAO\sim\Delta BAM\left(g.g\right)\)

=> \(\dfrac{AO}{AM}=\dfrac{AN}{BA}\Rightarrow AM.AN=AB.AO=2R^2\left(đpcm\right)\)

b) Vì tứ giác HANB nội tiếp

=> \(\widehat{HAB}=\widehat{HNB}\)

Ta có:

\(\left\{{}\begin{matrix}\widehat{MAH}=90^o-\widehat{HAB}\\\widehat{HBN}=90^o-\widehat{HNB}\end{matrix}\right.\) mà \(\widehat{HAB}=\widehat{HNB}\)

=> \(\widehat{MAH}=\widehat{HBN}\left(đpcm\right)\)

a) Để \(\dfrac{13}{x+5}\) là số tự nhiên

=> (x+5) ∈ ước nguyên dương của \(13=\left\{1;13\right\}\) 

Vậy \(x=\left\{-4;8\right\}\)
 

b) \(xy-y+2x=9\)

\(y\left(x-1\right)+2\left(x-1\right)=7\)

\(\left(x-1\right)\left(y+1\right)=7\)

Đề có lẽ thiếu điều kiện x,y; nếu x,y ∈ Z hoặc N thì e xét thuộc ước của 7 nha

\(E=6a+3-9a^2\)

\(=-\left(9a^2-6a-3\right)\)

\(=-\left(\left(3a\right)^2-2.3a+1-4\right)\)

\(=-\left(3a-1\right)^2+4\)

Vì \(-\left(3a-1\right)^2\le0\forall a\)

\(\Rightarrow E\le4\forall x\)

\(MaxE=4\Leftrightarrow x=\dfrac{1}{3}\)

a) \(\left(3x-y\right)^2-\left(y+3x\right)^2=\left(3x-y+y+3x\right)\left(3x-y-y-3x\right)\)

\(=6x.\left(-2y\right)=-12xy\)

b) \(\left(3y-1\right)^2+\left(3x+2\right)^2-2\left(3x+2\right)\left(3y-1\right)\)

\(=\left(3y-1-3x-2\right)^2=\left(3y-3x-3\right)^2=9\left(x-y-1\right)^2\)

c) \(\left(x+y\right)^2-2\left(x^2-y^2\right)+\left(x-y\right)^2\)

\(=\left(x+y-x+y\right)^2=4y^2\)

d) \(\left(x+y-z\right)^2+\left(y-z\right)^2+2\left(x+y-z\right)\left(y-z\right)\)

\(=\left(x+y-z+y-z\right)^2=\left(x+2y-2z\right)^2\)

e) \(\left(x^2+2xy\right)^2-\left(x^2-2xy\right)^2+4x^2\left(x^2-2xy\right)\)

\(=\left(x^2+2xy-x^2+2xy\right)\left(x^2+2xy+x^2-2xy\right)+4x^2\left(x^2-2xy\right)\)

\(=8x^3y+4x^2\left(x^2-2xy\right)=4x^2\left(2xy+x^2-2xy\right)=4x^4\)

f) \(\left(x-y-z\right)^2+\left(x+y+z\right)^2+\left(x-y-z\right)\left(2x+2y+2z\right)\)

\(=\left(x-y-z+x+y+z\right)^2=4x^2\)