Question

[Phạm Mỹ Châu]

cho a,b,c > 0 thỏa mãn \(a^2+b^2+c^2=3\)

CMR \(P=\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\ge\dfrac{3\sqrt{13}}{2}\)

Answer 1

[Lightning Farron]

Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)

Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)

\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)

\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)

Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)

\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)

\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*

Answer 2

[Lightning Farron]

B1:a)ĐK: \(x\ne 0;4;9\)

b)\(P=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\dfrac{1}{\sqrt{x}+1}\right)\)

\(=\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1+1}{\sqrt{x}+1}\right)\)

\(=\dfrac{x-9-x+4+x^{\dfrac{1}{2}}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{x^{\dfrac{1}{2}}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)\(=\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}\)

c)Vì \(x^{\dfrac{1}{2}}+1>0\forall x\) nên

\(P< 0< =>x-2x^{\dfrac{1}{2}}< 0\)

\(\Leftrightarrow x^{\dfrac{1}{2}}\left(x^{\dfrac{1}{2}}-2\right)< 0\)

\(\Leftrightarrow0< x< 4\)

Vậy 0<x<4 thì P<0

d)tA CÓ: \(\dfrac{1}{P}=\dfrac{x-2x^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}+1}=\dfrac{x-2x^{\dfrac{1}{2}}+1-1}{x^{\dfrac{1}{2}}+1}=\dfrac{\left(x^{\dfrac{1}{2}}-1\right)^2-1}{x^{\dfrac{1}{2}}+1}\ge-1\)

"=" khi x=1

B2:

a)\(A=x^2-2xy+y^2+4x-4y-5\)

\(=\left(x-y\right)^2+4\left(x-y\right)-5\)

\(=\left(x-y\right)^2-1+4\left(x-y\right)-4\)

\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)

\(=\left(x-y+5\right)\left(x-y-1\right)\)

b)\(P=x^4+2x^3+3x^2+2x+1\)

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

\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)

\(=\left(x^2+x+1\right)^2\ge0\forall x\)

Vậy MinP=0

c)\(Q=x^6+2x^5+2x^4+2x^3+2x^2+2x+1\)

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

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

\(=0\left(x^4+x^3+2x^2+x+3\right)+4=4\)

Vậy x^2+x=1 thì Q=4

B3:a)\(2xy+x+y=83\)

\(\Leftrightarrow x\left(2y+1\right)+\dfrac{1}{2}\left(2y+1\right)=\dfrac{167}{2}\)

\(\Leftrightarrow2x\left(2y+1\right)+1\left(2y+1\right)=167\)

\(\Leftrightarrow\left(2x+1\right)\left(2y+1\right)=167\)

Mà \(Ư\left(167\right)=\left\{\pm1;\pm167\right\}\)

\(\Leftrightarrow\left(x;y\right)=\left(-84;-1\right);\left(-1;-84\right);\left(0;83\right);\left(83;0\right)\)

Vậy...

b)\(y^2+2xy-3x-2=0\)

\(\Leftrightarrow x^2+y^2+2xy-x^2-3x-2=0\)

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

\(\Leftrightarrow\left(x+y\right)^2=\left(x+1\right)\left(x+2\right)\)

Vì \(x;y\in Z\) nên VT là số chính phương VP là tích 2 số nguyên liên tiếp

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2\end{matrix}\right.\)

Vậy...

B5:\(B=\dfrac{x^2+x+1}{x^2-x+1}\)

\(\Leftrightarrow x^2\left(B-1\right)+x\left(-B-1\right)+\left(B-1\right)=0\)

\(\Delta=\left(-B-1\right)^2-4\left(B-1\right)\left(B-1\right)\)

\(=-\left(B-3\right)\left(3B-1\right)\)

pt có nghiệm khi \(\Delta\ge0\)

\(\Leftrightarrow\left(B-3\right)\left(3B-1\right)\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}B-3\le0\\3B-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}B\le3\\B\ge\dfrac{1}{3}\end{matrix}\right.\)

Min B=1/3 khi x=-1; Max B=3 khi x=1