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

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

\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)

Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

1) Đặt \(B=x^2+y^2+z^2\)

\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)

Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)

\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)

Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)

\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

4) a) Giả sử \(a\ge b\ge c\). Ta có

\(a< b+c\Leftrightarrow2a< a+b+c=2\Rightarrow a< 1\Rightarrow b< 1;c< 1\)

b) Từ câu a), ta suy ra: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)

\(\Leftrightarrow ab+bc+bc>1+abc\)

Mặt khác, ta cũng có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow4>a^2+b^2+c^2+2\left(1+abc\right)\)

\(\Rightarrow4>a^2+b^2+c^2+2+2abc\)

\(\Rightarrow2-2abc>a^2+b^2+c^2\Rightarrow a^2+b^2+c^2< 2\left(1-abc\right)\)

1) Ghét hình, làm ''đại''

\(A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\)

Ta có: \(A^2=\left(\dfrac{xy}{z}\right)^2+\left(\dfrac{yz}{x}\right)^2+\left(\dfrac{xz}{y}\right)^2+2\left(x^2+y^2+z^2\right)\)

Mặt khác:

\(2\left[\left(\dfrac{xy}{z}\right)^2+\left(\dfrac{yz}{x}\right)^2+\left(\dfrac{xz}{y}\right)^2\right]\)

\(=\left[\left(\dfrac{xy}{z}\right)^2+\left(\dfrac{yz}{x}\right)^2\right]+\left[\left(\dfrac{xy}{z}\right)^2+\left(\dfrac{zx}{y}\right)^2\right]+\left[\left(\dfrac{yz}{x}\right)^2+\left(\dfrac{xz}{y}\right)^2\right]\ge2\left(x^2+y^2+z^2\right)\)

\(\Rightarrow A^2\ge3\left(x^2+y^2+z^2\right)=3\)

\(\Rightarrow A\ge\sqrt{3}\)

\(Min_A=\sqrt{3}\)khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

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2) Đặt tính chia cũng ra

Nhận thấy thương là \(ax^2+nx+1\)

\(\left(x^2-2x+1\right)\left(ax^2+nx+1\right)=P\left(x\right)\)

Nhân vô rồi đồng nhất thức với \(P\left(x\right)\), ta được \(\left\{{}\begin{matrix}a=3\\b=-4\end{matrix}\right.\)

Thay vô tìm nghiệm

6) \(3a^2+3b^2=10ab\)

\(3a^2-10ab+3b^2=0\)

\(3a^2-ab-9ab+3b^2=0\)

\(a\left(3a-b\right)-3b\left(3a-b\right)=0\)

\(\left(a-3b\right)\left(3a-b\right)=0\)

\(\left[{}\begin{matrix}a-3b=0\Leftrightarrow a=3b\\3a-b=0\Leftrightarrow3a=b\end{matrix}\right.\)

Vì a>b>0 nên chọn TH1

Thay vào P

\(P=\dfrac{a-b}{a+b}=\dfrac{3b-b}{3b+b}=\dfrac{2b}{4b}=\dfrac{1}{2}\)

\(13x^2+y^2-4xy-16x+2y+2022\)

\(=\left(y-2x\right)^2+2\left(y-2x\right).1+1+9x^2-12x+4+2017\)

\(=\left(y-2x+1\right)^2+\left(3x-2\right)^2+2017\)

Vậy: Min là 2017 khi \(x=\dfrac{2}{3};y=\dfrac{1}{3}\)

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

\(=x^2-6x+9+x^2+8x+16\)

\(=2x^2+2x+25\)

\(=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{49}{2}\)

Vậy: Min là \(\dfrac{49}{2}\) khi \(x=\dfrac{-1}{2}\)