Bài 5: Bảng căn bậc hai

Với x,y,z dương

Ta có:(x-y)²\(\ge0\forall x;y\)

=>x²+y²\(\ge\)2xy

Dấu = xảy ra khi x=y

Tương tự y²+z²\(\ge\)2yz

z²+x²\(\ge\)2zx

Cộng vế với vế 3 BĐT =>2(x²+y²+z²)\(\ge\)2(xy+yz+zx)

<=>x²+y²+z²\(\ge\)xy+yz+zx

<=>\(\dfrac{3}{xy+yz+zx}\ge\dfrac{3}{x^2+y^2+z^2}\)

Dấu = xảy ra khi và chỉ khi x=y=z

=>\(\dfrac{3}{xy+yz+zx}+\dfrac{2}{x^2+y^2+z^2}\ge\dfrac{5}{x^2+y^2+z^2}\)

Áp dụng BĐT bunhiacopski:

\(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{3^3}\right)\le\left(\dfrac{x+y+z}{3}\right)^2=\dfrac{1}{3^2}=\dfrac{1}{9}\)(Do x+y+z=1)

Dấu = xảy ra khi và chỉ khi \(\dfrac{x}{3}=\dfrac{y}{3}=\dfrac{z}{3}\)<=>x=y=z

=>\(\dfrac{5}{x^2+y^2+z^2}=\dfrac{5}{3\cdot\left(x^2+y^2+z^2\right)\left(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}\right)}\ge\dfrac{5}{3\cdot\dfrac{1}{9}}=15\)

=>\(\dfrac{3}{xy+yz+zx}+\dfrac{2}{x^2+y^2+z^2}\ge15\)(đpcm)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=y=z\\z+y+z=1\end{matrix}\right.\)<=>x=y=z=\(\dfrac{1}{3}\)

\(Q=\dfrac{2}{2x-1}\sqrt{8x^2\left(1-4x+4x^2\right)}\)

\(=\dfrac{2}{2x-1}\cdot2x\sqrt{2\left(1-4x+4x^2\right)}\)

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

\(=\dfrac{4x\sqrt{2-8x+8x^2}}{2x-1}\)

\(=\dfrac{4x\sqrt{\left(\sqrt{2}-\sqrt{8}x\right)^2}}{2x-1}\)

\(=\dfrac{4x\cdot\left(\sqrt{2}-\sqrt{8}x\right)}{2x-1}\)

\(=\dfrac{4x\cdot\left(\sqrt{2}-2\sqrt{2}x\right)}{2x-1}\)

\(=\dfrac{4\sqrt{2}x-8\sqrt{2}x^2}{2x-1}\)

điều kiện \(x>\dfrac{1}{2}\)

Q = \(\dfrac{2}{2x-1}\sqrt{8x^2\left(1-4x+4x^2\right)}\) = \(\dfrac{2}{2x-1}\sqrt{\left(\sqrt{8}x\right)^2\left(2x-1\right)^2}\)

Q = \(\dfrac{2}{2x-1}\left(\sqrt{8}x\right)\left(2x-1\right)\) \(\left(vìx>\dfrac{1}{2}\right)\)

Q = \(2\sqrt{8}x\)

đúng rồi bạn phải là \(\sqrt{12}-\sqrt{18}\)

mik thử làm xem

\(\dfrac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)

\(=\dfrac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{-6}\)

\(=\dfrac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{-6}\)

\(=-\dfrac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{6}\)

từ bước 1 ra bước 2 sai rồi bn ơi

\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)

= \(\dfrac{\left(a\sqrt{b}+b\sqrt{a}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)

= \(\dfrac{a\sqrt{ab}-ab+ab-b\sqrt{ab}}{a-b}\)

= \(\dfrac{\sqrt{ab}\left(a-b\right)}{a-b}=\sqrt{ab}\)

\(\left(\sqrt{ab}-\sqrt{\dfrac{a}{b}}+\dfrac{1}{a}\sqrt{4ab}+\dfrac{1}{b}\sqrt{\dfrac{b}{a}}\right):\left(1+\dfrac{2}{a}-\dfrac{1}{b}+\dfrac{1}{ab}\right)=\left(\sqrt{ab}-\sqrt{\dfrac{a}{b}}+\dfrac{2\sqrt{b}}{\sqrt{a}}+\dfrac{1}{\sqrt{ab}}\right):\left(\dfrac{ab+2b-a+1}{ab}\right)=\dfrac{ab-a+2b+1}{\sqrt{ab}}\cdot\dfrac{ab}{ab+2b-a+1}=\sqrt{ab}\)

\(x\ne 1;-1\)

\(B=\dfrac{2011}{x+1}+\dfrac{2012}{x-1}\) có nghĩa khi và chỉ khi \(\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm1}\)

x và x ≠ -1

\(\dfrac{2}{\sqrt{7}-5}+\dfrac{2}{\sqrt{7}+5}\)

\(=\dfrac{2\sqrt{7}+10+2\sqrt{7}-10}{7-25}=\dfrac{4\sqrt{7}}{-18}=\dfrac{-2\sqrt{7}}{9}\)

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

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

Cân bằng hệ số:

\(25x^2+36y^2\ge60xy\Rightarrow5x^2+\dfrac{36}{5}y^2\ge12xy\)

\(4x^2+9z^2\ge-12zx\Rightarrow3x^2+\dfrac{27}{4}z^2\ge-9zx\)

\(16y^2+25z^2\ge-40yz\Rightarrow\dfrac{4}{5}y^2+\dfrac{5}{4}z^2\ge-2yz\)

Cộng vế với vế:

\(8x^2+8y^2+8z^2\ge12xy-9zx-2yz\)

\(\Leftrightarrow8x^2+8y^2+8z^2+16xy+16yz+16zx\ge28xy+14yz+7zx\)

\(\Leftrightarrow8\left(x+y+z\right)^2\ge7\left(4xy+2yz+zx\right)\)

\(\Leftrightarrow4xy+2yz+zx\le\dfrac{8}{7}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{6}{7};\dfrac{5}{7};-\dfrac{4}{7}\right)\)