\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)

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

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

\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)

\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)

\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)

\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)

\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)

Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)

Ta có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4bc-4ac\)

\(\Leftrightarrow0=2a^2+2b^2+2c^2-2ab-2bc-2ac\)

\(\Leftrightarrow0=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\)

\(\Leftrightarrow0=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

Mà \(\left\{\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)

\(\Rightarrow a=b=c\) ( đpcm )

\(x^2+2x\sqrt{x-\frac{1}{x}}+3x+1=0\)

ĐK: \(x-\frac{1}{x}\ge0\)

\(+x=0\text{ thì }pt\text{ thành }0=1\text{ (vô lí)}\)

\(+\text{Xét }x\ne0;\text{ }pt\Leftrightarrow x+2\sqrt{x-\frac{1}{x}}=3+\frac{1}{x}\)

\(\Leftrightarrow\left(x-\frac{1}{x}\right)+2\sqrt{x-\frac{1}{x}}-3=0\)

Đặt \(\sqrt{x-\frac{1}{x}}=t\ge0;\text{ }pt\text{ thành }t^2+2t-3=0\)

Áp dụng BĐT Cô - si cho vế trái ta có

\(\Rightarrow\left\{\begin{matrix}\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge\frac{2a}{c}\\\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\\\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{2c}{b}\end{matrix}\right.\)

Cộng theo từng vế ta có:
\(\frac{a^2}{b^2}+\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{c^2}{a^2}\ge\frac{2a}{c}+\frac{2b}{a}+\frac{2c}{b}\)

\(\Rightarrow2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\)

\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\) ( đpcm )

Ta gọi 3 số lần lượt là a , b , c

Theo đề bài ta có :
\(\left\{\begin{matrix}\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}\\a^2+b^2+c^2=24309\end{matrix}\right.\)

Ta có \(\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}\)

\(\Leftrightarrow\frac{a^2}{\left(\frac{2}{5}\right)^2}=\frac{b^2}{\left(\frac{3}{4}\right)^2}=\frac{c^2}{\left(\frac{1}{6}\right)^2}\)

\(\Leftrightarrow\frac{a^2}{\frac{4}{25}}=\frac{b^2}{\frac{9}{16}}=\frac{c^2}{\frac{1}{36}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a^2}{\frac{4}{25}}=\frac{b^2}{\frac{9}{16}}=\frac{c^2}{\frac{1}{36}}=\frac{a^2+b^2+c^2}{\frac{2701}{3600}}=\frac{24309}{\frac{2701}{3600}}=32400\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\frac{2}{5}}=32400\\\frac{b}{\frac{3}{4}}=32400\\\frac{c}{\frac{1}{6}}=32400\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a=32400.\frac{2}{5}=12960\\b=32400.\frac{3}{4}=24300\\c=32400.\frac{1}{6}=5400\end{matrix}\right.\)

\(\Rightarrow A=12960+24300+5400=42660\)

Vậy số A = 42660

Áp dụng BĐT Cô - si cho 3 bộ số không âm

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)

\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)

\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)

\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)

Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)

Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}x+y\ge2\sqrt{xy}\\\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\sqrt{xy.\frac{1}{xy}}\)

\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\) ( đpcm )

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}x+y+z\ge3\sqrt{xyz}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt{\frac{1}{xyz}}\end{matrix}\right.\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\sqrt{xyz.\frac{1}{xyz}}\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( đpcm )

Bài 1)

Vận tốc trung bình trên cả quãng đường AC \(\Leftrightarrow\) \(V_{tb}=\frac{AC}{t_1+t_2}\)

Theo đề bài ta có :

\(V_{tb}=\frac{v_1+v_2}{2}\)

Mà \(V_{tb}=\frac{AC}{t_1+t_2}\)

\(\Rightarrow\frac{AC}{t_1+t_2}=\frac{v_1+v_2}{2}\)

\(\Rightarrow2AC=\left(v_1+v_2\right)\left(t_1+t_2\right)\)

\(\Rightarrow2AC=v_1\left(t_1+t_2\right)+v_2\left(t_1+t_2\right)\)

\(\Rightarrow2AC=v_1.t_1+v_1.t_2+v_2.t_1+v_2.t_2\)

Mà \(\left\{\begin{matrix}v_1.t_1=AB\\v_2.t_2=BC\end{matrix}\right.\)

\(\Rightarrow2AC=AB+v_1.t_2+v_2.t_1+BC\)

\(\Rightarrow2AC=AC+v_1.t_2+v_2.t_1\)

\(\Rightarrow AC=v_1.t_2+v_2.t_1\)

\(\Leftrightarrow AB+BC=v_1.t_2+v_2.t_1\)

Mà \(\left\{\begin{matrix}AB=v_1.t_1\\BC=v_2.t_2\end{matrix}\right.\)

\(\Rightarrow v_1.t_1+v_2.t_2=v_1.t_2+v_2.t_1\)

\(\Rightarrow v_1.t_1-v_2.t_1=v_1.t_2-v_2.t_2\)

\(\Rightarrow t_1\left(v_1-v_2\right)=t_2\left(v_1-v_2\right)\)

\(\Rightarrow t_1=t_2\) ( đpcm )

Chiểu cao tối thiểu của gương để người đó nhìn thấy toàn thể ảnh mình trong gương là HK

Xét tam giác OA'B'

K là trung điểm của OA'

H là trung điểm của OB'

\(\Rightarrow\) HK là đường trung bình của tam giác OA'B'

\(\Rightarrow HK=\frac{1}{2}A'B'\)

Mà \(AB=A'B'=1,7m\)

\(\Rightarrow HK=\frac{1}{2}.1,7m\)

\(\Rightarrow HK=0,85m=85cm\)

Vậy chiều cao tối thiểu của gương để người đó nhìn thấy toàn thể ảnh của mình trong gương là 85cm