HOC24
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Bài học
\(\left(x^2-4x\right)\sqrt{x^2+2x-3}\ge0\)
ĐK : \(\left[{}\begin{matrix}x\le-3\\x\ge1\end{matrix}\right.\)
\(\Leftrightarrow x^2-4x\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le0\end{matrix}\right.\)
Kết hợp với điều kiện \(\Rightarrow\left[{}\begin{matrix}x\le-3\\x\ge4\\x=1\end{matrix}\right.\)
\(A=3sin^2x+6cos^2=3sin^2x+6\left(1-sin^2x\right)\)
\(=6-3sin^2x\)
Do : \(0\le sin^2x\le1\Rightarrow\left\{{}\begin{matrix}6-3sin^2x\ge3\\6-3sin^2x\le6\end{matrix}\right.\)
Ta có :
\(a=\frac{-F_h}{m}=\frac{-10^4}{2000}=-5m/s^2\)
\(s=\frac{v^2-v_0^2}{2a}=\frac{0^2-20^2}{2.-5}=40m\)
Để d tiếp xúc với \(\left(C\right)\) thì
\(d_{\left(I;d\right)}=R\)
\(\Leftrightarrow\frac{\left|m\right|}{\sqrt{3^2+4^2}}=1\)
\(\Leftrightarrow\left|m\right|=5\Leftrightarrow\left[{}\begin{matrix}m=5\\m=-5\end{matrix}\right.\)
Bài 1 : \(VT=a^2+b^2+c^2+3abc=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)+3abc\left(a+b+c\right)}{a+b+c}\ge\frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc}{a+b+c}\)
\(=\frac{a^3+b^3+c^3+3abc+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+6abc}{a+b+c}\)
\(\ge\frac{2ab\left(a+b\right)+2bc\left(b+c\right)+2ca\left(c+a\right)+6abc}{a+b+c}\)
\(=\frac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{a+b+c}=6\)
Có sai sót gì xin cmt bên dưới ạ
\(VT=\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=\frac{a^2}{ca}+\frac{b^2}{ab}+\frac{c^2}{bc}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca}\)
Ta cần chứng minh :
\(\frac{\left(a+b+c\right)^2}{ab+bc+ca}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\frac{3}{ab+bc+ca}+2\ge\frac{9}{a+b+c}\)
Đặt \(a+b+c=t\)
\(\Leftrightarrow\frac{6}{t^2-3}+2\ge\frac{9}{t}\)
\(\Leftrightarrow2t^3-6t+6t-9t^2+27\ge0\)
\(\Leftrightarrow\left(2t+3\right)\left(t-3\right)^2\ge0\) ( Đúng )
Vậy ...
Giả sử \(a\ge b\ge c\)
\(\Rightarrow3a^2\ge a^2+b^2+c^2=3\Rightarrow a\ge1\)
Do đó ta có : \(a+b+c\ge1\)
\(\Rightarrow\left(a+b+c\right)^2\ge a+b+c\)
Áp dụng BĐT Cô-si ta có :
\(\left\{{}\begin{matrix}\frac{a}{c}+\frac{b}{a}+\frac{b}{a}\ge3\sqrt[3]{\frac{b^2}{ac}}=\frac{3b}{\sqrt[3]{abc}}\\\frac{b}{a}+\frac{c}{b}+\frac{c}{b}\ge3\sqrt[3]{\frac{c^2}{ab}}=\frac{3c}{\sqrt[3]{abc}}\\\frac{c}{b}+\frac{a}{c}+\frac{a}{c}\ge3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge\frac{a+b+c}{\sqrt[3]{abc}}\)
Ta cần chứng minh : \(\frac{a+b+c}{\sqrt[3]{abc}}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(\frac{a+b+c}{3}\right)^2\ge\sqrt[3]{abc}\)
Luôn đúng vì : \(\left(\frac{a+b+c}{3}\right)^2\ge\frac{a+b+c}{3}\ge\sqrt[3]{abc}\)
Dấu \("="\) xảy ra khi \(a=b=c=1\)
PTTQ của \(\Delta\) : \(4x+3y-24=0\)
Khoảng cách từ \(O\left(0;0\right)\) đến đen ta là :
\(d_{\left(O,\Delta\right)}=\frac{\left|4.0+3.0-24\right|}{\sqrt{3^2+4^2}}=\frac{24}{5}=4,8\)
Chọn A
\(VT=\frac{1-cosx}{sinx}\left[\frac{\left(1+cosx\right)^2}{sin^2x}-1\right]\)
\(=\frac{1-cosx}{sinx}.\left[\frac{2\left(1+cosx\right)-sin^2x}{sin^2x}-1\right]\)
\(=\frac{2\left(1-cos^2x\right)}{sin^3x}-\frac{2\left(1-cosx\right)}{sinx}\)
\(=\frac{2}{sinx}-\frac{2-2cosx}{sinx}\)
\(=\frac{2cosx}{sinx}=2cotx\)
\(y=\sqrt{a^2-2a+5}+\sqrt{a^2+2a+5}\)
\(=\sqrt{\left(1-a\right)^2+2^2}+\sqrt{\left(a+1\right)^2+2^2}\ge\sqrt{\left(1-a+a+1\right)^2+\left(2+2\right)^2}=\sqrt{20}\)
Dấu = xảy ra khi \(a=0\)