Ôn tập cuối năm môn Đại số - Hỏi đáp

\(VT=\frac{b^2c^2}{b+c}+\frac{a^2c^2}{a+c}+\frac{a^2b^2}{a+b}\ge\frac{\left(ab+bc+ca\right)^2}{2\left(a+b+c\right)}\ge\frac{3abc\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\sqrt{x+2017}-y^3=\sqrt{y+2017}-x^3\)

\(\Leftrightarrow\left(\sqrt{x+2017}-\sqrt{y+2017}\right)+\left(x^3-y^3\right)=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x^2+xy+y^2\right)\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow P=x^2-3x^2+12x-x^2+2018\)

\(=-3x^2+12x+2018=2030-3\left(x-2\right)^2\le2030\)

\(x_1=0\Rightarrow m=\frac{2019}{2}\Rightarrow x_2=\frac{2019}{2}\) thỏa mãn

Để \(0< x_1< 1< x_2\)

\(\Leftrightarrow y\left(0\right).y\left(1\right)< 0\)

\(\Leftrightarrow\left(2m-2019\right)\left(m-2018\right)< 0\)

\(\Rightarrow\frac{2019}{2}< m< 2018\)

Vậy \(\frac{2019}{2}\le m< 2018\)

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

TH1: \(\left\{{}\begin{matrix}x+y=0\\2x-5y=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=1\\2x-5y=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{12}{7}\\y=-\frac{5}{7}\end{matrix}\right.\)

\(\sqrt{3}\sin x+\cos x=\sqrt{2}\Leftrightarrow\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\sin x\cos\frac{\pi}{6}+\cos x\sin\frac{\pi}{6}=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\sin\left(x+\frac{\pi}{6}\right)=\sin\frac{\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{6}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k2\pi\\x=\frac{7\pi}{12}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

Chú ý rằng: sin45⁰ = cos45⁰, sin40⁰ = cos50⁰, sin50⁰ = cos40⁰

Ta được:

\(\dfrac{\cos50^0-\cos45^0+\cos50^0}{\cos40^0-\cos45^0+\cos50^0}-\dfrac{6\times3\left(\dfrac{\sqrt{3}}{3}+\tan15^0\right)}{3\left(1-\dfrac{\sqrt{3}}{3}\tan15^0\right)}\)

\(=1-6\left(\dfrac{\tan30^0+\tan15^0}{1-\tan30^0\times\tan15^0}\right)\)

\(=1-6\tan45^0=-5\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\) \(\frac{a}{b}+1=\frac{c}{d}+1\)

\(\Rightarrow\) \(\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)

\(\Rightarrow\)\(\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm\right)\)

Cách nêu tính chất đặc trưng:

A=\(\left\{x/\left(x^2+2x-3\right)\left(x^2-13x+42\right)\right\}\)

B=\(\left\{\frac{2x+1}{2^{x+1}},x\in N,0\le x\le4\right\}\)