a/ Dựng \(AH\perp BC\left(H\in BC\right)\)

Xét tg vuông ACH có

\(\cos C=\dfrac{CH}{AC}=\dfrac{CH}{b}\Rightarrow CH=b\cos C\)

Xét tg vuông ABH có

\(\cos B=\dfrac{BH}{AB}=\dfrac{BH}{c}\Rightarrow BH=c\cos B\)

\(\Rightarrow CH+BH=BC=a=b\cos C+c\cos B\)

b/

Đặt \(\widehat{BAH}=\alpha;\widehat{CAH}=\beta\)

\(\Rightarrow\cos A=\cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\)

\(=\dfrac{AH}{c}.\dfrac{AH}{b}-\dfrac{BH}{c}.\dfrac{CH}{b}=\dfrac{AH^2-BH.CH}{bc}=\)

\(=\dfrac{2AH^2-2BH.CH}{2bc}=\dfrac{c^2-BH^2+b^2-CH^2-2BH.CH}{2bc}=\)

\(=\dfrac{b^2+c^2-\left(BH+CH\right)^2}{2bc}=\dfrac{b^2+c^2-a^2}{2bc}\)

\(x^4+2x^2-3\\ =\left(x^4-x^2\right)+\left(3x^2-3\right)\\ =x^2\left(x^2-1\right)+3\left(x^2-1\right)\\ =\left(x^2-1\right)\left(x^2+3\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)

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

\(=x^4+3x^2-x^2-3\)

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

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

Bạn xem lại đề nhé.

Đặt: \(3x^2-5x-7=0\) 

\(\Delta=\left(-5\right)^2-4\cdot3\cdot\left(-7\right)=109>0\)

\(x_1=\dfrac{-\left(-5\right)+\sqrt{109}}{2\cdot3}=\dfrac{5+\sqrt{109}}{6}\)

\(x_2=\dfrac{-\left(-5\right)-\sqrt{109}}{2\cdot3}=\dfrac{5-\sqrt{109}}{6}\) 

=> \(3x^2-5x-7=\left(x-\dfrac{5+\sqrt{109}}{6}\right)\left(x-\dfrac{5-\sqrt{109}}{6}\right)\)

Đặt: \(\dfrac{1}{2x-y}=a;\dfrac{1}{x+y}=b\left(2x\ne y;x\ne-y\right)\)

Hpt trở thành:

\(\left\{{}\begin{matrix}3a-6b=1\\a-b=0\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}3a-6a=1\\a=b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-3a=1\\a=b\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{3}\\b=a=-\dfrac{1}{3}\end{matrix}\right.\) 

\(=>\left\{{}\begin{matrix}\dfrac{1}{2x-y}=\dfrac{1}{-3}\\\dfrac{1}{x+y}=\dfrac{1}{-3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=-3\\x+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=-6\\x+y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-6}{3}=-2\\-2+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3+2=-1\end{matrix}\right.\)

mik cần gấp

\(\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x^2+1}}+\dfrac{4}{\sqrt{y^2+1}}=9\\\dfrac{1}{\sqrt{x^2+1}}-\dfrac{1}{\sqrt{y^2+1}}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x^2+1}}+\dfrac{4}{\sqrt{y^2+1}}=9\\\dfrac{4}{\sqrt{x^2+1}}-\dfrac{4}{\sqrt{y^2+1}}=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x^2+1}}+\dfrac{4}{\sqrt{y^2+1}}+\dfrac{4}{\sqrt{x^2+1}}-\dfrac{4}{\sqrt{y^2+1}}=9+3\\\dfrac{1}{\sqrt{x^2+1}}-\dfrac{1}{\sqrt{y^2+1}}=\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{12}{\sqrt{x^2+1}}=12\\\dfrac{1}{\sqrt{y^2+1}}=1-\dfrac{3}{4}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+1=1\\y^2+1=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2=0\\y^2=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{15}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x+3y=-xy\\\dfrac{8}{x}-\dfrac{6}{y}=5\end{matrix}\right.\left(x;y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\8y-6x=5xy\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=-5xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=5\left(2x+3y\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=10x+15y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\-4x=23y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\cdot\dfrac{-23}{4}y+3y=\dfrac{-23}{4}y\cdot y\\x=\dfrac{-23}{4}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y^2-43y=0\\x=\dfrac{-23}{4}y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y\left(23y-43\right)=0\\x=\dfrac{-23}{4}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=0\left(ktm\right)\\y=\dfrac{43}{23}\left(tm\right)\end{matrix}\right.\\x=\dfrac{-23}{4}\cdot\dfrac{43}{23}=\dfrac{-43}{4}\end{matrix}\right.\)