\(5x^2-9x+30=0\\ \Leftrightarrow\left(\sqrt{5}x\right)^2+2\cdot\sqrt{5}\cdot\dfrac{9\sqrt{5}}{10}x+\dfrac{81}{20}+\dfrac{519}{20}=0\\ \Leftrightarrow\left(\sqrt{5}x+\dfrac{9\sqrt{5}}{10}\right)^2+\dfrac{519}{20}\)

mà \(\left(\sqrt{5}x+\dfrac{9\sqrt{5}}{10}\right)^2\ge0\forall x\Rightarrow\left(\sqrt{5}x+\dfrac{9\sqrt{5}}{10}\right)^2+\dfrac{519}{20}>0\forall x\)

=> pt vô nghiệm

Vậy phương trình vô nghiệm

Biển (Hồ) Caspi

\(\dfrac{x+2}{x-2}-\dfrac{x-3}{x+2}=5\\ \Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}=5\\ \Leftrightarrow\dfrac{x^2+4x+4-x^2+5x-6}{\left(x^2-4\right)}=5\\ \Leftrightarrow9x-2=5\left(x^2-4\right)\\ \Leftrightarrow5x^2-20=9x-2\\ \Leftrightarrow5x^2-9x-18=0\\ \Leftrightarrow\left(5x+6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+6=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{6}{5};3\right\}\)

a) \(B=\dfrac{4x^3-8x^2+3x-6}{2x^2-3x-2}=\dfrac{\left(4x^2+3\right)\left(x-2\right)}{\left(2x+1\right)\left(x-2\right)}=\dfrac{4x^2+3}{2x+1}\)

b) *mình tìm x ∈ Z để B ∈ Z nha

\(B\in Z\Rightarrow\dfrac{4x^2+3}{2x+1}\in Z\\ \Leftrightarrow\dfrac{4x^2+2x-2x-1+4}{2x+1}=\dfrac{4x\left(2x+1\right)-\left(2x+1\right)+4}{2x+1}=4x-1+\dfrac{4}{2x+1}\in Z\\ \Rightarrow\left\{{}\begin{matrix}x\in Z\\\dfrac{4}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\Leftrightarrow x\in\left\{-\dfrac{5}{2}\left(ktm\right);-\dfrac{3}{2}\left(ktm\right);-1\left(tm\right);0\left(tm\right);\dfrac{1}{2}\left(ktm\right);\dfrac{3}{2}\left(ktm\right)\right\}\end{matrix}\right.\)

Vậy x ∈ {-1; 0}

- "Tảng đá này nặng thật."

- "Chai rượu này nặng thật."

a) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=\sqrt{3^2\cdot15^2}=\left|3\cdot15\right|=45\)

b) \(\sqrt{9-4\sqrt{5}}+2=\sqrt{5-4\sqrt{5}+4}+2=\sqrt{\left(\sqrt{5}-2\right)^2}+2=\left|\sqrt{5}-2\right|+2=\sqrt{5}\)