Viết cách giải luôn ạ!

\(MN//BC\Rightarrow\widehat{ABC}=\widehat{AMN}=45^0\left(\text{đồng vị}\right)\\ \text{Theo tính chất góc ngoài: }\widehat{ACx}=\widehat{BAC}+\widehat{ABC}\\ \Rightarrow\widehat{BAC}=130^0-45^0=85^0\)

Chọn B

B

\(\dfrac{-15}{2}=\dfrac{-3}{\left|-4x+5\right|}\)

\(\Leftrightarrow\left|4x-5\right|=\dfrac{6}{15}=\dfrac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-5=\dfrac{2}{5}\\4x-5=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{27}{5}\\4x=\dfrac{23}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\\x=\dfrac{23}{20}\end{matrix}\right.\)

Em tách ra 1-2 bài/1 câu hỏi để mọi người hỗ trợ nhanh nhất nha!

Bài 3: 

a: Ta có: \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)

\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x}{\sqrt{x}-1}\)

\(Ba\rightarrow Ba^{2+}+2e\\ K\rightarrow K^++e\\ O+2e\rightarrow O^{2-}\\ N\rightarrow N^{5+}+5e\)