giúp mk câu b bài 1 vs ạ
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ĐKXĐ: \(-1\le x\le3\)
Đặt \(\sqrt{x+1}+\sqrt{3-x}=t\ge\sqrt{x+1+3-x}=2\)
\(\Rightarrow4+2\sqrt{-x^2+2x+3}=t^2\)
\(\Rightarrow\sqrt{-x^2+2x+3}=\dfrac{t^2-4}{2}\) (1)
Phương trình trở thành:
\(t-\dfrac{t^2-4}{2}=2\)
\(\Leftrightarrow2t-t^2=0\Rightarrow\left[{}\begin{matrix}t=0\left(loại\right)\\t=2\end{matrix}\right.\)
Thế vào (1):
\(\Rightarrow\sqrt{-x^2+2x+3}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
8 Her telephone number isn't known by me
9 The children will be brought home by my students
10 đúng r
11 We were given more information by her
12 All the workers of the plan were being instructed by the chief engineer
\(a)P=\left(\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}+\dfrac{1}{1-x}\right).\left(\dfrac{x^2}{x+1}+1\right).\left(x\ne1;x\ne-1\right).\\ P=\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}.\\ P=\dfrac{x^2-x}{x-1}.\dfrac{1}{x+1}.\\ P=\dfrac{x\left(x-1\right)}{x-1}.\dfrac{1}{x+1}.\\ P=x.\dfrac{1}{x+1}.\\ P=\dfrac{x}{x+1}.\)
\(P=\dfrac{1}{4}.\Rightarrow\dfrac{x}{x+1}=\dfrac{1}{4}.\\ \Leftrightarrow4x-x-1=0.\\ \Leftrightarrow3x-1=0.\\ \Leftrightarrow x=\dfrac{1}{3}\left(TM\right).\)
a: \(\widehat{P}=180^0-45^0-35^0=100^0\)
b: Số đo góc ngoài tại đỉnh N là:
\(\widehat{P}+\widehat{M}=100^0+45^0=145^0\)