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\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
Bài 1 :
\(A=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\\ =\sqrt{3}+2+2-\sqrt{3}=4\)
Bài 2 :
a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)
b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)
Bài 3 :
\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)
Bài 1:
Ta có: \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}+2-\sqrt{3}\)
=4
Bài 2:
a) Ta có: \(P=3x-\sqrt{x^2-10x+25}\)
\(=3x-\sqrt{\left(x-5\right)^2}\)
\(=3x-\left|x-5\right|\)
\(=\left[{}\begin{matrix}3x-x+5\left(x\ge5\right)\\3x+x-5\left(x< 5\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}2x+5\\4x-5\end{matrix}\right.\)
b) Thay x=2 vào P, ta được:
\(P=4\cdot2-5=8-5=3\)
Bài 3:
Ta có: \(M=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\pm1\)
B1: A = \(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
= \(\sqrt{4+2.2\sqrt{3}+3}-\sqrt{4-2.2\sqrt{3}+3}\)
= \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
= \(2+\sqrt{3}+2-\sqrt{3}=4\)
Vậy A = 4
b) P = 3x-\(\sqrt{x^2-10x+25}\) = \(3x-\sqrt{\left(x-5\right)^2}\)
= \(3x-\left|x-5\right|\)
Thay x = 2 vào biểu thức trên ta có :
P = 6 - 3 = 3
Vậy tại x = 2 thì P = 3
B3 : \(M=\dfrac{\sqrt{x^2-2x+1}}{x-1}\left(x\ne1\right)\)
= \(\dfrac{\left|x-1\right|}{x-1}\)
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