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Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Học tốt
Bài làm:
a) \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(B=2+\sqrt{5}-\sqrt{5}+2\)
\(B=4\)
Bài 1:
a) \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\frac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b) \(\sqrt{48}-6\sqrt{\frac{1}{3}}+\frac{\sqrt{3}-3}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}\)
\(=\sqrt{3}+1\)
c) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right)\div\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\div\frac{\sqrt{5}+\sqrt{2}}{5-2}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
Bài 2:
đk: \(x\ge1\)
Ta có: \(\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\frac{x+1}{16}}=5\)
\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x-1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow-3\sqrt{x-1}=5\)
\(\Rightarrow\sqrt{x-1}=-\frac{5}{3}\) (vô lý)
=> PT vô nghiệm
Câu 1:
\(A=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
Câu 2:
\(\Leftrightarrow\left|2x-3\right|=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=2\sqrt{3}\\2x-3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}+3}{2}\\x=\dfrac{-2\sqrt{3}+3}{2}\end{matrix}\right.\)