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\(B=\sqrt{x+\sqrt{x^2-1}}-\sqrt{x-\sqrt{x^2-1}}\)
\(B^2=x+\sqrt{x^2-1}+x-\sqrt{x^2-1}-2\sqrt{\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)}\)
\(B^2=2x-2\sqrt{x^2-x^2+1}\)
\(B^2=2x-2\)
\(\Rightarrow B=\sqrt{2x-2}\)
\(C=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-1}\left(ĐK:x\ge1\right)\)
\(C=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{x-1}\)
\(C=\sqrt{x-1}+1-\sqrt{x-1}=1\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
Câu 1: Sửa lạ đề chút nhé : 4x + 1 -> 4x -1
Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
=> \(\sqrt{2}.A\)= \(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)
= \(\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}-1\right)^2}\)
= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)
Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)
nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2
=> \(A=2:\sqrt{2}=\sqrt{2}\)
Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)
=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)
=> \(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
=> \(\sqrt{4+\sqrt{9-4\sqrt{2}}}=\sqrt{2}+1\)
=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)
=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)
\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)
\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\)
\(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\) \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)
\(A=\left(\dfrac{4x+4}{2\sqrt{2x^3}-8}-\dfrac{\sqrt{2x}}{2x+2\sqrt{2x}+4}\right)\)\(\left(\dfrac{1+2\sqrt{2x^3}}{1+\sqrt{2x}}\right)\)
\(=\left[\dfrac{4x+4-\sqrt{2x}\left(\sqrt{2x}-2\right)}{\left(\sqrt{2x}-2\right)\left(2x+2\sqrt{2x}+4\right)}\right]\)\(.\dfrac{\left(1+\sqrt{2x}\right)\left(2x-2\sqrt{2x}+4\right)}{1+\sqrt{2x}}\)
Làm tiếp nhé :>>
Đặt \(D=\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\) (D >/ 0 với mọi 1/2 < x)
\(\Rightarrow D^2=2\sqrt{4x-1}-2\sqrt{4x^2-4x+1}=2\sqrt{4x-1}-2\left|2x-1\right|=2\sqrt{4x-1}-2\left(1-2x\right)=4x-2+2\sqrt{4x-1}\)
\(\Rightarrow D=\sqrt{D^2}=\sqrt{4x-2+2\sqrt{4x-1}}=\left|\sqrt{4x-1}+1\right|=\sqrt{4x-1}+1\)