Bài 1: Rút gọn: \(A=\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
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12 tháng 9 2023
a: ĐKXĐ: x>=0; x<>4
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\cdot\dfrac{\sqrt{x}-2+2}{2}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)
TH1: M>=căn M
=>M^2>=M
=>M^2-M>=0
=>5*căn x-1>=0
=>x>=1/25 và x<>4
TH2: M<căn M
=>5căn x-1<0
=>x<1/25
Kết hợp ĐKXĐ, ta được: 0<=x<1/25
26 tháng 9 2021
a) \(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)
c) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{3-2\sqrt{2}}-1}{\sqrt{3-2\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}\)
26 tháng 9 2021
\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=1-\dfrac{1}{\sqrt{a}}< 1\\ c,a=3-2\sqrt{2}\Leftrightarrow\sqrt{a}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}=\dfrac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\)
TN
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
2 tháng 11 2019
Đặt biểu thức trên là A
\(A^2=m+2\sqrt{m-1}+m-2\sqrt{m-1}-2\sqrt{\left(m+2\sqrt{m-1}\right)\left(m-2\sqrt{m-1}\right)}\)
\(A^2=2m-2\sqrt{m^2-4\left(m-1\right)}=2m-2\sqrt{m^2-4m+4}\)
\(A^2=2m-2\sqrt{\left(m-2\right)^2}=2m-2\left(m-2\right)=2m-2m+4=4\)
\(\Rightarrow A=\pm2\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 4 2021
\(=\dfrac{\left(\sqrt{m}+2\right)\left(\sqrt{m}-2\right)}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}-\dfrac{5}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}-\dfrac{\sqrt{m}+3}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}\)
\(=\dfrac{m-4-5-\sqrt{m}-3}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}=\dfrac{m-\sqrt{m}-12}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}\)
\(=\dfrac{\left(\sqrt{m}-4\right)\left(\sqrt{m}+3\right)}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}=\dfrac{\sqrt{m}-4}{\sqrt{m}-2}\)
NU
28 tháng 8 2021
\(1,ĐKx\ge5\)
\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)
\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)
\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)
2a,ĐK \(x\ge0;x\ne9\)
,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
3 tháng 10 2021
\(a,M=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\\ M=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,M< \dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}< 0\\ \Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}< 0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}< 0\\ \Leftrightarrow\sqrt{a}-4< 0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a< 16\\ \Leftrightarrow0< a< 16\left(kết.hợp.ĐKXĐ\right)\)
\(A=\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
\(=\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\)
\(=\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)
\(=\sqrt{m-1}+1+\left|\sqrt{m-1}-1\right|\)