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a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )
a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
em làm luôn
\(P=\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{x-1}\)
b) thì em chưa làm đc :((
b, \(x=24-16\sqrt{2}=24-2.8.\sqrt{2}=24-8\sqrt{8}\)
\(=24-2.4\sqrt{8}=4^2-2.4\sqrt{8}+\left(\sqrt{8}\right)^2=\left(4-\sqrt{8}\right)^2\)
*, làm tiếp bước Q làm : \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(4-\sqrt{8}\right)^2}=\left|4-\sqrt{8}\right|=4-\sqrt{8}\)( vì \(4-\sqrt{8}>0\))
hay \(\frac{1}{4-\sqrt{8}-1}=\frac{1}{3-\sqrt{8}}=3+\sqrt{8}\)
Vậy với \(x=24-16\sqrt{2}\)thì \(P=3+\sqrt{8}\)
Đề có vấn đề theo tôi đề như sau :
\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)
Rheo tôi đề như vậy
mong xem lại đề
a) \(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-3+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{23\sqrt{x}+3x-20}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
a) \(ĐKXĐ:x\ne-\sqrt{3}\)
\(=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
b) \(=\dfrac{1-\sqrt{a^3}}{1-\sqrt{a}}=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
tại sao câu a không tìm điều kiện ?