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\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
a, Với \(x\ge0;x\ne1\)
\(Q=\left(\frac{x-1}{\sqrt{x}-1}-\frac{x\sqrt{x}-1}{x-1}\right):\left(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
a) đk : \(x\ge0\) ; \(x\ne1\)
A=\(\left(\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}-\frac{x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) \(=\frac{1-\sqrt{x}}{x+1}\)
b) đk : \(x\ne0;x\ne1\)
B=\(\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\left(\frac{-2\sqrt{x}}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\frac{-4x}{\left(x-1\right)^3}\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
1: Ta có: \(Q=\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-2\sqrt{x}+x\sqrt{x}-x\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x-1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\cdot\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
2: Ta có: \(\frac{1}{Q}=4\sqrt{x}-4\)
\(\Leftrightarrow Q=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(4\sqrt{x}-4\right)\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1=4x\sqrt{x}-4\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1-4x\sqrt{x}+4=0\)
\(\Leftrightarrow x-3x\sqrt{x}-\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-\left(3x\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(x\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left[\sqrt{x}-3\left(x+\sqrt{x}+1\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-3x-3\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(-3x-2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)=0\)(vì \(-3x-2\sqrt{x}-3\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}=1\)
hay x=1(không thỏa mãn ĐKXĐ)
Vậy: Không có giá trị nào của x thỏa mãn \(\frac{1}{Q}=4\sqrt{x}-4\)
Lời giải:
a)
\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)
b)
\(B=\left(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x}=\frac{x-1}{x}\)
