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Bài 3:
a: \(A=\frac{1}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{2\sqrt{x}}{4-x}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-2-\sqrt{x}-2-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=-\frac{2}{\sqrt{x}-2}\)
b: Thay x=3 vào A, ta được: \(A=-\frac{2}{\sqrt3-2}=\frac{2}{2-\sqrt3}=2\left(2+\sqrt3\right)=4+2\sqrt3\)
Bài 2:
a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)
\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)
b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
Bài 1:
a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)
\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)
Bài 2:
a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)
\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)
b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
Bài 1:
a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)
\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b) \(\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
a)\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x+1}\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\sqrt{\left(\sqrt{y}-1\right)^{2^2}}}{\sqrt{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
\(A=\frac{4xy}{y^2-x^2}:\left(\frac{1}{y^2+2xy+x^2}-\frac{x^3+y^3}{x^4-y^4}\right)\left(x\ne\pm y;y\ne0\right)\)
\(\Leftrightarrow A=\frac{4xy}{\left(y^2-x^2\right)\left(y^2+x^2\right)}:\left(\frac{1}{\left(y+x\right)^2}-\frac{x^3+y^3}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right)\)
