\(\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)
\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\)
\(\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\) RÚT GỌN
\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\) (ĐK: \(x\ne4;x>0\))
\(=\left[\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\left[\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{-2\sqrt{x}+\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left[\dfrac{-\left(\sqrt{x}+2\right)^2+\left(\sqrt{x}-2\right)^2-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{-x-4\sqrt{x}-4+x+4\sqrt{x}+4-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{-4x}\)
\(=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)}{-4x}\)
\(=-\dfrac{3\sqrt{x}+6-x-2\sqrt{x}}{4x}\)
\(=-\dfrac{\sqrt{x}-x+6}{4x}\)
\(\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\) (ĐK: \(x\ge0;x\ne1;x\ne\dfrac{1}{9}\))
\(=\left[\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\right]\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
A=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) ( ĐKXĐ x>0;x≠4)
P=\(\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{8\sqrt{x}}{x-4}\right):\left(2-\dfrac{2\sqrt{x}+3}{\sqrt{x}+2}\right)\)
E=\(\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\) (ĐKXĐ x≥0;x≠4)
RÚT GỌN GIÚP MÌNH VỚI A
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
Ch.minh
a) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) = \(\dfrac{3\sqrt{x}-1}{x-1}\)
b) (\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)-\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{1}{x-4}\) = -4\(\sqrt{x}\)
a.
Với \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) có:
\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{x-1}\\ =\dfrac{x+\sqrt{x}}{x-1}-\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}\\ =\dfrac{x+\sqrt{x}}{x-1}-\dfrac{x-2\sqrt{x}+1}{x-1}\\ =\dfrac{x+\sqrt{x}-x+2\sqrt{x}-1}{x-1}\\ =\dfrac{3\sqrt{x}-1}{x-1}=VP\)
b.
Với \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\) có:
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{1}{x-4}\right)\\ =(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}).\left(\dfrac{x-4}{1}\right)\\ =(\dfrac{x-2\sqrt{x}}{x-4}-\dfrac{x+2\sqrt{x}}{x-4}).\left(x-4\right)\\ =\left(\dfrac{x-2\sqrt{x}-x-2\sqrt{x}}{x-4}\right)\left(x-4\right)\\ =\dfrac{-4\sqrt{x}\left(x-4\right)}{x-4}\\ =-4\sqrt{x}=VP\)
Tìm x
1) \(\sqrt{\dfrac{3x-1}{x+2}}=2\)
2)\(\sqrt{\dfrac{5x-7}{2x- 1}}=2\)
3)\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
4) \(\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-21}{9-x}\dfrac{1}{\sqrt{x}+3}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}+12}{x-9}\)
\(N=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{6}{x-1}\)
\(M=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+22}{x-4}\)
d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)
\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
A=\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}A=\dfrac{\sqrt{x}+2.\left(\sqrt{x}+2\right)-3.\left(\sqrt{x}-2\right)+12}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}A=\dfrac{\sqrt{x}+2\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}A=\dfrac{22}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)
a : \(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
b : \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)với x ≥ 0 x ≠ 10
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)với x ≥ 0 x ≠ 9
d : \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)với x ≥ 0 x ≠ 9
a: ĐKXĐ: x>0
\(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
b: ĐKXĐ: x>=0; x<>16
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{x+16}{x+16}\cdot\dfrac{\sqrt{x}+2}{x-16}=\dfrac{\sqrt{x}+2}{x-16}\)
c: ĐKXĐ: x>=0; x<>25
\(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
\(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
d: \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{-3}{\sqrt{x}-3}\)
chứng minh p=\(\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)(x>0)không phụ thuộc vào x
\(P=\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt[4]{x}\left(\sqrt[4]{x}-1\right)}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{x}}+1\right)^2}}{1+\sqrt{x}}\)
\(=\left(-\sqrt[4]{x}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{1}{\sqrt{x}}+1}{1+\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}}}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}=0\)
\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
RÚT GONJ
2)
ĐK: \(x\ge0;x\ne4\)
Biểu thức trở thành:
\(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+2\sqrt{a}+3\sqrt{a}+6}{a-4}-\dfrac{a-2\sqrt{a}-\sqrt{a}+2}{a-4}-\dfrac{4\sqrt{a}-4}{a-4}\\ =\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{a-4}\\ =\dfrac{4\sqrt{a}+8}{a-4}\\ =\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\\ =\dfrac{4}{\sqrt{a}-2}\)
1:
\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+2\sqrt{x}-8-x-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)
\(=\dfrac{-2\sqrt{x}-11}{-4}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(2\sqrt{x}+11\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-3\right)}\)
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}
\)
\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)
\(\left(1-\dfrac{4}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
TÌM ĐKXĐ VÀ RÚT GỌN
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=-\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)
\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\) (ĐK: \(x\ge0\))
\(=\dfrac{\left(\sqrt{x}\right)^3+1^3}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=x-\sqrt{x}+1\)
______________
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: \(x\ge0;x\ne9\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)