Ta có: \(\dfrac{312}{x}-\dfrac{312}{x+4}=\dfrac{1}{2}\)
\(\Leftrightarrow x\left(x+4\right)=624x+2496-624x\)
\(\Leftrightarrow x^2+4x-2496=0\)
\(\Leftrightarrow\left(x+2\right)^2=2500\)
\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-52\end{matrix}\right.\)
Rút gọn biểu thức sau :
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\) ( x ≥ 0 ; x ≠ 16 )
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(B=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}}\)
\(A=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
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 : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
\(\dfrac{4}{\left(2+\dfrac{2}{1+\dfrac{4}{5}}\right)x-\left(1+\dfrac{4}{2+\dfrac{1}{1+\dfrac{7}{8}}}\right)}+\dfrac{1}{\left(2+\dfrac{1}{3+\dfrac{1}{4}}\right)}=4+\dfrac{2}{1+\dfrac{8}{9}}\)
Tìm x biết: (viết kết quả dưới dạng hổn số)
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 : \(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
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
\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
RÚT GONJ
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
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
M = \(1:(\dfrac{1}{2+\sqrt{x}}+\dfrac{\dfrac{3x}{2}}{4-x}-\dfrac{2}{4-2\sqrt{x}}):\dfrac{1}{4-2\sqrt{x}}\)
a) Rút gọn M
b) Tìm giá trị của x để M = 20
