Rút gọn:
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(B=\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2015}+\sqrt{2017}}\)
Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
4) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
5) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
6) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp mình với, cần gấp ạ
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
4) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
5) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp vs ạ
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
3: Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{x-1}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
4.
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)-\left(\dfrac{1}{x+\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
a. Rút gọn A.
b. Tính x khi \(A=\dfrac{1}{2}\)
5. CMR
\(\left(\dfrac{\sqrt{30}}{\sqrt{3}}-\dfrac{\sqrt{20}}{\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right).\sqrt{4+\sqrt{15}}=2\)
nhanh lên nha
Rút gọn
\(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
B1. Với \(x\ge0,x\ne4.Chobiểuthức\)
\(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{2-\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{1}{x\sqrt{x}+27}\)
a, tính giá trị biểu thức khi B= 1/4
b, Rút gọn A
c, Tìm giá trị của x để A>1/2
d, Với C= B : A. Tìm GTLN C
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
A= \(\dfrac{3}{\sqrt{7}-2}\) + \(\sqrt{\left(\sqrt{7}-3\right)}^2\)
B= \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right)\):\(\left(\dfrac{\sqrt{x}+1}{x-1}\right)\)
Rút gọn A,B
Rút gọn:
1) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
2)\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
3) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
4) \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
Mng giúp e vs ạ, cần gấp :<
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
Rút gọn các biểu thức sau:
a. A = \(\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
b. B = \(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\) (x > 0 ; x ≠ 1)
\(a,A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2^2-\sqrt{3}^2}\)
\(=\dfrac{4}{1}=4\)
Vậy \(A=4\)
\(b,B=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) với \(x>0,x\ne1\)