A= \(\dfrac{2\sqrt{a}}{\sqrt{a}+3}\)+\(\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)+\(\dfrac{3+7\sqrt{a}}{9-a}\)
\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
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:
\(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}+1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+2\sqrt{x}-8-x-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)
\(=\dfrac{-2\sqrt{x}-11}{-4}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(2\sqrt{x}+11\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-3\right)}\)
Rút gọn : \(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3+7\sqrt{a}}{9-a}\) với a≥0,a≠9
\(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3+7\sqrt{a}}{9-a}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{3+7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{2a-6\sqrt{a}+a+4\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{3a-9\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{3\sqrt{a}}{\sqrt{a}+3}\)
A= \(\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)\) ĐK:(a≥0, a≠9)
B= \(\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)\) ĐK:(a≥0, a≠9)
C= \(\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)\) ĐK:(a>0, a≠1)
D= \(\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}\) ĐK:(a≥0, a≠1)
E= \(\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}\) ĐK:(a≥0, a≠4)
Giúp mìnk với nha !!!
\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)
\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
\(E=\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}=\dfrac{a+\sqrt{a}+2+\sqrt{a}-2}{a-4}=\dfrac{a+2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-2}\)
28. A=\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
a. rút gọn A
b. tính A với x = \(7-4\sqrt{3}\)
c. tìm x khi A=3
a:
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
b: Khi x=7-4căn 3 thì
\(A=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)
c: A=3
=>căn x-2=1
=>x=9(loại)
\(a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)\)
\(=\dfrac{x-3\sqrt{x}-x+9}{x-9}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{9-x+x-9-x+4\sqrt{x}-4}\)
\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-2}{4\sqrt{x}-4-x}\)
\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(x-4\sqrt{x}+4\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}\)
\(c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn đề bài.
Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1
Rút gọn các biểu thức
a)\(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\)
b)\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
Help me !!!
\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-3\left(\sqrt{a}-3\right)-a+2}{a-9}\)
\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\)
\(=\dfrac{11}{a-9}\)
\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(\text{đ}k\text{x}\text{đ}:a\ge0;a\ne9\right)\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a-3}\right)\left(\sqrt{a+3}\right)}-\dfrac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\dfrac{a-2}{\left(\sqrt{a}+3\right)\left(\sqrt{a-3}\right)}\\ =\dfrac{a+3\sqrt{a}-\left(3\sqrt{a}-9\right)-\left(a-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{11}{\left(\sqrt{a}-3\right)\left(\sqrt{a+3}\right)}\)
\(b,\dfrac{a+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\ =\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}\\ =\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Rút gọn biểu thức:
\(\dfrac{\sqrt{a-2}+2}{3}\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
Đk:\(a>2\)
\(\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
Đặt \(b=\sqrt{a-2}\Leftrightarrow a=b^2+2\)
Biểu thức \(\Leftrightarrow\dfrac{b+2}{3}\left(\dfrac{b}{3+b}+\dfrac{b^2+2+7}{11-b^2-2}\right):\left(\dfrac{3b+1}{b^2-3b}-\dfrac{1}{b}\right)\)
\(=\dfrac{b+2}{3}\left[\dfrac{b}{3+b}-\dfrac{b^2+9}{b^2-9}\right]:\left[\dfrac{3b+1}{b\left(b-3\right)}-\dfrac{b-3}{b\left(b-3\right)}\right]\)
\(=\dfrac{b+2}{3}.\dfrac{b\left(b-3\right)-b^2-9}{\left(b-3\right)\left(3+b\right)}:\dfrac{3b+1-\left(b-3\right)}{b\left(b-3\right)}\)
\(=\dfrac{b+2}{3}.\dfrac{-3\left(b+3\right)}{\left(b-3\right)\left(3+b\right)}.\dfrac{b\left(b-3\right)}{2\left(b+2\right)}\)
\(=-\dfrac{b}{2}\)
\(=\dfrac{\sqrt{a-2}}{-2}\)
a) \(\sqrt{4x^2-9}=2\sqrt{x+3}\)
b) \(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c) \(\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4\)
d)\(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
1.Chứng minh:\(\dfrac{a+\sqrt{2+\sqrt{5}.}\sqrt{\sqrt{9-4\sqrt{5}}}}{3\sqrt{2-\sqrt{5}}.\sqrt[3]{\sqrt{9+4\sqrt{5}-}3\sqrt{a^2}+\sqrt[3]{a}}}\)=\(-\sqrt[3]{a}-1\)
2.Rút gọn: \(\left(\dfrac{a^3\sqrt[]{a}-2a^3\sqrt{b}+\sqrt[3]{a^2}-\sqrt[3]{b}}{\sqrt[3]{a^2-\sqrt[3]{ab}}}+\dfrac{\sqrt[3]{a^2b}-\sqrt[3]{ab^2}}{\sqrt[3]{a}-\sqrt[3]{b}}\right)1\dfrac{1}{\sqrt[3]{a^2}}\)