tính : \(\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1+\sqrt{a}}\right)\left(1-\frac{1}{1-\sqrt{a}}\right)\) với \(a=\frac{1}{9}\)
Những câu hỏi liên quan
Rút gọn1.Aleft(frac{2sqrt{x}}{sqrt{x}+3}+frac{sqrt{x}}{sqrt{x}-3}-frac{3x+3}{x-9}right):left(frac{2sqrt{x}-2}{sqrt{x}-3}-1right)2.Bleft(frac{sqrt{a}+1}{sqrt{ab}+1}+frac{sqrt{ab}+sqrt{a}}{sqrt{ab}-1}-1right):left(frac{sqrt{a}+1}{sqrt{ab}+1}-frac{sqrt{ab}+sqrt{a}}{sqrt{ab}-1}+1right)3.Cleft(frac{2x-1+sqrt{x}}{1-x}+frac{2xsqrt{x}+x-sqrt{x}}{1+xsqrt{x}}right).left(frac{left(x-sqrt{x}right)left(1-sqrt{x}right)}{2sqrt{x}-1}right)
Đọc tiếp
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
Tính tổng sau:Afrac{1}{left[sqrt[3]{2}right]}+frac{1}{left[sqrt[3]{3}right]}+frac{1}{left[sqrt[3]{4}right]}+frac{1}{left[sqrt[3]{5}right]}+frac{1}{left[sqrt[3]{6}right]}+frac{1}{left[sqrt[3]{7}right]}+frac{1}{left[sqrt[3]{9}right]}+...+frac{1}{left[sqrt[3]{2012^3-1}right]}(trong tổng trên không có các số dạng frac{1}{left[sqrt[3]{n}right]} với n là lập phương 1 số nguyên,ví dụ:1 và 8)
Đọc tiếp
Tính tổng sau:
\(A=\frac{1}{\left[\sqrt[3]{2}\right]}+\frac{1}{\left[\sqrt[3]{3}\right]}+\frac{1}{\left[\sqrt[3]{4}\right]}+\frac{1}{\left[\sqrt[3]{5}\right]}+\frac{1}{\left[\sqrt[3]{6}\right]}+\frac{1}{\left[\sqrt[3]{7}\right]}+\frac{1}{\left[\sqrt[3]{9}\right]}+...+\frac{1}{\left[\sqrt[3]{2012^3-1}\right]}\)
(trong tổng trên không có các số dạng \(\frac{1}{\left[\sqrt[3]{n}\right]}\) với n là lập phương 1 số nguyên,ví dụ:1 và 8)
Ta có từ n3 + 1 đến (n + 1)3 - 1 có
(n + 1)3 - 1 - n3 - 1 + 1 = 3n2 + 3n số có phần nguyên bằng n
Áp dụng vào cái ban đầu ta có
\(=\frac{3.1^2+3.1}{1}+\frac{3.2^2+3.2}{2}+...+\frac{3.2011^2+3.2011}{2011}\)
= 3.1 + 3 + 3.2 + 3 + ...+ 3.2011 + 3
= 3.2011 + 3(1 + 2 +...+ 2011)
= 6075231
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
left(frac{x-3sqrt{x}}{x-9}-1right):left(frac{9-x}{x+sqrt{x}-6}-frac{sqrt{x}-3}{sqrt{x}-2}-frac{sqrt{x}-2}{sqrt{x}+3}right)left(frac{sqrt{a}+1}{sqrt{a}-1}-frac{sqrt{a}-1}{sqrt{a}+1}+4sqrt{a}right)left(sqrt{a}-frac{1}{sqrt{a}}right)left(frac{2sqrt{x}+x}{xsqrt{x}-1}-frac{1}{sqrt{x}+1}right):left(1-frac{sqrt{x}+2}{x+sqrt{x}+1}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}+frac{1}{2-sqrt{x}}
Đọc tiếp
\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}+\frac{1}{2-\sqrt{x}}\)
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
Đúng 0
Bình luận (0)
d) ĐK: x>=0; x khác 4.
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}.\)
\(D=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(D=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Rút gọn các biểu thức
\(A=\left(1+\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\frac{a+\sqrt{a}}{a-1}\frac{\sqrt{a}}{a-\sqrt{a}}\right)\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(C=\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{a-1}\right):\frac{a}{2+2\sqrt{a}}\)
tính \(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
a) rút gọn A
b) tính A với \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\)\(\left[\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right]\left[\frac{a+1}{\sqrt{a}}\right]\)
\(A=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\) \(\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\left(a+1\right)}{a-1}\)
ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(a=2\left(16-15\right)\)
\(a=2\)
khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)
vậy.....
Đúng 0
Bình luận (0)
\(\left(\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}\right):\left(1+\sqrt{\frac{a+1}{a-1}}\right)\)
\(\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a^3}+3\right)\left(a-b\right)}\right)\)
Rút gọn 2 biểu thức trên?
Ai giúp mình với, tks nhiều
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Đúng 0
Bình luận (0)
Qleft(frac{2}{2+2sqrt{a}}+frac{1}{2-2sqrt{a}}-frac{a^2+1}{1-a^2}right)left(1+frac{1}{a}right)left(frac{1}{2left(1+sqrt{a}right)}+frac{1}{2left(1-sqrt{a}right)}-frac{a^2+1}{left(1-sqrt{a}right)left(1+sqrt{a}right)left(1+aright)}right)left(frac{a+1}{a}right)left(frac{left(1-sqrt{a}right)left(1+aright)+left(1+sqrt{a}right)left(1+aright)-2left(a^2+1right)}{2left(1-aright)left(1+aright)}right)left(frac{a+1}{a}right)left(frac{1+a-sqrt{a}-asqrt{a}+1+a+sqrt{a}+asqrt{a}-2a^2-2}{2left(1-aright)left(1+arig...
Đọc tiếp
\(Q=\left(\frac{2}{2+2\sqrt{a}}+\frac{1}{2-2\sqrt{a}}-\frac{a^2+1}{1-a^2}\right)\left(1+\frac{1}{a}\right)\)
\(=\left(\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+1}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a\right)}\right)\left(\frac{a+1}{a}\right)\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+a\right)+\left(1+\sqrt{a}\right)\left(1+a\right)-2\left(a^2+1\right)}{2\left(1-a\right)\left(1+a\right)}\right)\left(\frac{a+1}{a}\right)\)
\(=\left(\frac{1+a-\sqrt{a}-a\sqrt{a}+1+a+\sqrt{a}+a\sqrt{a}-2a^2-2}{2\left(1-a\right)\left(1+a\right)}\right)\left(\frac{a+1}{a}\right)\)
\(=\left(\frac{2a-2a^2}{2\left(1-a\right)\left(1+a\right)}\right)\)
\(=\frac{a}{a}\)= 1
1. Chứng minh sqrt[3]{3+sqrt[3]{3}}+sqrt[3]{3-sqrt[3]{3}} 2sqrt[3]{3}2. a) Tính Afrac{2b.sqrt{x^2-1}}{x-sqrt{x^2-1}} với xfrac{1}{2}left(sqrt{frac{a}{b}}+sqrt{frac{b}{a}}right)left(a,b0right)
b) Tính Bfrac{xy-sqrt{x^2-1}.sqrt{y^2-1}}{xy+sqrt{x^2-1}.sqrt{y^2-1}} với xfrac{1}{2}left(a+frac{1}{a}right);yfrac{1}{2}left(b+frac{1}{b}right)left(a,bge1right)3. Cho x,y thỏa mãn xyge0. Tính Bleft(left|sqrt{xy}+frac{x}{2}+frac{y}{2}right|-left|xright|right)+left(left|sqrt{xy}-frac{x}{2}-frac{y}{2}right|...
Đọc tiếp
1. Chứng minh \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
2. a) Tính \(A=\frac{2b.\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(a,b>0\right) \)
b) Tính \(B=\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right);y=\frac{1}{2}\left(b+\frac{1}{b}\right)\left(a,b\ge1\right)\)
3. Cho x,y thỏa mãn \(xy\ge0\). Tính \(B=\left(\left|\sqrt{xy}+\frac{x}{2}+\frac{y}{2}\right|-\left|x\right|\right)+\left(\left|\sqrt{xy}-\frac{x}{2}-\frac{y}{2}\right|-\left|y\right|\right)\)
4. Cho \(\frac{2x+2\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(x+1\right)^2}=\frac{A}{\sqrt{x}-2}+\frac{B\sqrt{x}+C}{x+1}+\frac{D\sqrt{x}+E}{\left(x+1\right)^2}\). Tìm các số A,B,C,D,E để đẳng thức trên là đúng với mọi x
V=\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\)
W= \(\left(\frac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\frac{1-3\sqrt{a}+a}{a\sqrt{a}-1}-\frac{1}{\sqrt{a}-1}\right):\frac{a+1}{1-\sqrt{a}}\)
ĐKXĐ:...
\(V=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1-2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\right)=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(W=\left(\frac{\sqrt{a}-1}{a+\sqrt{a}+1}-\frac{a-3\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{1}{\sqrt{a}-1}\right).\left(\frac{1-\sqrt{a}}{a+1}\right)\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2-a+3\sqrt{a}-1-\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{1-\sqrt{a}}{a+1}\right)\)
\(=\left(\frac{-\left(a+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\left(\frac{-\left(\sqrt{a}-1\right)}{a+1}\right)=\frac{1}{a+\sqrt{a}+1}\)
Đúng 0
Bình luận (0)