\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
Những câu hỏi liên quan
Chứng minh các đẳng thức sau: a) left(1-a^2right):left[left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)left(frac{1
+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)right]+1frac{2}{1-a}b) left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right):left(frac{a}{sqrt{ab}+b}
+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)sqrt{b}-sqrt{a}c) frac{sqrt{a}+sqrt{b}-1}{a
+sqrt{ab}}+frac{sqrt{a}-sqrt{b}}{2sqrt{ab}}left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a
+sqrt{ab}}right)frac{sqrt{a}}{a}d) left(frac{asqrt{a}+bsqrt{b}}{sqrt{a}...
Đọc tiếp
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Chứng minh các đẳng thức sau:
a) left(1-a^2right):left(left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right).left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)right)+1frac{2}{1-a}
b) left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right):left(frac{a}{sqrt{ab}+b}+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)sqrt{b}-sqrt{a}
c) frac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}}+frac{sqrt{a}-sqrt{b}}{2sqrt{ab}}.left(frac{sqrt{b}}{a-sqrt{ab}}+frac{sqrt{b}}{a+sqrt{ab}}right)frac{sqrt{a}}{a}
Đọc tiếp
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
Pleft(frac{sqrt{a}-sqrt{b}}{sqrt{a}+sqrt{b}}-frac{sqrt{a}+sqrt{b}}{sqrt{a}-sqrt{b}}right)sqrt{frac{1}{a}-frac{1}{b}}left(frac{left(sqrt{a}-sqrt{b}right)^2}{a-b}-frac{left(sqrt{a}+sqrt{b}right)^2}{a-b}right).sqrt{frac{b-a}{ab}}frac{a-2sqrt{ab}+b-a-2sqrt{ab}-b}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{a-b}.sqrt{frac{b-a}{ab}}frac{-4sqrt{ab}}{2017-2018}.sqrt{frac{2018-2017}{ab}}4sqrt{ab}.sqrt{frac{1}{ab}}sqrt{frac{16ab}{ab}}4
Đọc tiếp
\(P=\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)\sqrt{\frac{1}{a}-\frac{1}{b}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}-\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{a-b}\right).\sqrt{\frac{b-a}{ab}}\)
\(=\frac{a-2\sqrt{ab}+b-a-2\sqrt{ab}-b}{a-b}.\sqrt{\frac{b-a}{ab}}\)
\(=\frac{-4\sqrt{ab}}{a-b}.\sqrt{\frac{b-a}{ab}}\)\(=\frac{-4\sqrt{ab}}{2017-2018}.\sqrt{\frac{2018-2017}{ab}}\)
\(=4\sqrt{ab}.\sqrt{\frac{1}{ab}}\)\(=\sqrt{\frac{16ab}{ab}}\)\(=4\)
sao tổng lại lớn hơn hiệu
Thu gọn biểu thức
a, A = \(\frac{2\sqrt{3-\sqrt{3+\sqrt{3+\sqrt{48}}}}}{\sqrt{6}-2}\)
b, B = \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
Chứng minh :
\(B=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
C/Minh đẳng thức:
a) left(frac{sqrt{a}+2}{a+2sqrt{a}+1}-frac{sqrt{a}-2}{a-1}right).frac{sqrt{a}+1}{sqrt{a}}frac{2}{a-1} (với a0, b0, a≠b)
b)frac{2}{sqrt{ab}}:left(frac{1}{sqrt{a}}-frac{1}{sqrt{b}}right)^2-frac{a+b}{left(sqrt{a}-sqrt{b}right)^2}-1 (với a0, b0,a≠b)
c) frac{2sqrt{a}+3sqrt{b}}{sqrt{ab}+2sqrt{a}-3sqrt{b}-6}-frac{6-sqrt{ab}}{sqrt{ab}+2sqrt{a}+3sqrt{b}+6}frac{a+9}{a-9} (với a≥0, b≥0,a≠9)
Đọc tiếp
C/Minh đẳng thức:
a) \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{a-1}\) (với a>0, b>0, a≠b)
b)\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) (với a>0, b>0,a≠b)
c) \(\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}=\frac{a+9}{a-9}\) (với a≥0, b≥0,a≠9)
Đề bài: Rút gọn biểu thức:
1. frac{sqrt{a^2+x^2}+sqrt{a^2-x^2}}{sqrt{a^2+x^2}-sqrt{a^2-x^2}}-sqrt{^{ }frac{a^4}{x^4}-1}
2. left(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right) . left(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)
3. left(frac{3}{sqrt{1+x}}sqrt{1-x}right) : left(frac{3}{sqrt{1-x^2}}+1right)
4. left(sqrt{a}+frac{b-sqrt{ab}}{sqrt{a}+sqrt{b}}right) : left(frac{a}{sqrt{ab+b}}+frac{b}{sqrt{ab}-a}-frac{a+b}{sqrt{ab}}right)
5. frac{sqrt{a}+sqrt{b}-1}{a+sqrt{ab}} + frac{sqrt{a}-sqrt{b}}{2...
Đọc tiếp
Đề bài: Rút gọn biểu thức:
1. \(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{^{ }\frac{a^4}{x^4}-1}\)
2. \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\) . \(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
3. \(\left(\frac{3}{\sqrt{1+x}}\sqrt{1-x}\right)\) : \(\left(\frac{3}{\sqrt{1-x^2}}+1\right)\)
4. \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\) : \(\left(\frac{a}{\sqrt{ab+b}}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)
5. \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) .\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
Các bạn giúp tớ nhé, hứa sẽ tick, tớ cảm ơn!!!!
1.
Đặt \(\sqrt{a^2+x^2}=m,\sqrt{a^2-x^2}=n\Rightarrow x^2=\frac{m^2-n^2}{2}\)
\(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}-1}=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{(a^2+x^2)(a^2-x^2)}{x^4}}\)
\(=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\frac{\sqrt{(a^2+x^2)(a^2-x^2)}}{x^2}\)
\(=\frac{m+n}{m-n}-\frac{mn}{\frac{m^2-n^2}{2}}=\frac{(m+n)^2}{m^2-n^2}-\frac{2mn}{m^2-n^2}=\frac{m^2+n^2}{m^2-n^2}\)
\(=\frac{2a^2}{2x^2}=\frac{a^2}{x^2}\)
2.
\(=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)
\(=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})\)
\(=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)=(\sqrt{a}+1)^2(\sqrt{a}-1)^2\)
\(=(a-1)^2\)
3.
\(=\frac{3(1-x)}{\sqrt{1+x}.\sqrt{1-x}}:\frac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{3(1-x)}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{3+\sqrt{1-x^2}}=\frac{3(1-x)}{3+\sqrt{1-x^2}}\)
Đúng 0
Bình luận (0)
4. Bạn xem lại đề xem đã đúng chưa?
5.
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{\sqrt{b}(a+\sqrt{ab})+\sqrt{b}(a-\sqrt{ab})}{(a-\sqrt{ab})(a+\sqrt{ab})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{2a\sqrt{b}}{a^2-ab}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}.\frac{1}{a-b}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{1}{a+\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}=\frac{1}{\sqrt{a}}\)
Đúng 0
Bình luận (2)
Rút gọn:
\((\frac{2}{\sqrt{a}-\sqrt{b}}-\frac{2\sqrt{a}}{a\sqrt{a}+b\sqrt{b}}.\frac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}):4\sqrt{ab}\)
Lời giải:
ĐKXĐ:..................
\(\left(\frac{2}{\sqrt{a}-\sqrt{b}}-\frac{2\sqrt{a}}{a\sqrt{a}+b\sqrt{b}}.\frac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\right):4\sqrt{ab}=\left(\frac{2(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}-\frac{2\sqrt{a}}{(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)}.\frac{a-\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\right):4\sqrt{ab}\)
\(=\left(\frac{2\sqrt{a}+2\sqrt{b}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}-\frac{2\sqrt{a}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\right).\frac{1}{4\sqrt{ab}}=\frac{2\sqrt{b}}{a-b}.\frac{1}{4\sqrt{ab}}=\frac{1}{2\sqrt{a}(a-b)}\)
Đúng 0
Bình luận (0)
tth, Trần Thanh Phương, Nguyễn Văn Đạt, Nguyễn Việt Lâm, Nguyễn Huy Thắng, Akai Haruma, Nguyễn Thị Ngọc Thơ, Nguyễn Thanh Hằng, Lê Thị Thục Hiền, Sakura, Nguyễn Huy Tú, Akai Haruma, Nguyễn Huy Thắng, Ribi Nkok Ngok, Mysterious Person, soyeon_Tiểubàng giải, Võ Đông Anh Tuấn, Phương An, Trần Việt Linh,...
Đúng 0
Bình luận (0)
Các bn xem bài này mk làm đúng khônga)left(frac{asqrt{a}+bsqrt{b}}{sqrt{a}+sqrt{b}}-sqrt{ab}right)left(frac{sqrt{a}+sqrt{b}}{a-b}right)^21VTleft(frac{asqrt{a}+bsqrt{b}-sqrt{ab}left(sqrt{a}+sqrt{b}right)}{sqrt{a}+sqrt{b}}right)left(frac{sqrt{a}+sqrt{b}}{a-b}right)^2 left(frac{asqrt{a}+bsqrt{b}-asqrt{b}-bsqrt{b}}{sqrt{a}+sqrt{b}}right)left(frac{sqrt{a}+sqrt{b}}{a-b}right)^2 left(frac{left(asqrt{a}-asqrt{b}right)+left(asqrt{b}-bsqrt{b}right)}{sqrt{a}+sqrt{b}}right)left(frac{sqrt{a}+sqrt{b}}{a...
Đọc tiếp
Các bn xem bài này mk làm đúng không
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
VT=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(a\sqrt{a}-a\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a+\sqrt{b}}}\right)\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
= \(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\frac{a-b}{a-b}=1\Rightarrow\left(=VP\right)\)
b)\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\)
VT=\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\sqrt{a}+\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
=\(a+\sqrt{ab}-\sqrt{ab}-b=a-b\Rightarrow\left(=VP\right)\)
Đây là đề chứng minh hả !
Phần a , b đúng r
Nhưng phần b chỗ \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\) = a - b
Dùng hằng đẳng thức thức 3 như vậy sẽ hay hơn !
Chúc bạn học tốt!
Đúng 0
Bình luận (0)
Thu gọn:
\(A=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
10 tikk