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30 tháng 10 2020
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
M
11 tháng 8 2020
A= \(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{\sqrt{b}}{\sqrt{a}.\sqrt{a}-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-\sqrt{b}.\sqrt{b}}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{a}.\sqrt{a}.\sqrt{b}-\sqrt{b}.\sqrt{b}\sqrt{a}\right)\)
A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
A = b-a
B = \(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}}{a-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-1}\)
B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}\left(a+\sqrt{a}\right)}{a^2-a}\right).\frac{a-1}{\sqrt{a}+1}\)
B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}.\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
\(B=\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}-\frac{a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B= \(\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)-a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B= \(\left(\frac{\left(\sqrt{a}+1\right)\left(a\sqrt{a}-a\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B = \(\frac{\left(\sqrt{a}+1\right)a\left(\sqrt{a}-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(B=\frac{a\left(\sqrt{a}^2-1^2\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(B=\frac{a\left(a-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
B = \(\frac{a-1}{\sqrt{a}+1}\)
15 tháng 7 2020
Bài 1:
b) Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\frac{\sqrt{2\left(4+\sqrt{7}\right)}}{\sqrt{2}}-\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}-\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\frac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Bài 2:
a) Ta có: \(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)\)
\(=a-\sqrt{a}-a-\sqrt{a}\)
\(=-2\sqrt{a}\)
b) Ta có: \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{ab}-\sqrt{ab}=0\)
d) Ta có: \(\frac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\sqrt{a}+\sqrt{b}-\left(\sqrt{a}+\sqrt{b}\right)\)
=0
Bài 3:
a) ĐKXĐ: x≥0
Ta có: \(\frac{\sqrt{27x}}{\sqrt{3}}=6\)
\(\Leftrightarrow\frac{\sqrt{27}\cdot\sqrt{x}}{\sqrt{3}}=6\)
\(\Leftrightarrow3\cdot\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{3}=2\)
hay \(x=4\)(thỏa mãn)
Vậy: S={4}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-1\end{matrix}\right.\Leftrightarrow x\ge0\)
Ta có: \(\sqrt{x+1}=3-\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=\left(3-\sqrt{x}\right)^2\)
\(\Leftrightarrow x+1=9-6\sqrt{x}+x\)
\(\Leftrightarrow x+1-9+6\sqrt{x}-x=0\)
\(\Leftrightarrow-8+6\sqrt{x}=0\)
\(\Leftrightarrow6\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=\frac{8}{6}=\frac{4}{3}\)
hay \(x=\frac{16}{9}\)(thỏa mãn)
Vậy: \(S=\left\{\frac{16}{9}\right\}\)
19 tháng 8 2018
Bài 3 : Áp dụng BĐT Bu - nhi - a cốp xki ta có :
\(A=\sqrt{x-2}+\sqrt{4-x}\le\sqrt{\left(1^2+1^2\right)\left(x-2+4-x\right)}=\sqrt{2.2}=2\)
Vậy GTLN của A là 2 . Dấu \("="\) xảy ra khi \(x=3\)
\(B=\sqrt{6-x}+\sqrt{x+2}\le\sqrt{\left(1^2+1^2\right)\left(6-x+x+2\right)}=\sqrt{2.8}=4\)
Vậy GTLN của B là 4 . Dấu \("="\) xảy ra khi \(x=2\)
\(C=\sqrt{x}+\sqrt{2-x}\le\sqrt{\left(1^2+1^2\right)\left(x+2-x\right)}=\sqrt{2.2}=2\)
Vậy GTLN của C là 2 . Dấu \("="\) xảy ra khi \(x=1\)
ND
19 tháng 8 2018
Bài 2:
a .\(\dfrac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
\("="\Leftrightarrow a=b\)
b. \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\Leftrightarrow a+b< \left(\sqrt{a}+\sqrt{b}\right)^2\Leftrightarrow a+b< a+b+2\sqrt{ab}\left(a,b>0\right)\)
\(c.a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\) ( t nghĩ là > thôi )
d. \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
\(\Leftrightarrow2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ca}+a\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)
\("="\Leftrightarrow a=b=c\)
e. \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
\(\Leftrightarrow\dfrac{a+b}{2}-\dfrac{a+b+2\sqrt{ab}}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a+2b-a-b-2\sqrt{ab}}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{4}\ge0\) ( đúng)
\("="\Leftrightarrow a=b\)
13 tháng 12 2022
a: \(=2\sqrt{2}+30\sqrt{2}-3\sqrt{2}+6\sqrt{2}=26\sqrt{2}\)
b: \(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}+\sqrt{3}+\dfrac{5}{2}\sqrt{3}=-\dfrac{9}{2}\sqrt{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2019
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
1 tháng 6 2021
a)
\(\left(\sqrt{a+\sqrt{b}}\ne\sqrt{a-\sqrt{b}}\right)^2\)
\(=a+\sqrt{b}\ne2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}+a-\sqrt{b}\)
\(=2a\ne2\sqrt{a^2-b}=2\left(a\ne\sqrt{a^2}-b\right)\)
\(\Rightarrow\sqrt{a+\sqrt{b}}\ne\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\ne\sqrt{a^2}-b\right)}\)
\(\Rightarrowđpcm\)
1 tháng 6 2021
b)
\(\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}\ne}\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2\)
\(=\frac{a+\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a+\sqrt{a^2-b}}{2}.\frac{a-\sqrt{a^2-b}}{2}}+\frac{a-\sqrt{a^2-b}}{2}\)
\(=\frac{a}{2}+\frac{\sqrt{a^2-b}}{2}\ne\sqrt[2]{\frac{a^2-a^2+b}{2.2}}+\frac{a}{2}-\frac{\sqrt{a^2-b}}{2}\)
\(=a\ne2\frac{\sqrt{b}}{2}=a\ne\sqrt{b}\)
\(\Rightarrow\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\ne\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\sqrt{a\ne\sqrt{b}}\)
\(\Rightarrowđpcm\)
phân tích thành n. tử
\(a\sqrt{b}-b\sqrt{a}\)
\(=\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)\)
Có sai sót gì xin bỏ qua~!