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a) Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\cdot\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\frac{1}{1+\sqrt{a}}\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{1+\sqrt{a}}\)
\(=1+\sqrt{a}\) Bằng 1 kiểu gì đây._.?
a) Xin lỗi sửa lại phần a:
Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=1\)
b) Ta có: \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{6}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)=3-2=1\)
Đề của bạn bị sai, mình sửa lại đề ở dưới nhé!
\(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}+a+1\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(a+1\right)\left(a+\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(a+1\right)^2-\left(\sqrt{a}\right)^2}\)
\(=\frac{\left(a^2-\sqrt{a}\right)\left(a-\sqrt{a}+1\right)-\left(a^2+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{\left(a-1\right)^2-a}\)
\(=\frac{a^3-a^2\sqrt{x}+a^2-a\sqrt{a}+a-\sqrt{a}-a^3-a^2\sqrt{a}-a^2-a\sqrt{a}-a-\sqrt{a}+\left(a+1\right)\left[\left(a+1\right)^2-a\right]}{a^2+2a+1-a}\)
\(=\frac{-2a^2\sqrt{a}-2a\sqrt{a}-2\sqrt{a}+\left(a+1\right)\left(a^2+2a+1-a\right)}{a^2+a+1}\)
\(=\frac{-2\sqrt{a}\left(a^2+a+1\right)+\left(a+1\right)\left(a^2+a+1\right)}{a^2+a+1}\)
\(=\frac{\left(a^2+a+1\right)\left[-2\sqrt{x}+\left(x+1\right)\right]}{a^2+a+1}\)
\(=x-1-2\sqrt{x}\)
\(=\left(\sqrt{x}-1\right)^2\)
(Chúc you học giỏi nhoa!)
Ui ui mình ghi lộn, xin lỗi nhoa, vì x dễ làm hơn nên mình ghi lộn a thành x, mong bạn thông cảm hihi!
\(VT=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+a+1\)
\(=a-\sqrt{a}-a-\sqrt{a}+a+1\)
\(=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2=VP\)
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: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
cau c í mk thấy bn chép sai đề nên mk sửa lại đề rồi bạn xem lại đề rồi so với bài làm của mk nha có j ko hiểu thì ib mk nha
\(a)VT = \dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{a + \sqrt a }}{{\sqrt a }}\\ = \dfrac{{a + 2\sqrt a + 1 - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a }}\\ = \dfrac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)}} + \sqrt a + 1\\ = \dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a - 1}} + \sqrt a + 1\\ = \sqrt a - 1 + \sqrt a + 1\\ = 2\sqrt a = VP (đpcm) \)
\(b)VT = \dfrac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\\ = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - \left( {x - 2\sqrt {xy} + y} \right)\\ = x - \sqrt {xy} + y - x + 2\sqrt {xy} - y\\ = \sqrt {xy} (đpcm)\\ c)VT = \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\ = \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \sqrt a - \sqrt b .\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}\\ = \dfrac{{a - b}}{{a - b}} = 1 (đpcm)\\ d)VT = \left[ {\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2} + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}} \right]:\sqrt b \\ = \dfrac{{a - 2\sqrt {ab} + b + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}:\sqrt b \\ = \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \left( {\sqrt a - \sqrt b } \right):\sqrt b \\ = \sqrt a + \sqrt b - \sqrt a + \sqrt b :\sqrt b \\ = \dfrac{{2\sqrt b }}{{\sqrt b }} = 2 (đpcm) \)
Câu c đề sai (đã sửa)
\(M=\left(\frac{a-2\sqrt{a}+1}{a+1}\right):\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)-\left(a+1\right)}\right]\)
\(M=\left[\frac{\left(\sqrt{a}-1\right)^2}{a+1}\right]:\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(M=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\left[\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(M=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(M=\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(\sqrt{a}-1\right)\left(a+1\right)}{\left(\sqrt{a}-1\right)^2}=\sqrt{a}+1\)
\(M>1\Leftrightarrow\sqrt{a}-1>1\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>4\)
\(M=\sqrt{3-2\sqrt{2}}-1\)
\(M=\sqrt{\left(\sqrt{2}-1\right)^2}-1=\sqrt{2}-1-1=\sqrt{2}-2\)
\(VT=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)+a+1\)
\(=a-\sqrt{a}-a-\sqrt{a}+a+1\)
\(=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2=VP\)