\(\sqrt{2a}-\sqrt{18^3}+4\sqrt{\dfrac{a}{2}}\)
\(\sqrt{\dfrac{a}{1+2b+b^2}}\cdot\sqrt{\dfrac{4a+8ab+4ab^2}{225}}\)
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LC
27 tháng 9 2018
a) ...= \(\dfrac{1}{4}\).\(6\sqrt{5}\) +\(2\sqrt{5}\) - \(3\sqrt{5}\) +5
= \(\dfrac{3}{2}\sqrt{5}\) -\(\sqrt{5}\) +5
=5 - \(\dfrac{1}{2}\sqrt{5}\)
d) ...= \(\sqrt{\dfrac{a}{\left(1+b\right)^2}}\) . \(\sqrt{\dfrac{4a\left(1+b\right)^2}{15^2}}\)
= \(\sqrt{\dfrac{4a^2\left(1+b\right)^2}{\left(1+b\right)^2.15^2}}\) = \(\sqrt{\dfrac{4a^2}{15^2}}\)= \(\dfrac{2a}{15}\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
9 tháng 6 2018
d) \(\dfrac{2}{5}\sqrt{50x}-\dfrac{3}{4}\sqrt{8x}\)
\(=\dfrac{2}{5}.5\sqrt{2x}-\dfrac{3}{4}\sqrt{8x}\)
\(=\dfrac{2\sqrt{2x}}{1}-\dfrac{3\sqrt{2x}}{2}\)
\(=\dfrac{4\sqrt{2x}-3\sqrt{2x}}{2}\)
\(=\dfrac{\sqrt{2x}}{2}\)
8 tháng 6 2018
c) \(3y^2.\sqrt{\dfrac{x^4}{9y^2}}=\sqrt{\dfrac{9y^4x^4}{9y^2}}=\dfrac{\sqrt{9y^2x^4}}{\sqrt{1}}=\sqrt{\left(3yx^2\right)^2}=3yx^2\)
T
5 tháng 8 2017
Bài này đưa về giải hệ phương trình
\(\left\{{}\begin{matrix}a-b+4ab=1\\a^2+b^2=2\end{matrix}\right.\) với \(a,b\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b+4ab=1\left(1\right)\\\left(a-b\right)^2+2ab=2\left(2\right)\end{matrix}\right.\)
Từ pt (1) suy ra \(a-b=1-4ab\Rightarrow\left(a-b\right)^2=1+16a^2b^2-8ab\)
Do đó
\(\left(2\right)\Rightarrow1+16a^2b^2-8ab+2ab=2\)
\(\Leftrightarrow16a^2b^2-6ab-1=0\)
Xem đây là pt bậc 2 với ab tìm được \(\left[{}\begin{matrix}ab=\dfrac{1}{2}\\ab=-\dfrac{1}{8}\end{matrix}\right.\)
- TH1: \(ab=\dfrac{1}{2}\Rightarrow a-b=-1\)
Có \(\left\{{}\begin{matrix}a-b=-1\\ab=\dfrac{1}{2}\end{matrix}\right.\) tìm được \(\left\{{}\begin{matrix}a=\dfrac{-1+\sqrt{3}}{2}\\b=\dfrac{1+\sqrt{3}}{2}\end{matrix}\right.\) (thỏa mãn a,b>0)
Từ đó tìm x
Tương tự cho TH còn lại
10 tháng 7 2022
a: \(=ab+2\cdot\sqrt{\dfrac{b}{a}\cdot ab}-\sqrt{ab\cdot\left(\dfrac{a}{b}+\dfrac{1}{\sqrt{ab}}\right)}\)
\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)
\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)
b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)
\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)
\(=a^3nb-a^3b^3+a^4\)
2 tháng 7 2021
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
ai gúp m vs
\(\sqrt{2a}-\sqrt{18^3}+4\sqrt{\dfrac{a}{2}}=\sqrt{2}.\sqrt{a}-54\sqrt{2}+2\sqrt{2}.\sqrt{a}=3\sqrt{2}.\sqrt{a}-54\sqrt{2}\)
\(\sqrt{\dfrac{a}{1+2b+b^2}}.\sqrt{\dfrac{4a+8ab+4ab^2}{225}}=\sqrt{\dfrac{a}{\left(b+1\right)^2}}.\sqrt{\dfrac{4a\left(1+2b+b^2\right)}{225}}=\dfrac{\sqrt{a}}{\left|b+1\right|}.\dfrac{\sqrt{4a\left(b+1\right)^2}}{15}=\dfrac{\sqrt{a}}{\left|b+1\right|}.\dfrac{2\sqrt{a}\left|b+1\right|}{15}=\dfrac{2a}{15}\)