Tính hợp lý:
\(\sqrt{12}+\sqrt{27}-\sqrt{3}\)
Giúp mk nha
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25 tháng 9 2021
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
HN
21 tháng 1 2017
\(\sqrt{12}+\sqrt{27}-\sqrt{3}\)
\(=\sqrt{4.3}+\sqrt{9.3}-\sqrt{3}\)
\(=2\sqrt{3}+3\sqrt{3}-\sqrt{3}=4\sqrt{3}\)
31 tháng 10 2021
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
15 tháng 7 2019
\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}.\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)\)
\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{5\sqrt{6}-12}{18\sqrt{2}}\)
\(E=\frac{36\sqrt{2}}{18\sqrt{6}}+\frac{12\sqrt{3}}{18\sqrt{6}}+\frac{\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{3}}\)
\(E=\frac{36\sqrt{2}+12\sqrt{3}+\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{6}}\)
\(E=\frac{51\sqrt{2}}{18\sqrt{6}}\)
\(E=\frac{17\sqrt{2}}{6\sqrt{6}}\)
\(E=\frac{17\sqrt{2}}{2.3\sqrt{2}.\sqrt{3}}\)
\(E=\frac{17}{\sqrt{2}.3\sqrt{2}.\sqrt{3}}\)
\(E=\frac{17}{6\sqrt{3}}\)
\(E=\frac{17\sqrt{3}}{18}\)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
BN
1
11 tháng 7 2017
\(\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{\left(9\sqrt{5}+9\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
chúc bn hc tốt
A
3
\(\sqrt{12}+\sqrt{27}-\sqrt{3}=\sqrt{4.3}+\sqrt{9.3}-\sqrt{3}\)
\(=\sqrt{4}.\sqrt{3}+\sqrt{9}.\sqrt{3}-\sqrt{3}=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\)
\(=\left(2+3-1\right)\sqrt{3}=4\sqrt{3}\)