P = \(1-(\sqrt{45}-\sqrt{20}-\sqrt{3})(\sqrt{20}-\sqrt{45}-\sqrt{3})\)
K
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17 tháng 6 2019
\(1-\left(\sqrt{45}-\sqrt{20}-\sqrt{3}\right)\left(\sqrt{20}-\sqrt{45}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{45}-\sqrt{20}-\sqrt{3}\right)\left(\sqrt{20}-\sqrt{45}-\sqrt{3}\right)\)
\(=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)\)
= 1 + 3
= 2
17 tháng 6 2019
\(1-\left(\sqrt{45}-\sqrt{20}-\sqrt{3}\right)\cdot\left(\sqrt{20}-\sqrt{45}-\sqrt{30}\right)\)
= \(1-\left(\sqrt{3}+\left(\sqrt{20}-\sqrt{45}\right)\right)\cdot\left(\sqrt{3}-\left(\sqrt{20}-\sqrt{45}\right)\right)\)
=\(1-\left(\sqrt{3}^2-\left(\sqrt{20}-\sqrt{45}\right)^2\right)\)
=\(1-\left(3-\left(20-2\cdot\sqrt{20}\cdot\sqrt{45}+45\right)\right)\)
=\(1-\left(3-\left(65-2\cdot\sqrt{900}\right)\right)\)
=\(1-\left(3-\left(65-2.30\right)\right)\)
=\(1-\left(3-5\right)\)
=3
27 tháng 7 2017
\(2\sqrt{20}-\sqrt{45}+3\sqrt{18}+3\sqrt{32}-\sqrt{50}\\ =4\sqrt{5}-3\sqrt{5}+9\sqrt{2}+12\sqrt{2}-5\sqrt{2}\\ =\sqrt{5}+16\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{8}+\sqrt{72}\)
\(=2\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=-\sqrt{5}+12\sqrt{2}\)
d) \(\dfrac{3}{\sqrt{3}+1}\)
\(=\dfrac{3\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\dfrac{3\left(\sqrt{3}-1\right)}{2}\)
\(=\dfrac{3\sqrt{3}-3}{2}\)
e) \(\dfrac{2}{\sqrt{10}-\sqrt{7}}\)
\(=\dfrac{2\left(\sqrt{10}+\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}\)
\(=\dfrac{2\left(\sqrt{10}+\sqrt{7}\right)}{3}\)
\(=\dfrac{2\sqrt{10}+2\sqrt{7}}{3}\)
HN
3
23 tháng 6 2021
\(\dfrac{1}{3}\sqrt{45}-\sqrt{20}+\sqrt{9+4\sqrt{5}}\)
= \(\dfrac{1}{3}.3.\sqrt{5}-2\sqrt{5}+\sqrt{\left(2+\sqrt{5}\right)^2}\)
= \(\sqrt{5}-2\sqrt{5}+2+\sqrt{5}=2\)
AT
23 tháng 6 2021
\(\dfrac{1}{3}\sqrt{45}-\sqrt{20}+\sqrt{9+4\sqrt{5}}=\dfrac{1}{3}\sqrt{9.5}-\sqrt{4.5}+\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(\dfrac{1}{3}.3\sqrt{5}-2\sqrt{5}+\sqrt{\left(2+\sqrt{5}\right)^2}=\sqrt{5}-2\sqrt{5}+\left|2+\sqrt{5}\right|\)
\(=\sqrt{5}-2\sqrt{5}+2+\sqrt{5}=2\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)
\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)
\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)
\(=\left(10-4+6\right)\sqrt{2}\)
\(=12\sqrt{2}\)
b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)
\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)
\(=\left(8-15+15-3\right)\sqrt{5}\)
\(=5\sqrt{5}\)
c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)
\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)
\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)
AH
Akai Haruma
Giáo viên
15 tháng 10 2023
Lời giải:
a. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}=4+3.\sqrt{\frac{1}{9}}.\sqrt{x-5}$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
b. Sửa đoạn 4x-45 thành 4x-20.
ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{4}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\frac{1}{3}\sqrt{x-5}-\frac{2}{3}\sqrt{x-5}=4$
$\Leftrightarrow \frac{5}{3}\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=\frac{12}{5}$
$\Leftrightarrow x-5=\frac{144}{25}=5,76$
$\Leftrightarrow x=10,76$ (tm)
10 tháng 7 2021
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
10 tháng 7 2021
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
10 tháng 7 2021
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
10 tháng 7 2021
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
S
13 tháng 6 2019
\(A=2\sqrt{5}-\sqrt{45}+2\sqrt{20}=2\sqrt{5}-\sqrt{3^2.5}+2\sqrt{2^2.5}=2\sqrt{5}-3\sqrt{5}+4\sqrt{5}=3\sqrt{5}\)
\(B=\left(\sqrt{18}-\frac{1}{2}\cdot\sqrt{32}+12\sqrt{2}\right):\sqrt{2}=\left(3\sqrt{2}-\frac{1}{2}\cdot4\sqrt{2}+12\sqrt{2}\right):\sqrt{2}\)
\(=13\sqrt{2}:\sqrt{2}=13\)
\(C=\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\cdot\sqrt{3}=\left(2\sqrt{3}+6\sqrt{3}-3\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
\(D=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
P = \(1\)- (\(3\sqrt{5}\) -\(2\)\(\sqrt{5}\) -\(\sqrt{3}\) )(\(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\))
= 1- (\(\sqrt{5}-\sqrt{3}\) )(\(-\sqrt{5}-\sqrt{3}\) )
= 1+ ( \(\sqrt{5}-\sqrt{3}\) )( \(\sqrt{5}+\sqrt{3}\) )
= 1+ (5-3)
= 3