Thực hiện phép tính:
a. \(\sqrt{4\left(11+6\sqrt{2}\right)}-\sqrt{9\left(11-6\sqrt{2}\right)}\)
b.\(\sqrt{600+200\sqrt{5}}+\sqrt{54-18\sqrt{5}}\)
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26 tháng 8 2021
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
26 tháng 9 2021
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
26 tháng 9 2021
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
TT
8 tháng 7 2023
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
3 tháng 8 2023
a: =(2căn 3-8căn 3)(căn 3-1)
=-6căn 3*(căn 3-1)
=-18+6căn 3
b: \(=\dfrac{6-2\sqrt{5}}{\sqrt{5}-3}-\sqrt{5}+2\)
=-2-căn 5+2=-căn 5
c: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)
13 tháng 7 2016
@.@ Trời ơi, nhiều thế ^^
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)\)
\(=\left(\sqrt{2}.\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)=2\sqrt{5}-2-6+\frac{6}{\sqrt{5}}=\frac{16\sqrt{5}}{5}-8\)
b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}=\frac{75\sqrt{2}+50\sqrt{2}-45\sqrt{2}}{\sqrt{10}}=\frac{80\sqrt{2}}{\sqrt{10}}=\frac{80}{\sqrt{5}}=16\sqrt{5}\)c) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)
\(=2+\sqrt{2}+2-\sqrt{2}=4\)
d) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)}^2\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
e) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=1+\sqrt{2}-\sqrt{2}+1=2\)g) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
HP
3 tháng 9 2021
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
6 tháng 6 2021
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
24 tháng 11 2021
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
bạn chỉ cần để ý tách hằng là oke
chỉ cho này \(11+6\sqrt{2}=3^3+2.3.\sqrt{2}+\sqrt{2}^2=\left(3+\sqrt{2}\right)^2\)
và cái đằng sau nữa cũng tương tự \(11-6\sqrt{2}=\left(3-\sqrt{2}\right)^2\)
biểu thức \(< =>\sqrt{4}.\left(3+\sqrt{2}\right)^2-\sqrt{9}.\left(3-\sqrt{2}\right)^2\)ok ?
câu b tự làm đi
bạn bỏ cái mũ 2 của dòng thứ 2 từ dưới lên , mình đánh lộn