thực hiện phép tính:
A=\(\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)
B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\)
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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
24 tháng 8 2021
a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)
\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)
c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)
24 tháng 11 2021
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
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: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)
28 tháng 6 2021
a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
12 tháng 7 2021
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\sqrt{3}+2+\sqrt{2}+1-\sqrt{2}-\sqrt{3}\)
=3
b) Ta có: \(B=\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left[\sqrt{3}+1-3\left(2+\sqrt{3}\right)+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right]\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{5}{2}\left(3+\sqrt{3}\right)\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left(-5-2\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}=\dfrac{1}{2}\)
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...
8 tháng 6 2021
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)
\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)
\(=-5\sqrt{2}+6\sqrt{2}=\sqrt{2}\)
b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)
\(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4}{\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}+1-4\sqrt{3}\)
\(=3-4\sqrt{3}\)
\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)