1.Thực hiện phép tính:
a.\(45-\left[\left(18-15\right)^2+2^4\right]\)
b. \(A=50+\left\{30-2\left[14-48:4^2\right]\right\}\)
c. \(65.59+65.42-13.5\)
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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: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
HQ
Hà Quang Minh
Giáo viên
10 tháng 10 2023
a) \(\left( { - 3} \right).\left( { - 2} \right).\left( { - 5} \right).4\)\( = \left[ {\left( { - 3} \right).\left( { - 2} \right)} \right].\left( { - 5} \right).4\)\( = 6.\left( { - 5} \right).4 = - 30.4 = - 120\).
b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).
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
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
19 tháng 10 2021
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
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: \(B=\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left[\sqrt{3}+1-3\left(2+\sqrt{3}\right)+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right]\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{5}{2}\left(3+\sqrt{3}\right)\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left(-5-2\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)
\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}=\dfrac{1}{2}\)
14 tháng 5 2022
a: \(=\left(x+2\right)^2\cdot\dfrac{2x-1}{3\left(x+2\right)}=\dfrac{\left(x+2\right)\left(2x-1\right)}{3}\)
b: \(=\dfrac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x\left(x-1\right)}=\dfrac{-2\left(x+1\right)}{\left(x-1\right)\left(x+2\right)}\)
\(=45-\left(3^2+2^4\right)=45-\left(9+16\right)=45-25=20\)
b)\(=50+\left(30-2\left(14-3\right)\right)=50+\left(30-22\right)=50+8=58\)