Thực hiện phép tính:
a) \(\left( { - 3} \right).\left( { - 2} \right)\left( { - 5} \right).4\)
b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\).
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a) \(26+173+74+27\)
\(=\left(26+74\right)+\left(173+27\right)\)
\(=100+200\)
\(=300\)
b) \(75\cdot37+89\cdot46+75\cdot52-89\cdot21\)
\(=75\cdot\left(37+52\right)+89\cdot\left(46-21\right)\)
\(=75\cdot89+89\cdot25\)
\(=89\cdot\left(75+25\right)\)
\(=89\cdot100\)
\(=8900\)
c) \(2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\)
\(=2^{7-2}+5^{4-3}\cdot2^4-3\cdot2^5\)
\(=2^5+5\cdot2^4-3\cdot2^5\)
\(=2^4\cdot\left(2+5-3\cdot2\right)\)
\(=2^4\cdot\left(7-6\right)\)
\(=2^4\)
\(=16\)
d) \(100:\left\{250:\left[450-\left(4\cdot5^3-2^2\cdot25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4\cdot5^3-4\cdot5^2\right)\right]\right\}\)
\(=100:\left[250:\left(450-4\cdot5^2\cdot4\right)\right]\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
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
a) \(\left( { + 4} \right).\left( { - 8} \right)\) là tích của hai số nguyên khác dấu nên mang dấu âm. Vậy \(\left( { + 4} \right).\left( { - 8} \right) < 0\)
b) \(\left( { - 3} \right).4\) là tích của hai số nguyên khác dấu nên mang dấu âm. Vậy\(\left( { - 3} \right).4 < 4\)
c) \(\left( { - 5} \right).\left( { - 8} \right)\) là tích của hai số nguyên âm nên \(\left( { - 5} \right).\left( { - 8} \right) = 5.8\)
\(\left( { + 5} \right).\left( { + 8} \right)\) là tích của hai số nguyên dương nên \(\left( { + 5} \right).\left( { + 8} \right) = 5.8\)
Vậy \(\left( { - 5} \right).\left( { - 8} \right) = \left( { + 5} \right).\left( { + 8} \right)\).
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: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
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
a) \(6 - 8 = 6 + \left( { - 8} \right) = - \left( {8 - 6} \right) = - 2\)
b) \(3 - \left( { - 9} \right) = 3 + 9 = 12\)
c) \(\left( { - 5} \right) - 10 = \left( { - 5} \right) + \left( { - 10} \right)\)\( = - \left( {5 + 10} \right) = - 15\)
d) \(0 - 7 = 0 + \left( { - 7} \right) = - 7\)
e) \(4 - 0 = 4 + 0 = 4\) (vì số đối của 0 là 0)
g) \(\left( { - 2} \right) - \left( { - 10} \right) = \left( { - 2} \right) + 10\)\( = 10 - 2 = 8\).
\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)
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) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).