(25^2×x-1)-(54^x)=15000
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a, 568 + 568 x 135 - 568 x 36
= 568 x ( 1 + 135 - 36)
= 568 x 100
56 800
b, 17 x 49 + 49 x 37 + 54 x 26 + 54 x 25
= 49 x ( 17 + 37) + 54 x ( 26 + 25 )
= 49 x 54 + 54 x 51
= 54 x ( 49 + 51)
= 54 x 100
= 5 400
c, 14 x 35 x 5 + 10 x 25 x 7 + 20 x 70
= ( 2 x 35) x ( 7 x 5) + ( 10 x 7) x 25 + 20 x 70
= 70 x 35 + 70 x 25 + 20 x 70
= 70 x ( 35 + 25 + 20)
= 70 x 80
= 5 600
a) 568 + 568 x 135 - 568 x 36
= 568 x ( 1+1+135-1-36 )
=568 x 100
=5680
b) 17 x 49 + 49 x 37 + 54 x 26 + 54 x 25
= 49 x ( 17 + 37 ) + 54 x ( 26 + 25 )
= 49 x 54 + 54 x 51
= 54 x ( 49 + 51 )
= 54 x 100
= 5400
c) 14 x 35 x 5 + 10 x 25 x 7 + 20 x 70
= 70 x 35 + 10 x 25 x 7+ 20 x 70
= 70 x 35 + 250 x 7 + 20 x 70
= 70 x 35 + 70 x 25 + 20 x 70
= 70 x (35+25+20)
= 70 x 80
= 5600
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\(2,=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\\ 3,=\left(3x-5\right)\left(x+1\right)\\ 4,sai.đề\\ 5,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ 6,=\left(x+3\right)\left(x+5\right)\)
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Bài giải
a, \(81+243+19=\left(81+19\right)+243=100+243=343\)
b, \(2\text{ x }25\text{ x }2\text{ x }16\text{ x }4=25\text{ x }4\text{ x }2\text{ x }2\text{ x }16=25\text{ x }8\text{ x }32=200\text{ x }32=6400\)
c, \(32\text{ x }47+32\text{ x }54-32=32\text{ x }\left(47+54-1\right)=32\text{ x }100=3200\)
a,81+19+243
=100+243
b,(2x25)x2x(16x4)
=50x2x64
=100x64
=6400
c,32x47+32x54-32x1
=32x(47+54-1)
=32x100
=3200
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\(\dfrac{x}{3}=\dfrac{7}{25}-\dfrac{1}{5}\Leftrightarrow\dfrac{25x}{75}=\dfrac{21}{75}-\dfrac{15}{75}\Leftrightarrow25x=21-15\Rightarrow25x=6\Rightarrow x=0,24\)
a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)
hay \(x=\dfrac{6}{25}\)
Vậy: \(x=\dfrac{6}{25}\)
b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)
hay \(x=\dfrac{5}{99}\)
Vậy: \(x=\dfrac{5}{99}\)
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a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}