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\(\left(0,125\right)^3\cdot512\)
\(=\left(0,125\right)^3\cdot2^9\)
\(=\left(0,125\right)^3\cdot8^3\)
\(=\left(0,125\cdot8\right)^3\)
\(=1^8=1\)
\(\left(0,25\right)^4\cdot1024\)
\(=\left(0,25\right)^4\cdot2^{10}\)
\(=\left(0,625\right)^2\cdot32^2\)
\(=\left(0,625\cdot32\right)^2\)
\(=20^2=400\)
\(\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)
\(\left(0,25\right)^4.1024=\left(0,25\right)^4.2^{10}=\left(0,25\right)^4.2^8.2^2\)
= \(\left(0,25\right)^4.4^44\)
= \(\left(0,25.4\right)^4.4\)
= \(1^4.4\)
= 1.4=4
a) (0.25)^3*32
= (0.5)^5*0.5*2^5
=1^5*0.5
=1*0.5
=0.5
b)(-0.125)^3*80^4
=(-0.125)^3*80^3*80
=(-0.125*80)^3*80
=(-10)^3
=-1000*80
=-80000
c) 8^2*4^5/2^20
=(23)2*(22)5*2^20
=2^6*2^10*2^20
=2^36
d)81^11*3^17/27^10*9^15
=((34)11*3^17)/(33)10*(32)15
=(3^44*3^17)/(3^30*3^30)
=3^61/3^60
=3
Bài b) dòng thứ 3 từ dưới đếm lên, phải là, (-10)^3*80 nha, gấp quá mình ghi nhầm ;)
a) \(\sqrt{25}-\sqrt{\dfrac{49}{4}}+\sqrt{0,25}\)
\(=5-3,5+0,5\)
\(=2\)
b) \(\left(-\dfrac{25}{24}-\dfrac{31}{42}\right)-\left(\dfrac{-7}{27}-\dfrac{3}{42}\right)\)
\(=\dfrac{-25}{24}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}\)
\(=\left(\dfrac{-25}{24}+\dfrac{7}{27}\right)+\left(-\dfrac{31}{42}+\dfrac{3}{42}\right)\)
\(=-\dfrac{169}{216}+-\dfrac{2}{3}=-1\dfrac{97}{216}\)
c) \(\dfrac{103}{10}-\left(9,5-0,25.18\right):0,5\)
\(=10,3-\left(9,5-0,25.18\right):0,5\)
\(=10,3-5:0,5\)
\(=0,3\)
d) \(\left(10^3+15.10^2+5^3\right):7\)
\(=\left(10^3+10^2\right)\left(15+5^3\right):7\)
\(=\left(1100.140\right):7\)
\(=22000\)
a) (1 / 7.7) 7 = 1
b) (0,125.8)3=1
c)( 0,25.0,25.32)2=22=4
d) (90/15) 3 = 6 3
) ( 790/79)4=104
f) (3/0,75)2=82
\(a,\frac{1}{3}x+0.25=\frac{5}{7}\)
\(\Leftrightarrow\frac{1}{3}x=\frac{13}{28}\)
\(\Leftrightarrow x=\frac{39}{28}\)
vậy...
\(b,\frac{11}{12}x+0,25=\frac{5}{6}\)
\(\Leftrightarrow\frac{11}{12}x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{7}{11}\)
vậy.....
\(c,\left(\frac{-1}{3}\right)^2+\frac{2}{3}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{9}+\frac{2}{3}x=\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{5}{36}\)
\(\Leftrightarrow x=\frac{5}{24}\)
vậy......
\(d,\left(3x+2\right)^3=-\frac{8}{125}\)
\(\Leftrightarrow3x+2=-\frac{2}{5}\)
\(\Leftrightarrow3x=-\frac{12}{5}\)
\(\Leftrightarrow x=-\frac{4}{5}\)
vậy.......
\(\frac{1}{3x}+0,25=\frac{5}{7}\)
\(\frac{1}{3x}+\frac{1}{4}=\frac{5}{7}\)
\(\frac{1}{3x}=\frac{13}{28}\)
\(3x=\frac{28}{13}\)
\(x=\frac{28}{39}\)
\(\frac{11}{12x}+0,25=\frac{5}{6}\)
\(\frac{11}{12x}+\frac{1}{4}=\frac{5}{6}\)
\(\frac{11}{12x}=\frac{7}{12}\)
\(x=\frac{11}{12}:\frac{7}{12}\)
\(x=\frac{7}{11}\)
\(\left(-\frac{1}{3}\right)^2+\frac{2}{3x}=\frac{1}{4}\)
\(\frac{1}{9}+\frac{2}{3x}=\frac{1}{4}\)
\(\frac{2}{3x}=\frac{5}{36}\)
\(x=\frac{2}{3}:\frac{5}{36}\)
\(x=\frac{5}{24}\)
\(\left(3x+2\right)^3=\left(-\frac{8}{125}\right)\)
\(\left(3x+2\right)^3=\left(-\frac{2}{5}\right)^3\)
\(\Rightarrow3x+2=-\frac{2}{3}\)
\(3x=-\frac{8}{3}\)
\(x=-\frac{9}{8}\)
a,=\(\frac{1}{2}\)
b,-80000
c,\(\frac{1}{16}\)
d,\(1,271734748x10^{29}\)
Ta có :
\(\left(0,25\right)^3.4^3=\left(0,25\times4\right)^3=1^3=1\)