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29 tháng 9 2018

((3\(^2\)))\(^2\) - ((-5\(^2\)))\(^2\) + ((-2\(^3\)))\(^2\)

= 81 - 625 + 64

= -544+ 64

= -480

2\(^4\) + 8[(-2)\(^2\) :\(\dfrac{1}{2}\)]\(^0\) - 2\(^{-2}\). 4 + (-2)\(^2\)

= 16+ 8.1 - \(\dfrac{1}{4}\). 4 + 4

= 16+ 8- 1+4

= 27

2\(^4\) + 3(\(\dfrac{1}{2}\))\(^0\) + 2\(^{-2}\).8 + [(-2)\(^3\). \(\dfrac{1}{2^4}\)].2 - \(\dfrac{1}{2}\)

= 16 + 3.1 +\(\dfrac{1}{4}\).8 + [(-8).\(\dfrac{1}{16}\)].2 -\(\dfrac{1}{2}\)

= 16 + 3+ 2 + \(\dfrac{-1}{2}\).2- \(\dfrac{1}{2}\)

= 21 + (-1)- \(\dfrac{1}{2}\)

= 20-\(\dfrac{1}{2}\) = \(\dfrac{40}{2}\) - \(\dfrac{1}{2}\)= \(\dfrac{39}{2}\)

\(\dfrac{15^{10}.5^{10}}{75^{10}}\) + \(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}\)

= \(\dfrac{\left(15.5\right)^{10}}{75^{10}}\) + \(\dfrac{\left(0,4.2\right)^5}{\left(0.4\right)^6}\)

= \(\dfrac{75^{10}}{75^{10}}\) + \(\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)

= 1 + \(\dfrac{2^5}{0,4}\) = 1+ 80 = 81

\(\dfrac{2^{13}.9^4}{6^3.8^3}\)

= \(\dfrac{2^{13}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}\) = \(\dfrac{2^{13}.3^8}{2^3.3^3.2^9}\)

= \(\dfrac{2^4.3^5}{2^3}\) = 2.3\(^5\) = 486

2 tháng 2 2021

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

2 tháng 2 2021

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

24 tháng 8 2021

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

 

24 tháng 8 2021

\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

 

14 tháng 7 2021

mấy câu còn lại nữa kìa bn

28 tháng 5 2022

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

3 tháng 8 2023

a) \(...=\dfrac{19}{8}:\dfrac{15}{4}x\dfrac{8}{3}=\dfrac{19}{8}x\dfrac{4}{15}x\dfrac{8}{3}=\dfrac{76}{45}\)

b) \(...=\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{17}{6}=\dfrac{3}{2}x\dfrac{3}{7}x\dfrac{6}{17}=\dfrac{27}{119}\)

c) \(...=\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}=\dfrac{14}{3}-\dfrac{7}{4}x\dfrac{5}{12}=\dfrac{14}{3}-\dfrac{35}{48}=\dfrac{14x16}{48}-\dfrac{35}{48}=\dfrac{224}{48}-\dfrac{35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

3 tháng 8 2023

\(a,2\dfrac{3}{8}:3\dfrac{3}{4}\times2\dfrac{2}{3}\\ =\dfrac{2\times8+3}{8}:\dfrac{3\times4+3}{4}\times\dfrac{2\times3+2}{3}\\ =\dfrac{19}{8}:\dfrac{15}{4}\times\dfrac{8}{3}\\ =\dfrac{19\times4\times8}{8\times15\times3}=\dfrac{76}{45}\)

\(b,1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\\ =\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{2\times6+5}{6}\\ =\dfrac{3}{2}\times\dfrac{3}{7}\times\dfrac{6}{17}\\ =\dfrac{54}{238}=\dfrac{27}{119}\)

\(c,4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\\ =\dfrac{4\times3+2}{3}-\dfrac{1\times4+3}{4}:\dfrac{2\times5+2}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}.\dfrac{5}{12}\\ =\dfrac{14}{3}-\dfrac{35}{48}\\ =\dfrac{14\times16-35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

27 tháng 10 2018

a) = 4. 5/4 + 25. [ 3/2 : (5/4)2] : 27/8

= 5 + 25. 12/5: 27/8

=5 +160/9

=205/9

b) = 8+ 3- 1+2.8

=11-1+2.8

=10+2.8

=10+ 16

= 26

c)= 3+1+1/4:2

= 4+ 0,125

=4,125

3 tháng 9 2021

a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)

b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)

c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)

d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)

b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)

\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)

hay x=1

7 tháng 9 2023

Bài 1:

a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)

b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{10}\)

c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)

\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)

\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)

\(=1\)

Bài 2:

a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)

b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)

c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)

d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)

Bài 3:

a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)

\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)

\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)

\(=9.\dfrac{2}{3}.-1\)

\(=6.-1=-6\)

b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)

\(#Wendy.Dang\)

7 tháng 9 2023

Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))

8 tháng 11 2018

câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)

\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)

\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)

\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)

\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)

\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)

\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)

\(B=1^{2010}-1^{2009}=1-1=0\)

8 tháng 11 2018

câu 2

a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)

\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)

\(\Leftrightarrow2x=\dfrac{31}{12}\)

\(\Leftrightarrow x=\dfrac{31}{24}\)

b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)