Tính:
a) \(0,\left(3\right)+3\dfrac{1}{3}+0,4\left(2\right)\)
b) \(\left[0,\left(5\right).0,\left(2\right)\right]:\left(3\dfrac{1}{3}:\dfrac{33}{25}\right)\)
Tính:
a) (-0,4)2 - (-0,4)3 . (-3)
b) \(\left(1\dfrac{3}{4}\right)^3\) - \(\left(1\dfrac{3}{4}\right)^2\) + (-1,031)0
c) \(\left(\dfrac{2}{3}\right)^3\) - 4. \(\left(-1\dfrac{3}{4}\right)^2\) + \(\left(-\dfrac{2}{3}\right)^3\)
Lời giải:
a. $=0,16-(-0,064).(-3)=0,16-0,192=-0,032$
b. $=(1\frac{3}{4})^2(1\frac{3}{4}-1)+1=(1\frac{3}{4})^2.\frac{3}{4}+1$
$=\frac{147}{64}+1=\frac{211}{64}$
c.
$=(\frac{2}{3})^3-4(\frac{-7}{4})^2-(\frac{2}{3})^3$
$=-4(\frac{-7}{4})^2=\frac{-49}{4}$
Tính:
a) (-0,4)2 - (0,4)3 . (-3)
b) \(\left(1\dfrac{3}{4}\right)^3\) - \(\left(1\dfrac{3}{4}\right)^2\) + (-1,031)0
c) \(\left(\dfrac{2}{3}\right)^3\) - 4. \(\left(-1\dfrac{3}{4}\right)^2\) + \(\left(-\dfrac{2}{3}\right)^3\)
a) = 0,16 - 0,064 . (-3)
= 0,16 + 0,192
= 0,352
b) = (7/4)³ - (7/4)² + 1
= 343/64 - 49/16 + 1
= 147/64 + 1
= 211/64
c) = 8/27 - 4.(-7/4)² - 8/27
= -4.49/16
= -49/4
a)\(2\dfrac{3}{3}.4.\left(-0,4\right)+1\dfrac{3}{5}.1,75+\left(-7,2\right):\dfrac{9}{11}\)
b)\(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}.\left(-5\right)^0\)
a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)
\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)
\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)
\(=\dfrac{-54}{5}\)
b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)
\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)
\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)
\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)
Kết quả học tập kì I của lớp 6A xếp thành 3 loại:giỏi,khá,trung bình.Số học sinh giỏi chiếm \(\dfrac{1}{3}\) số học sinh của lớp,số học sinh khá chiếm 40% số học sinh cả lớp,số học sinh trung bình là 12 em.Tính số học sinh lớp 6A và tỉ số phần trăm của học sinh giỏi so với học sinh cả lớp.
a) \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
b) \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
c) \(x\left(x+3\right)^3-\dfrac{x}{4}\left(x+3\right)=0\)
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
Bài 1:
a,\(3^7\) : \(3^5\)- \(\left(\dfrac{5}{17}\right)^0\) b,\(\left(\dfrac{5}{2}\right)^{13}\) : \(\left(\dfrac{1}{2}+2\right)^3\) c, 8.\(\left(\dfrac{1}{4}\right)^3\) +\(\left(\dfrac{2}{27}\right)^0\) - \(\dfrac{1}{8}\)
Bài 2 :
a, \(\dfrac{3^4.4^4}{6^4}\) b,\(\dfrac{15^3}{10^3}\) c, \(\dfrac{4^2.12^5}{9^2.2^{10}}\) d, \(\dfrac{6^2+5.2^2+4}{15}\)
Bài 3 :
a, \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\) b,\(\dfrac{6^6+6^3.3^3+3^6}{-73}\)
Mọi người giúp mình nhé mình sẽ cho bạn 1 like
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) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)
\(#Wendy.Dang\)
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 :))
Võ Ngọc Phương
Bài 3b, kết quả -(3)6 = - 729 em nhá chứ không phải (-3)6
1 Thự hiện phép tính:
a, 4. (\(1\dfrac{1}{4}^2\))+25.\(\left[\left(\dfrac{3}{4}\right)^{ }2:\left(\dfrac{5}{4}\right)^{ }2\right]:\left(\dfrac{3}{2}\right)^3\)
b,\(2^3+3\left(\dfrac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\dfrac{4}{2}\right].8\)
c,3-\(\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
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
a)\(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\) b)\(\left(0,\left(3\right)+\dfrac{\text{|}-2\text{|}}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)
a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)
\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)
\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)
\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)
\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)
b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)
\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)
Thực hiện phép tính:
1, \(\left(\dfrac{-1}{2}\right)^2.\left|+8\right|-\left(-\dfrac{1}{2}\right)^3:\left|-\dfrac{1}{16}\right|\)
2, \(\left|-0,25\right|-\left(-\dfrac{3}{2}\right)^2:\dfrac{1}{4}+\dfrac{3}{4}.2017^0\)
3, \(\left|\dfrac{2}{3}-\dfrac{5}{6}\right|.\left(3,6:2\dfrac{2}{5}\right)^3\)
4, \(\left|\left(-0,5\right)^2+\dfrac{7}{2}\right|.10-\left(\dfrac{29}{30}-\dfrac{7}{15}\right):\left(-\dfrac{2017}{2018}\right)^0\)
5, \(\dfrac{8}{3}+\left(3-\dfrac{1}{2}\right)^2-\left|\dfrac{-7}{3}\right|\)
`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`
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)