a: Ta có: \(0,\left(3\right)+\dfrac{10}{3}+0,4\left(2\right)\)

\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{4}{9}\)

\(=\dfrac{33}{9}+\dfrac{4}{9}=\dfrac{37}{9}\)

b: Ta có: \(\dfrac{4}{9}+1.2\left(31\right)-0,\left(13\right)\)

\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)

\(=\dfrac{440}{990}+\dfrac{1219}{990}-\dfrac{130}{990}\)

\(=\dfrac{139}{90}\)

c: Ta có: \(2,\left(4\right)\cdot\dfrac{3}{11}\)

\(=\dfrac{22}{9}\cdot\dfrac{3}{11}\)

\(=\dfrac{2}{3}\)

d: Ta có: \(-0,\left(3\right)+\dfrac{1}{3}\)

\(=-\dfrac{1}{3}+\dfrac{1}{3}\)

=0

2.Tính

a) 3/7 nhân 4=12/7              b) 2/3 nhân 1=2/3       c) 6/11 nhân 0=0

d)2 nhân 4/7=8/7                e) 1 nhân 3/8=3/8      g) 0 nhân 13/18=0

Chúc e học tốt

2.Tính

a) 3/7 nhân 4=12/7             b) 2/3 nhân 1=2/3     c) 6/11 nhân 0=0

d)2 nhân 4/7=8/7             e) 1 nhân 3/8=3/8        g) 0 nhân 13/18=0

a) = 12/7

b) = 2/3

c) = 0

d) = 8/7

e) = 3/8

g) = 0

a) \(=\dfrac{\left(-1\right)^4}{3^4}=\dfrac{1}{81}\)

b) \(=\dfrac{\left(-9\right)^3}{4^3}=\dfrac{-729}{64}\)

c) \(=\left(-\dfrac{2}{10}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\)

d) \(=1\)

\(a,=\dfrac{1}{81}\\ b,=\dfrac{729}{64}\\ c,=0,04\\ d,=1\)

\(a,=\sqrt{3}+4\sqrt{3}+20\sqrt{3}-10\sqrt{3}=15\sqrt{3}\\ b,=4\sqrt{5}+\sqrt{5}-1-\dfrac{20\left(\sqrt{5}-1\right)}{4}\\ =5\sqrt{5}-1-5\sqrt{5}+5=4\\ c,=\dfrac{6\sqrt{13}+6+6\sqrt{13}-6}{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}=\dfrac{12\sqrt{13}}{12}=\sqrt{13}\\ d,=\left(\sin^238^0+\cos^238^0\right)+\left(\tan67^0-\tan67^0\right)=1+0=1\)

a: \(=\sqrt{3}+4\sqrt{3}+4\cdot5\sqrt{3}-10\sqrt{3}\)

\(=15\sqrt{3}\)

b: \(=2\cdot2\sqrt{5}+\sqrt{5}-1-5+5\sqrt{5}\)

=-6

a: =20/60+12/60-15/60=17/60

b: =7/8-1/4-2/5=5/8-2/5=25/40-16/40=9/40

\(a,\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{20}{60}+\dfrac{12}{60}-\dfrac{15}{60}=\dfrac{17}{60}\\ b,\dfrac{7}{8}-\left(\dfrac{1}{4}+\dfrac{2}{5}\right)=\dfrac{7}{8}-\dfrac{13}{20}=\dfrac{9}{40}\)

a)

\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)

b)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\  = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)

`1`

`a, 1/2 +1/3= 3/6 + 2/6 =5/6`

`d, 1/3 +3/5= 5/15 + 9/15=14/15`

`c,4/5 +1/2= 8/10 + 5/10= 13/10`

`2`

`a,1/2 +1/4=2/4 +1/4=3/4`

`b, 2/3 +1/6 = 4/6+1/6=5/6`

`c, 7/12 +1/2=7/12+ 6/12= 13/12`

`3`

Giải

Cả `2` ngày đi tất cả số quãng đường là :

`1/4 +1/2 =1/4+ 2/4= 3/4 ( quãng đường)`

đ/s...

`@ yL`

1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)

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

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

2: \(x^2-8x+7=0\)

=>\(x^2-x-7x+7=0\)

=>\(x\left(x-1\right)-7\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

1:

a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)

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

\(=x^2+1\)

1)

\(a,7x^2\cdot(2x^3+3x^5)\\=7x^2\cdot2x^3+7x^2\cdot3x^5\\=14x^5+21x^7\\---\\b,(x^3-x^2+x-1):(x-1)(dkxd:x\ne 1)\\=[x^2(x-1)+(x-1)]:(x-1)\\=(x-1)(x^2+1):(x-1)\\=x^2+1\)

2)

\(x^2-8x+7=0\\\Leftrightarrow x^2-x-7x+7=0\\\Leftrightarrow x(x-1)-7(x-1)=0\\\Leftrightarrow (x-1)(x-7)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

\(\text{#}Toru\)