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

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

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

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

\(a,3^3+3^2-\left(2^7:2^5+7^8:7^7\right)=27+9-\left(2^2+7\right)=36-11=25\)

\(b,27.77+27.27-27=27\left(77+27-1\right)=27.103=2781\)

\(c,52:\left(43-30\right)+144-16=52:13+128=4+128=132\)

\(f,3^2.101-3^2.101^0=3^2\left(101-101^0\right)=9\left(101-1\right)=9.100=900\)

a.=27+9-(2^2+7^1)                                                 =36-(4+7)                                                               =36-11                                                                   =25

3³+3²-(2⁷:2⁵+7⁸:7⁷)

=27+9-(2²+7)

=36-(4+7)

=36-11

=25

b.bạn tự lm nha

c.52:(43-30)+144-16

=52:13+144-16

=4+144-16

=148-16

=132

d.174:{2[36+(4²-23)]}

=174:{2[36+(16-23)]}

=174:{2[36+-7]}

=174:{2.29}

=174:58

=3

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

a: =>0<x<0,25+0,8=1,05

hay x=1

b: =>7/7<x<7/3

=>1<x<7/3

hay x=2

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)

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

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

\(\text{#}Toru\)