a) (a-b)^3=-(b-a)^3

\(Taco:-\left(b-a\right)^3\)

=\(-\left(b-a\right)\left(b-a\right)\left(b-a\right)\)

\(=\left(a-b\right)\left(b-a\right)\left(b-a\right)\)

\(=-\left(a-b\right)\left(a-b\right)\left(b-a\right)\)

\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)=\left(a-b\right)^3\)

\(\left(-a-b\right)^2=\left(-a-b\right)\left(-a-b\right)\)

\(=-\left(a+b\right)\left(-a-b\right)\)

\(=\left(a+b\right)\left(a+b\right)\)

\(=\left(a+b\right)^2\)

Các giao điểm nối được là :

45 x 2 = 90 giao điểm

=> Ta có : n đường thẳng  nối được với n - 1 đường thẳng

=> Ta có : n(n - 1) = 90

          => n(n - 1)   = 10.9

n = 10         

=> Có 10 đường thẳng

\(D=\frac{5}{8}:\frac{9}{-2}+\frac{5}{8}:\frac{-9}{7}+1\frac{5}{8}\)

\(=\frac{5}{8}.\frac{-2}{9}+\frac{5}{8}.\frac{-7}{9}+\frac{13}{8}\)

\(=\frac{5}{8}.\left(\frac{-2}{9}+\frac{-7}{9}\right)+\frac{13}{8}\)

\(=\frac{-5}{8}+\frac{13}{8}\)

\(=1\)

D = \(\frac{5}{8}:\frac{9}{-2}+\frac{5}{8}:\frac{-9}{7}+1\frac{5}{8}\)

    = \(\frac{5}{8}.\frac{-2}{9}+\frac{5}{8}.\frac{-7}{9}+1+\frac{5}{8}\)

    = \(\frac{5}{8}.\left(\frac{-2}{9}+\frac{-7}{9}\right)+1+\frac{5}{8}\)

     = \(-\frac{5}{8}+1+\frac{5}{8}\)

     = \(-\frac{5}{8}+\frac{5}{8}+1\)

     = \(1\)

Trả lời

C=4/20-7/-8-1/3

=1/5-(-7/8)-1/3

=1/5+7/8-1/3

=24+105-40/120

=89/120

D=1/3-1/5-(-1/6)-1/8

=1/3-1/5+1/6-1/8

=(5/15-3/15)+1/6-1/8

=2/15+1/6-1/8

=16+20-15/120

=21/120

=7/40

\(C=\frac{4}{20}-\frac{7}{-8}-\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}+\frac{7}{8}-\frac{1}{3}\)

\(=\frac{24+105-40}{120}=\frac{89}{120}\)

\(D=\frac{1}{3}-\frac{1}{5}-\frac{-1}{6}-\frac{1}{8}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{6}-\frac{1}{8}\)

\(=\left(\frac{5}{15}-\frac{3}{15}\right)+\left(\frac{8}{48}-\frac{6}{48}\right)\)

\(=\frac{2}{15}+\frac{1}{24}\)

\(=\frac{48}{360}+\frac{15}{360}\)

\(=\frac{63}{360}=\frac{21}{120}=\frac{7}{40}\)

\(A=12+\frac{3}{5}+\frac{17}{24}+\frac{7}{5}\)

\(A=12+\left(\frac{3}{5}+\frac{7}{5}\right)+\frac{17}{24}\)

\(=12+2+\frac{17}{24}\)

\(=14+\frac{17}{24}\)

\(=\frac{336}{24}+\frac{17}{24}\)

=\(=\frac{353}{24}\)

Trả lời

A=12+3/5+17/24+7/5

=12+(3/5+7/5)+17/24

=12+10/5+17/24

=12+2+17/24

=14/1+17/24

=336/24+17/24

=353/24

B=2/3+1/5-5/3+3/5+1

=10+2-10+6+15/15

=23/15.

Học tốt nha !

a) (x-1)*(x+2)-(x-3)*(-x+4)=19

\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)

\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)

\(\Leftrightarrow2x^2-3x+7-19=0\)

\(\Leftrightarrow2x^2-3x-12=0\)

Đề sai??

b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)

\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)

\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)

\(\Leftrightarrow-30x^2+7x-4+17=0\)

\(\Leftrightarrow-30x^2+7x+13=0\)

???

Vẽ BH vuông góc với AC

Theo định lý Pythagore, ta có:

BC²=BH²+CH²=BH²+(AC-AH)²

=BH²+AH²+AC²-2AC.AH

Mà ta lại có:AH²+BH²=AB² (định lý Pythagore, tam giác ABH vuông tại H) 

và AH=1/2AB (do tam giác ABH là nửa tam giác đều)

Cho nên: BC²=AB²+AC²-2.1/2AB.AC=AB²+AC²-AB.AC (*)

Thay AB=28cm, AC=35cm vào (*), ta được:

BC²=1029=>BC=7\(\sqrt{21}\)cm

Vậy BC=7\(\sqrt{21}\)cm