a. Ta có: p + e + n = 58

Mà p = e, nên: 2p + n = 58 (1)

Theo đề, ta có: n = 20 (2)

Thay (2) vào (1), ta được: 2p + 20 = 58

=> p = 19

Vậy p = e = 19 hạt, n = 20 hạt.

Vậy số khối của A bằng: p + n = 20 + 19 = 39(đvC)

b. Ta có: p + e + n = 34

Mà p = e, nên: 2p + n = 34 (1)

Theo đề: p = 11 (2)

Thay (2) vào (1), ta được:

22 + n = 34

=> n = 12

Vậy p = e = 11 hạt, n = 12 hạt.

a) Ta có: \(p=\dfrac{58-20}{2}=19\left(hạt\right)\) \(\Rightarrow A=p+n=20+19=39\)

b) Ta có: \(e=11\left(hạt\right)\) \(\Rightarrow n=34-11\cdot2=12\left(hạt\right)\)

\(a,x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\\ =xy-xz+yz-xy+xz-yz\\ =\left(xy-xy\right)+\left(xz-xz\right)+\left(yz-yz\right)\\ =0+0+0\\ =0\left(dpcm\right)\)

\(b,x\left(y+z-yz\right)-y\left(z+x-zx\right)+z\left(y-x\right)\\ =xy+xz-xyz-yz-xy+xyz+yz-xz\\ =\left(xy-xy\right)+\left(xz-xz\right)+\left(xyz-xyz\right)+\left(yz-yz\right)\\ =0+0+0+0\\ =0\left(dpcm\right)\)

x:54 dư 12

=>x=54k+12

=>\(\frac{x}{9}\)=6k+1+\(\frac{3}{9}\)(dư 1)

y:72 dư 10

=>y=72k'+10

=>\(\frac{y}{9}\)=8k'+1+\(\frac{1}{9}\)(dư 1)

=>(x+y) :9 dư 3+1=4

mình nhầm sửa chỗ 6k+1+\(\frac{3}{9}\)(dư 1) thành (dư 3) nhé

b ) 5 x ( x - 1 ) - 3x ( x - 1 )

= ( 5x - 3x ) ( x - 1 )

= 2x ( x - 1 )

c ) x ( x + y ) - 5 x - 5y

= x ( x + y ) - 5 ( x + y )

= ( x - 5 ) (x + y )

d ) x ^ 2 + xy + x

= x ( x + y + 1 )

e ) x ( x - y ) + y ( y - x )

= x ( x - y ) - y ( x - y )

= ( x - y ) ( x -y )

= ( x - y ) ²

a, 5x(x-1)-3x(x-1)=(5x-3x)(x-1)=2x(x-1)

b, x(x+y)-5x-5y=x(x+y)-5(x+y)=(x-5)(x+y)

c, x²+xy+x=x(x+y+1)

d, x(x-y)+y(y-x)=x(x-y)-y(x-y)=(x-y)(x-y)=(x-y)²

b)5x(x-1)-3x(x-1)=x(x-1).(5-3)=x(x-1).2

c)x(x+y)-5x-5y=x(x+y)-(5x+5y)=x(x+y)-[5(x+y)[=(x+y).(x-5)

d)x^2+xy+x=x(x+y+1)

e)x(x-y)+y(y-x)=x(x-y)-(x-y).y=(x-y).(x-y)

a. 3xy( 4x + y - \(\dfrac{4}{3}\) )

b. 2x²( 3x + 1 )

c. (2x + 3 )( x - y )

d. xy( 1 - x )( x - 1 )

e. 6( 2x + 1 )( x + y )