a) 

(1/3-1/5)x1/4 = 1/3x1/4-1/5x1/4=1/12-1/20=1/30

(1/3-1/5)x1/4 = 2/15x1/4=1/30

b)

2/5 x 3/7 + 2/5 x 4/7= 6/35 + 8/35 = 14/35 = 2/5

2/5 x 3/7 + 2/5 x 4/7= (3/7+4/7) x 2/5 = 1 x 2/5 = 2/5

a: C1: (1/3-1/5)*1/4

=2/15*1/4=2/60=1/30

C2: (1/3-1/5)*1/4

=1/3*1/4-1/5*1/4

=1/12-1/20=1/30

b: C1: 2/5*3/7+2/5*4/7

=6/35+8/35=14/35=2/5

C2: 2/5*3/7+2/5*4/7

=2/5(3/7+4/7)

=2/5*1=2/5

a)x^m+4+x^m+3-x-1

=(x^m+4-x)+(x^m+3-1)

=x(x^m+3-1)+(x^m+3-1)

=(x+1)(x^m+3-1)

Với x=-2 ta có:... bn tự thay

b)x⁶-x⁴+2x³+2x²=x⁶-2x⁵+2x⁴+2x⁵-4x⁴+4x³+x⁴-2x³+2x²

=x⁴(x²-2x+2)+2x³(x²-2x+2)+x²(x²-2x+2)

=(x⁴+2x³+x²)(x²-2x+2)

=[x²(x²+2x+1)](x²-2x+2)

=x²(x+1)²(x²-2x+2)

Với x=-2 bn tự thay nhé h mk bận

2+6+12+20+...+90

Cách tính mk ko rành lắm xin các anh chị giúp đỡ 

đáp án là : 330

t nha

A = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 9 x 10

A x 3 = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3  + ... + 9 x 10 x 3

A x 3 = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 ( 5 - 2 ) + ... + 9 x 10 x ( 11 - 8 )

A x 3 = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 9 x 10 x 11 - 8 x 9 x 10

A x 3 = 9 x 10 x 11

A = 9 x 10 x 11 : 3

A = 330

  Vậy A = 330

Đặt A = 1.2 + 2.3 + 3.4 + 4.5 + ... + 9.10

=> 3A = = 1.2.3 + 2.3.( 4 - 1 ) + 3.4. ( 5 - 2 ) + ... + 9.10.( 11 - 8 )

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - 3.4.5 + ...... + 8.9.10 - 9.10.11

=> 3A = 9.10.11

=> A = 9.10.11 : 3

Vậy A = 330

  Vậy A = 330

Đặt S=

1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ............. + 9 x 10 

=>3S + 1.2.3+2.3.3+...+99.100.3

=>1.2.3+2.3(4-1)+3.4(5-2)+...+99.100(101-98)

=>1.2.3+2.3.4-1.2.3+3.4.5+-2.3.4+...+99.100.101-98.100.101

=>99.100.101=999900

=>S=333300

\(2,4\times1994\times2+1,6\times3996\times3+4,8\times4010\)
\(=4,8\times1994+4,8\times3996+4,8\times4010\)
\(=4,8\times\left(1994+3996+4010\right)\)
\(=4,8\times10000=48000\)

Ta có : \(B\text{=}4x^2-12x+9\)

\(B\text{=}\left(2x-3\right)^2\)

Với \(x\text{=}\dfrac{1}{2}\)

\(\Rightarrow B\text{=}\left(2.\dfrac{1}{2}-3\right)^2\)

\(B\text{=}\left(-2\right)^2\text{=}4\)

Ta có : \(A\text{=}5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5\left(x^2-9\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5x^2-45+4x^2+12x+9+x^2-12x+36\)

\(A\text{=}10x^2\)

Với \(x\text{=}-\dfrac{1}{5}\)

\(\Rightarrow A\text{=}10.\left(-\dfrac{1}{5}\right)^2\text{=}\dfrac{2}{5}\)

B = 4x² - 12x + 9

= (2x - 3)²

Tại x = 1/2 ta có:

B = (2.1/2 - 3)²

= (-2)²

= 4

-------------------

A = 5(x + 3)(x - 3) + (2x + 3)² + (x - 6)²

= 5x² - 45 + 4x² + 12x + 9 + x² - 12x + 36

= 10x²

Tại x = 1/5 ta có:

A = 10.(1/5)²

= 2/5

\(B=4x^2-12x+9\)

\(B=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

\(B=\left(2x-3\right)^2\)

Thay \(x=\dfrac{1}{2}\) vào B ta có: 

\(B=\left(2\cdot\dfrac{1}{2}-3\right)^2=4\)

_______________________

\(A=5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A=5\left(x^2-9\right)+4x^2+12x+9+x^2-12x+36\)

\(A=5x^2-45+5x^2+45\)

\(A=10x^2\)

Thay \(x=\dfrac{1}{5}\)vào A ta có:

\(A=10\cdot\left(-\dfrac{1}{5}\right)^2=\dfrac{2}{5}\)

Theo quy luật trên, trong biểu thức $B$ sẽ có nhân tử $(x-34)$

Mà với $x=34⇒x-34=0$

Nên $B=(x-1)(x-2)(x-3).....(x-34)(x-35)=(x-1).(x-2).(x-3).....0.(x-35)=0$

Vậy $B=0$

B= (X-1).(X-2).(X-3). ... .(X-34).(X-35) tại X = 34

B= (34-1).(34-2).(34-3)....(34-34).(34-35)

B= 33.32.31. ... .0.(-1)

B=0 

3x5+3x2+3x3

=3x(5+2+3)

=3x(7+3)

=3x10

=30

3 x 5 + 3 x 2 + 3 x 3 = 3 x ( 5 + 2 + 3 )

                               = 3 x  10  

                               =    30

Tích mk nha ^_^

3 x 5 + 3 x 2 + 3 x 3 

= 3 x ( 5 + 2 + 3 )

= 3 x 10

= 30