A=(7-2x)(7+2x)+(2x+7)²

    =49-4x²+4x²+28x+49

   = 98+28x

B=(4x-5)²-(2x-1)(8x-5)

  = 16x²-25-((8x(2x-1))-(5(2x-1)))

  = 16x²-25-((16x²+8x)-(10x+5))

  = 16x²-25-(16x²+8x-10x-5)

  = 16x²-25-16x²-8x+10x+5

   = -20+2x

a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)

\(=7-4x^2+4x^2+28x+49\)

\(=28x+56\)

b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)

\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)

\(=16x^2-40x+25-16x^2+18x-5\)

\(=-22x+20\)

c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)

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

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

\(=\left(10x-8\right)^2\)

\(=100x^2-160x+64\)

d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)

\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)

\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)

\(=-9b^2+6bc-c^2+9b^2+c^2\)

=6bc

3/5+4/6=3/5+2/3=9/15+10/15=19/15

7-1/3=20/3

5x3/8=15/8

12/35:2/5=12/35x5/2=60/70=6/7

\(\dfrac{3}{5}+\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{19}{15}\)
\(7-\dfrac{1}{3}=\dfrac{20}{3}\)
\(5x\dfrac{3}{8}=\dfrac{15}{8}\)
\(\dfrac{12}{35}:\dfrac{2}{5}=\dfrac{12}{35}x\dfrac{5}{2}=\dfrac{6}{7}\)

Bài 1:

a, \(\left(x-y+2z\right)^2=x^2+y^2+4z^2-2xy-4yz+4zx\)

b, \(\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)=\left(4x^2-9\right)\left(4x^2+9\right)=16x^4-81\)

a: \(A=\left|5x-3\right|-5\left(5-2x\right)\)

\(=\left|5x-3\right|-25+10x\)

TH1: x>=3/5

\(A=5x-3+10x-25=15x-28\)

TH2: x<3/5

\(A=3-5x+10x-25=5x-22\)

b: \(B=12-2x-\left|3-2x\right|\)

\(=-2x+12-\left|2x-3\right|\)

TH1: x>=3/2

\(B=-2x+12-\left(2x-3\right)\)

\(=-2x+12-2x+3=-4x+15\)

TH2: x<3/2

\(B=-2x+12-\left(3-2x\right)\)

\(=-2x+12-3+2x=9\)

1) 30x-30x^2-31

2)6x^4-2x^3-15x^2+23x-6

a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`

`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`

`=10(12x^2+25)-(120x^2+49)`

`=120x^2+250-120x^2-49`

`=201`

b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`

Lời giải:
a. 

$(2x+5)^3-(2x-5)^3-(120x^2+49)$

$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$

$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$

$=10(12x^2+25)-(120x^2+49)=250-49=201$

b.

$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$

$=7(1-10x)$

a) Ta có: \(\left(2x+5\right)^3-\left(2x-5\right)^3-\left(120x^2+49\right)\)

\(=8x^3+60x^2+150x+125-8x^3+60x^2-150x+125-120x^2-49\)

\(=201\)

b) Ta có: \(\left(4-5x\right)^2-\left(3+5x\right)^2\)

\(=\left(4-5x-3-5x\right)\left(4-5x+3+5x\right)\)

\(=7\left(-10x+1\right)\)

\(=-70x+7\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

= 6/20 = 3/10

= 342/57 = 6

= 85/15 = 17/3

= 14 x 10/7 = 140/7 = 20

= 14/5 x 13/6 = 182/30 = 91/15

= 6/20 = 3/10

= 342/57 = 6

= 85/15 = 17/3

= 14 x 10/7 = 140/7 = 20

= 14/5 x 13/6 = 182/30 = 91/15
tick cho mình nha