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Bài 1:

a: \(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-5+20x-4x^2-12x-9\)

\(=-30\)

b: \(B=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-\left(x^3+4x^2-246x\right)-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

\(=-175\)

d: \(D=25x^2-20x+4-36x^2-12x-1+11\left(x^2-4\right)-48+32x\)

\(=-11x^2-32x+3-48+32x+11x^2-44\)

=-89

a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

Do đó: A=C+D

\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-4x^2-12x-9-5+20x\)

\(=-30\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

=-175

A=C+D=-30-175=-205

b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

Do đó: B=E+F

\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)

\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)

\(=-18x^3+8x^2+7x-4\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)

\(=-95\)

\(B=-18x^3+8x^2+7x-99\)

 

a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

Do đó: A=C+D

\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-4x^2-12x-9-5+20x\)

\(=-30\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

=-175

A=C+D=-30-175=-205

b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

Do đó: B=E+F

\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)

\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)

\(=-18x^3+8x^2+7x-4\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)

\(=-95\)

\(B=-18x^3+8x^2+7x-99\)

12 tháng 7 2019

\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)

\(=6x^2+12x+x+2-6x^2+10x\)

\(=23x+2\)

12 tháng 7 2019

a) (6x + 1)(x + 2) - 2x(3x - 5)

= 6x2 + 12x + x + 2 - 6x2 + 10x

= (6x2 - 6x2) + (12x + x + 10x) + 2

= 23x + 2

b) (2x - 1)2 - (2x - 3)(2x + 3)

= 4x2 - 4x + 1 - 4x2 + 9

= (4x2 - 4x2) - 4x + (1 + 9)

= -4x + 10

c) (2x - 3)3  - (3x  + 1)(5 - 4x) - 16x2

= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2

= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5

= 8x3 - 40x2 + 43x - 5

d) (3x + 2) - (x - 5) - x(3x - 13)

= 3x  + 2 - x + 5 - 3x2 + 13x

= (3x - x + 13x) + (2 + 5) - 3x2

= 15x + 7 - 3x2

13 tháng 7 2018

Mình giải từ cuối lên , mình giải dần -)

n,  <=> x(2x-1)-3(2x-1)=0

<=> (x-3)(2x-1)=0

<=> x= 3 hoặc x= 1/2

m, <=> (x+2)(x2-3x+5)-x2(x+2)=0

<=> (x+2)(x2-3x+5-x2)=0

<=> (x+2)(5-3x)=0

=> x= -2 hoặc5/3

13 tháng 7 2018

trả lời chi tiết giúp mình với

28 tháng 3 2020

Copy có khác, ko đọc đc j!!! heheʌl

Câu 3:

1)

a) Ta có: 3x−2=2x−33x−2=2x−3

⇔3x−2−2x+3=0⇔3x−2−2x+3=0

⇔x+1=0⇔x+1=0

hay x=-1

Vậy: x=-1

b) Ta có: 3−4y+24+6y=y+27+3y3−4y+24+6y=y+27+3y

⇔27+2y=27+4y⇔27+2y=27+4y

⇔27+2y−27−4y=0⇔27+2y−27−4y=0

⇔−2y=0⇔−2y=0

hay y=0

Vậy: y=0

c) Ta có: 7−2x=22−3x7−2x=22−3x

⇔7−2x−22+3x=0⇔7−2x−22+3x=0

⇔−15+x=0⇔−15+x=0

hay x=15

Vậy: x=15

d) Ta có: 8x−3=5x+128x−3=5x+12

⇔8x−3−5x−12=0⇔8x−3−5x−12=0

⇔3x−15=0⇔3x−15=0

⇔3(x−5)=0⇔3(x−5)=0

Vì 3≠0

nên x-5=0

hay x=5

Vậy: x=5

29 tháng 3 2020

a) 3x - 2 = 2x - 3

\(\Leftrightarrow\) 3x - 2 - 2x + 3 = 0

\(\Leftrightarrow\) x + 1 = 0

\(\Rightarrow\) x = -1

b) 3 - 4y + 24 + 6y = y + 27 + 3y

\(\Leftrightarrow\) 3 - 4y + 24 + 6y - y - 27 - 3y = 0

\(\Leftrightarrow\) -2y = 0

\(\Rightarrow\) y = 0

c)7 - 2x = 22 - 3x

\(\Leftrightarrow\) 7 - 2x - 22 + 3x = 0

\(\Leftrightarrow\) -15 + x = 0

\(\Rightarrow\) x = 15

d) 8x - 3 = 5x + 12

\(\Leftrightarrow\) 8x - 3 - 5x - 12 = 0

\(\Leftrightarrow\)3x -15 = 0

\(\Leftrightarrow\) 3x = 15

\(\Rightarrow\) x = 5

e) x - 12 + 4x = 25 + 2x - 1

\(\Leftrightarrow\) x - 12 + 4x - 25 - 2x + 1 = 0

\(\Leftrightarrow\) 3x - 36 = 0

\(\Leftrightarrow\) 3x = 36

\(\Rightarrow\) x = 12

f ) x + 2x + 3x - 19 = 3x + 5

\(\Leftrightarrow\) x + 2x + 3x - 19 - 3x - 5 = 0

\(\Leftrightarrow\)3x - 24 = 0

\(\Leftrightarrow\) 3x = 24

\(\Rightarrow\) x = 8

g) 11+ 8x - 3 = 5x - 3 +x

\(\Leftrightarrow\)8x + 8 = 6x - 3

\(\Leftrightarrow\)8x - 6x = -3 - 8

\(\Leftrightarrow\)2x = -11

\(\Rightarrow\)x = \(-\frac{11}{2}\)

h) 4 - 2x +15 = 9x + 4 -2

\(\Leftrightarrow\)19 - 2x = 7x + 4

\(\Leftrightarrow\)-2x - 7x = 4 - 19

\(\Leftrightarrow\)-9x = -15

\(\Rightarrow\)x = \(\frac{15}{9}\) = \(\frac{5}{3}\)