Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

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

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)

\(=x^2-y^2\)

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

a) M = (x² + 3xy - 3x³) + (2y³ - xy + 3x³)

= x² + 3xy - 3x³ + 2y³ - xy + 3x³

= x² + (3xy - xy) + (-3x³ + 3x³) + 2y³

= x² + 2xy + 2y³

Tại x = 5 và y = 4

M = 5² + 2.5.4 + 2.4³

= 25 + 40 + 2.64

= 65 + 128

= 193

b) N = x²(x + y) - y(x² - y²)

= x³ + x²y - x²y + y³

= x³ + (x²y - x²y) + y³

= x³ + y³

Tại x = -6 và y = 8

N = (-6)³ + 8³

= -216 + 512

= 296

c) P = x² + 1/2 x + 1/16

= (x + 1/2)²

Tại x = 3/4 ta có:

P = (3/4 + 1/2)² = (5/4)² = 25/16

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)

\(=2x^2-2xy-y^2+2xy\)

\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)

\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)

\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(=5x^2-20xy-4y^2+20xy\)

\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)

\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)

\(=-xy\left(x+1\right)\)

\(=\dfrac{1}{2}.100\left(100+1\right)=50.101=5050\)

cái đoạn\(-xy\left(x+1\right)\)đổi x+1 thành y+1 nha mik đánh nhầm

`a)A=x(x+y)-x(y-x)`

`=x^2+xy-xy+x^2`

`=2x^2`

Thay `x=-3`

`=>A=2.9=18`

`b)B=4x(2x+y)+2y(2x+y)-y(y+2x)`

`=8x^2+4xy+4xy+2y^2-y^2-2xy`

`=8x^2+y^2+6xy`

Thay `x=1/2,y=-3/4`

`=>B=8*1/4+9/16-9/4`

`=2+9/16-9/4`

`=9/16-1/4=5/16`

Bài 1:

a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)

\(\Rightarrow4\left(x-2\right)-3x+4=0\)

\(\Rightarrow4x-8-3x+4=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)

\(\Rightarrow10x+35-15x-6=25\)

\(\Rightarrow-5x+29=25\)

\(\Rightarrow-5x=25-29\)

\(\Rightarrow-5x=-4\)

\(\Rightarrow x=\dfrac{4}{5}\)

c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Rightarrow x+4=0\)

\(\Rightarrow x=-4\)

d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Rightarrow-x-21=0\)

\(\Rightarrow x=-21\)

Bài 2:

a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(P=8x^2y-6y^2-9x^2y+12y^2\)

\(P=-x^2y+6y^2\)

Thay x = -1 ; y = 2 vào P ta được

\(P=-\left(-1\right)^2.2+6.2^2\)

\(P=-2+24=22\)

b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(Q=20x^3-12x^2y-4x^3-x^2y\)

\(Q=16x^3-13x^2y\)

Thay x = -1 ; y = 2 vào Q ta được

\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)

\(Q=-16-26\)

\(Q=-42\)

c) \(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)

\(H=2xy\)

Thay x = 1/4 ; y = 2012 vào H ta được

\(H=2.\dfrac{1}{4}.2012\)

\(H=1006\)

1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Leftrightarrow8x-16-6x+8=2\)

\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)

b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Leftrightarrow30x-20-15x-6+55-20x=25\)

\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)

\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)

2.

a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)

\(\Leftrightarrow x^2y-18y^2\)

tại x=-1 , y=2

ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)

vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2

b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)

\(\Leftrightarrow17x^3-13x^2y\)

tại x=-1,y=2

ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)

vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)

c)\(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)

\(\Leftrightarrow x^4+2xy-x^3\)

tại x=1/4 và y=2012

ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)

a. 8(x-2)-2(3x-4)=2

8x-16-6x-8 =2

(8x-6x)-(16-8)=2

2x-2 =2

2x =2+2

2x =4

x =\(\dfrac{4}{2}\)=2

trả lời hộ với mai thi rồi

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

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

\(=36x^2\)(1)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:

\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)

b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

\(=\left(x-1\right)^2=100^2=10000\)

A       = 2\(x^2\)y + \(xy\) - 3\(xy\)

Thay \(x\) = -2; y = 4 vào biểu thức A ta có: 

A      =  2\(\times\) (-2)² \(\times\) 4 + (-2) \(\times\) 4 - 3 \(\times\) (-2) \(\times\) 4

A     = 2 \(\times\) 4 \(\times\) 4 - 8 + 6 \(\times\) 4

A     = 8 \(\times\) 4 - 8 + 24

A     =  32 - 8 + 24

A    =  24 + 24

A    =    48

B = (2\(x^2\) + \(x\) - 1) - ( \(x^2+5x-1\) )

Thay \(x\) = - 2 vào biểu thức B ta có:

B = { 2\(\times\)(-2)² + (-2) - 1} - { (-2)² +5\(\times\)(-2) - 1}

B = { 2 \(\times\) 4  - 3} - { 4 - 10 - 1}

B = { 8 - 3} - { 4 - 11}

B = 5 - (-7)

B = 5 + 7

B = 12