a: Trường hợp 1: x<-1

A=3(-x-1)+2(-x+3)+2x-5

=-3x-3-2x+6+2x-5

=-3x-2

Trường hợp 2: -1<=x<3

A=3(x+1)+2(3-x)+2x-5

=3x+3+6-2x+2x-5

=3x+4

Trường hợp 3: x>=3

A=3(x+1)+2(x-3)+2x-5

=3x+3+2x-6+2x-5

=7x-8

b: Trường hợp 1: x<-7

B=-x-7+2(3-x)-x-4

=-2x-11+6-2x=-4x-5

TRường hợp 2: -7<=x<3

B=x+7+2(3-x)-x-4

=3+6-2x=-2x+9

Trường hợp 3: x>=3

B=x+7+2x-6-x-4=2x-3

c: Trường hợp 1: x<1

C=8(1-x)+2(3-x)-5x-3

=8-8x+6-2x-5x-3

=-15x+11

Trường hợp 2: 1<=x<3

C=8(x-1)+2(3-x)-5x-3

=8x-8+6-2x-5x-3

=x-5

TRường hợp 3: x>=3

C=8(x-1)+2(x-3)-5x-3

=8x-8+2x-6-5x-3

=5x-17

a: Trường hợp 1: x<-1

A=2(3-x)+3(-x-1)+2x-5

=6-2x-3x-3+2x-5

=-3x-2

Trường hợp 2: -1<=x<3

A=2(3-x)+3(x+1)+2x-5

=6-2x+3x+3+2x-5

=3x+4

TRường hợp 3: x>=3

A=2(x-3)+3(x+1)+2x-5

=2x-6+3x+3+2x-5

=7x-8

b: Trường hợp 1: x<-7

B=7-x+2(3-x)-x-4

=3-2x+6-2x=-4x+9

Trường hợp 2: -7<=x<3

B=x+7+2(3-x)-x-4

=3+6-2x=-2x+9

Trường hợp 3: x>=3

B=x+7+2x-6-x-4=2x-3

1,

\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

a: =18x^3y^2-12x^3y^3+6x^2y^2

b: (-3x+2)(5x^2-1/3x+4)

=-12x^3+x^2-12x+10x^2-2/3x+8

=-12x^3+11x^2-38/3x+8

c: =x^2-x-2+3x-x^2

=2x-2

d: =4x^2+12x+9-4x^2+25-(x-1)(x^2+12)

=12x+34-x^3-12x+x^2+12

=-x^3+x^2+46

a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)

\(=x^2-12x^2+16x+7x-5\)

\(=-11x^2+23x-5\)

b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)

\(=7x^3-35x-3x^3y^2+18x^2y^3\)

c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)

\(=10x^2-35x+8x-28\)

\(=10x^2-27x-28\)

a: =12x^3y^2-12x^3y^3+6x^2y^2

b: =\(\left(-3x+2\right)\left(5x^2-\dfrac{1}{3}x+4\right)\)

=-15x^3+x^2-12x+10x^2-2/3x+8

=-15x^3+11x^2-38/3x+8

c: =x^2-x-2+3x-x^2

=2x-2

1 ) a) \(4x^2-x^2+8x^2\)

\(=\left(4+8\right).x^2+x^2-x^2\)

\(=12.x^3\)

b) \(\frac{1}{2}.x^2.y^2-\frac{3}{4}.x^2.y^2+x^2.y^2\)

\(\left(\frac{1}{2}-\frac{3}{4}\right).x^2.x^2.x^2.+y^2+y^2+y^2\)

\(=-\frac{1}{4}.x^6+y^6\)

c) \(3y-7y+4y-6y\)

\(=\left(3-7+4-6\right).y.y.y.y\)

\(=-6.y^4\)

2) 

\(\left(-\frac{2}{3}.y^3\right)+3y^2-\frac{1}{2}.y^3-y^2\)

\(\left(-\frac{2}{3}+3-\frac{1}{2}\right).y^3.y^3-y\)

\(=\frac{25}{6}.y^5\)

b) \(5x^3-3x^2+x-x^3-4x^2-x\)

\(=\left(5-3-4\right).\left(x^3.x^2+x-x^3-x^2-x\right)\)

\(=-2.0=0\)

hông chắc

3)a)  \(5xy^2.\frac{1}{2}x^2y^2x\)

\(\left(5.\frac{1}{2}\right).x^2.x^2.x.y^2.y^2\)

\(=\frac{5}{2}.x^5.y^4\)

b) Tổng các bậc của đơn thức là

5+4 = 9

Hệ số của đơn thức là \(\frac{5}{2}\)

Phần biến là x;y

Thay x=1;y=-1 vào đơn thức

\(\frac{5}{2}.1^5.\left(-1\right)^4\)

\(\frac{5}{2}.1.\left(-1\right)\)

\(\frac{5}{2}.\left(-1\right)=-\frac{5}{2}\)

Vậy ....

chắc không đúng đâu uwu

a) \(A=\left(5x-3\right)^2-2\left(5x-3\right)\left(x+3\right)+\left(x+3\right)^2\) ( \(5x-3\) chứ sao lại \(5x+3\) )

\(\Leftrightarrow A=\left[\left(5x-3\right)-\left(x+3\right)\right]^2\)

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

\(\Leftrightarrow A=\left(4x-6\right)^2\)

\(\Leftrightarrow A=\left(4x\right)^2-2.4x.6+6^2\)

\(\Leftrightarrow A=16x^2-48x+36\)

b) \(x^3+5x^2+6x\)

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

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

\(=x\left[\left(x^2+3x\right)+\left(2x+6\right)\right]\)

\(=x\left[x\left(x+3\right)+2\left(x+3\right)\right]\)

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