Gọi chiều rộng là x

=>Chiều dài là x+5

Theo đề, ta có: (x+5-3)(x+2)=16

=>(x+2)^2=16

=>x+2=4

=>x=2

=>Chiều dài là 7m

Chu vi lúc đầu là (2+7)*2=18(m)

\(5-\left(6-x\right)=4\left(3-2x\right)\)

\(5-6+x=12-8x\)

\(-1+x=12-8x\)

\(x-1=12-8x\)

\(12+1=8x+1\)

\(8x=13-1\)

\(x=12:8\)

\(x=\dfrac{12}{8}=\dfrac{3}{2}\)

\(PT\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\dfrac{13}{9}\)

Vậy: \(S=\left\{\dfrac{13}{9}\right\}\)

Câu 3:

a. $y^2+2y+1=(y+1)^2$
b. $9x^2+y^2-6xy=(3x)^2-2.3x.y+y^2=(3x-y)^2$
c. $25a^2+4b^2+20ab=(5a)^2+2.5a.2b+(2b)^2$

$=(5a+2b)^2$
d. Sửa đề:

$x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+(\frac{1}{2})^2$

$=(x-\frac{1}{2})^2$

Câu 5:

a. $x(x-2)+x-2=0$

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

$\Leftrightarrow (x-2)(x+1)=0$

$\Leftrightarrow x-2=0$ hoặc $x+1=0$

$\Leftrightarrow x=2$ hoặc $x=-1$

b. 

$5x(x-3)-x+3=0$

$\Leftrightarrow 5x(x-3)-(x-3)=0$
$\Leftrightarrow (x-3)(5x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $5x-1=0$

$\Leftrightarrow x=3$ hoặc $x=\frac{1}{5}$

a: \(A=\left(a+3\right)\left(9a-8\right)-\left(a+2\right)\left(9a-1\right)\)

\(=9a^2-8a+27a-24-\left(9a^2-a+18a-2\right)\)

\(=9a^2+19a-24-9a^2-17a+2=2a-22\)

Thay a=-3 vào A, ta được:

\(A=2\cdot\left(-3\right)-22=-6-22=-28\)

b: \(Q=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)

\(=6x^2+23x-55-6x^2-23x-21\)

=-55-21

=-76

=>Q không phụ thuộc vào biến x

Câu a:

A = (a + 3).(9a - 8) - (2+ a)(9a - 1) (1)

Thay a = -3 vào biểu thức (1) ta có:

A = (-3 + 3).(9.3 - 8) - (2 - 3).(9.(-3) - 1)

A = 0.(27 - 8) - (-1).(-27 - 1)

A = 0 + (-28)

A = - 28 (đpcm)

a: \(3y^3-7y^2+5y-1\)

\(=3y^3-y^2-6y^2+2y+3y-1\)

\(=y^2\left(3y-1\right)-2y\left(3y-1\right)+\left(3y-1\right)\)

\(=\left(3y-1\right)\left(y^2-2y+1\right)=\left(3y-1\right)\left(y-1\right)^2\)

Ta có: \(2y^3-y^2-4y+3\)

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

\(=2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)=\left(y-1\right)\left(2y^2+y-3\right)\)

\(=\left(y-1\right)\left(2y^2+3y-2y-3\right)=\left(y-1\right)\left(2y+3\right)\left(y-1\right)=\left(y-1\right)^2\cdot\left(2y+3\right)\)

Ta có: \(B=\frac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)

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