`a)`

`A=-4x^5y^3+6x^4y^3-3x^2y^3z^2+4x^5y^3-x^4y^3+3x^2y^3z^2-2y^4+22`

`A=(-4x^5y^3+4x^5y^3)+(6x^4y^3-x^4y^3)-(3x^2y^3z^2-3x^2y^3z^2)-2y^4+22`

`A=5x^4y^3-2y^4+22`

        `->` Bậc: `7`

`b)B-5y^4=A`

`=>B=A+5y^4`

`=>B=5x^4y^3-2y^4+22+5y^4`

`=>B=5x^4y^3+3y^4+22`

a=2, b=0

a ) \(x^2.\frac{y^3}{5}=\frac{A}{35.\left(x+y\right)}\)

\(\Leftrightarrow5A=35.x^2.y^3.\left(x+y\right)\)

\(\Leftrightarrow A=7x^2y^3\left(x+y\right)\)

b ) \(\frac{x^2-4x+4}{x^2-4}=\frac{x-2}{A}\)

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

\(\Leftrightarrow A=\frac{\left(x-2\right)^2\left(x+2\right)}{\left(x-2\right)^2}=x+2\).

a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)

b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)

Th1: 2x+3 ≥ 0
Khi đó: |2x+3| =x+2
 (2x+3)= x+2
- 2x+3= x+2
-2x-x= 2-3
 x= -1
Th2: 2x+3 < 0
Khi đó: |2x+3|=x+2
 -(2x+3) = x +2
 -2x-3 = x+2
 -3x = 5
 x=-5/3

Vậy x= -1

      x= -5/3

Lớp 6 cugx học dạng v nè

`x/2=y/3 <=> x/8=y/12;

`y/4=z/5 <=> y/12=z/15.`

`<=> x/8=y/12=z/15=(x^2-y^2)/(64-144)=16/80=1/5`.

`@ x/8=1/5 <=> x= 8/5`.

`@ y/12=1/5 <=> y=12/5`.

`@ z/15=1/5 <=> y=15/5`.

Vậy...

Lời giải:

a. Đặt $\frac{x}{2}=\frac{y}{3}=a\Rightarrow x=2a; y=3a$

$x^2-y^2=(2a)^2-(3a)^2=-16$

$\Rightarrow -5a^2=-16\Rightarrow a=\pm \frac{4}{\sqrt{5}}$

Nếu $a=\frac{-4}{\sqrt{5}}$ thì:

$x=2a=\frac{-8}{\sqrt{5}}; y=3a=\frac{-12}{\sqrt{5}}; z=\frac{5}{4}y=-3\sqrt{5}$

Nếu $a=\frac{4}{\sqrt{5}}$ thì:

$x=2a=\frac{8}{\sqrt{5}}; y=3a=\frac{12}{\sqrt{5}}; z=\frac{5}{4}y=3\sqrt{5}$

b.

Nếu $x\geq \frac{-3}{2}$ thì:

$2x+3=x+2$

$\Leftrightarrow x=-1$

Nếu $x< \frac{-3}{2}$ thì:

$-2x-3=x+2$

$\Leftrightarrow -5=3x\Leftrightarrow x=\frac{-5}{3}$

Thử lại thấy 2 giá trị $-1, \frac{-5}{3}$ đều tm

c.

$f(x)=-3x+6=0$

$\Leftrightarrow -3x=-6\Leftrightarrow x=2$

Vậy $x=2$ là nghiệm của đa thức.