a. 

⇔ \(\left\{{}\begin{matrix}x-2y=4.3-5\\2x+y=3.3\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x-2y=7\\2x+y=9\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}-2x+4y=-14\\2x+y=9\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}5y=-5\\2x+y=9\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}y=-1\\2x-1=9\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}y=-1\\x=5\end{matrix}\right.\)

Vậy nghiệm của hpt là: (5;1)

Đặt C(x)=0

\(\Leftrightarrow-2x\left(2x-3\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow-4x^2+6x-2x+2=0\)

\(\Leftrightarrow-4x^2+4x+2=0\)

\(\Leftrightarrow4x^2-4x-2=0\)

\(\Leftrightarrow\left(2x-1\right)^2=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}\\2x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3}+1\\2x=-\sqrt{3}+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{2}\\x=\dfrac{-\sqrt{3}+1}{2}\end{matrix}\right.\)

Đặt Q(x)=0

\(\Leftrightarrow2\left(x-3\right)-\left(x-1\right)=0\)

\(\Leftrightarrow2x-6-x+1=0\)

\(\Leftrightarrow x=5\)

a. Bạn tự giải

b. \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\4x+2y=6m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\5x=10m-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2m-1\\y=-m+2\end{matrix}\right.\)

\(\dfrac{2}{x}-\dfrac{1}{y}=-1\Rightarrow\dfrac{2}{2m-1}-\dfrac{1}{-m+2}=-1\) (\(m\ne\left\{\dfrac{1}{2};2\right\}\))

\(\Leftrightarrow2\left(-m+2\right)-\left(2m-1\right)=\left(m-2\right)\left(2m-1\right)\)

\(\Leftrightarrow2m^2-m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\dfrac{3}{2}\end{matrix}\right.\)

a/ 3(1 - x) - 5(2x - 2) = 0 

    => 3 - 3x - 10x + 10 = 0

    => -13x = -13

    => x = 1

     Vậy x = 1

b/ |3x - 2| - 4 = 0 => |3x - 2| = 4  

    Suy ra 2 trường hợp:

    Vậy x = 2 , x = -2/3

c/ 2x - x³ = 0 => x.(2 - x²) = 0 

     => x = 0 

    hoặc 2 - x² = 0 => x² = 2 => x = \(\sqrt{2}\)  hoặc x = \(-\sqrt{2}\)

         Vậy \(x=\left\{0;\sqrt{2};-\sqrt{2}\right\}\)

d/ x(1 - 2x) + (2x² - x + 4) = 0 

     => x - 2x² + 2x² - x + 4 = 0

     => 4 = 0 (vô lí)

     Vậy vô nghiệm

đa thức trên có nghiệm khi x=-1

vì (-1)^3+2x^2+2x+1=0

nếu đúng thì bạn cho mình làm quen nha

Lời giải:
a.

a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$

$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$

$\Leftrightarrow 3x+1=5x-3$

$\Leftrightarrow 4=2x$

$\Leftrightarrow x=2$

b.

$(2x-1)(x+3)-(x-2)(x+3)=3x+1$

$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$

$\Leftrightarrow x^2+5x+1=3x+1$

$\Leftrightarrow x^2+2x=0$

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

$\Leftrightarrow x=0$ hoặc $x=-2$

c.

$x^2(x-1)-x(x-1)(x+1)=0$

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

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

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

$\Leftrightarrow x-1=0$ hoặc $-x=0$

$\Leftrightarrow x=1$ hoặc $x=0$

d.

$4x(x-5)-(2x-3)(2x+3)=9$

$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$

$\Leftrightarrow -20x=0$

$\Leftrightarrow x=0$

a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)

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

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow2x=4\)

hay x=2

b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)

\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)

\(\Leftrightarrow x^2+2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)

\(\Leftrightarrow4x^2-20x-4x^2+9=9\)

hay x=0