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( x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²- (x+4)(x-4)+3x²

<=>(x²-25)-(x+3)²+3(x-2)²=(x+1)²+(16-x²)+3x²

<=>x²-25-x²-6x-9+3x²-12x+12=x²+2x+1+16-x²+3x²

<=>3x²-18x-22=3x²+2x+17

<=>-20x=39

=>x=-39/20

Điều kiện xác định:

\(x\ne0;x+5\ne0\)

<=>\(x\ne0;x\ne-5\)

\(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

 \(=\frac{x^2+2x}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

=\(\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2x^2-50}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

=\(\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

\(=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x\left(x^2-x+5x-5\right)}{2x\left(x+5\right)}\)

\(\frac{x\left[x\left(x-1\right)+5\left(x-1\right)\right]}{2x\left(x+5\right)}=\frac{x\left(x-1\right)\left(x+5\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\)

*Với P=1 thì \(\frac{x-1}{2}\)=1

<=>x-1=2

<=>x=3

*Với P= -3 thì \(\frac{x-1}{2}=-3\)

<=>x-1 = -6

<=>x=-5

Mà x\(\ne\)5

nên với P=-3 thì không tìm được x

     với P = 1 thì x=3

a) x = 4,4.

b) x = 1,860147051.

c) x = 0

d) x = 4.

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a) x = 4,4

b) x = 1,860147051

c) x = 0

d) x = 4

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

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

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

    \(\Leftrightarrow-\left(x+2\right)=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

\(b,3x\left(x+5\right)-2\left(x+5\right)=3x^2\)

\(\Leftrightarrow3x^2+15x-2x-10=3x^2\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(15x-2x\right)-10=0\)

\(\Leftrightarrow13x-10=0\Leftrightarrow13x=10\Leftrightarrow x=\frac{10}{13}\)

ta có :  2 ( x - 1 ) - x ( 3 - x ) = x^2

 =>      2x - 2 - 3x

Câu a:

(x + 2)(x -3) - (x -2)(x + 5) = 0

x^2 + 2x - 3x - 6 - x^2 - 5x + 2x + 10 = 0

(x^2 - x^2) + (2x - 3x - 5x + 2x) + (10 - 6) = 0

0 + (-x - 5x + 2x) + 4 = 0

-6x + 2x + 4 = 0

-4x + 4 = 0

4x = 4

x = 4 : 4

x = 1

Vậy x = 1

Câu b:

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

2x^2 - 8x + 3x - 12+ x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20

(2x^2 + x^2) - (8x+5x-3x +2x) - (12 - 10) = 3x^2 - (12x + 5x) + 20

3x^2 - (13x - 3x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - (10x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 - 3x^2 + 17x - 20 = 0

(3x^2 - 3x^2) + (-12x + 17x) - (2 + 20) = 0

0 + 5x - 22 = 0

5x = 22

x = 22/5

Vậy x = 22/5