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

<=> -x² + 8x - 9 = 3x - x² - 2

<=> -x² + 8x = 3x - x² - 2 + 9

<=> -x² + 8x = 3x - x² - 7

<=> -x² + 8x - (-x² + 3x) = 3x - x² - 7 - (-x³ + 3x)

<=> 5x = 7

<=> x = 5/7

=> x = 5/7

\(\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)=\left(1-x\right)\)

\(-x^2+8x-9=3x-x^2-2\)

\(-x^2+8x=3x-x^2-7\)

\(5x=7\)

\(x=7:5\)

\(x=\frac{7}{5}\)

\(\left(9x-5\right)\left(2x-1\right)-\left(3x+2\right)\left(6x-7\right)=-1\)

\(\Rightarrow\left(18x^2-9x-10x+5\right)-\left(18x^2-21x+12x-14\right)=-1\)

\(\Rightarrow18x^2-19x+5-18x^2+9x+14=-1\)

\(\Rightarrow-10x+19=-1\)

\(\Rightarrow-10x=-20\)

\(\Rightarrow x=2\)

(9x - 5).(2x - 1) - (3x + 2).(6x - 7) = -1

<=> -10x + 19 = -1

<=> -10x = -1 - 19

<=> -10x = -20

<=> x = 10

=> x = 10

Góc C = 180 - 130 = 50 

Góc B = 360 - (80+120+50) =110

D = (y² + 2)(y - 4) - (2y² + 1)(y - 2)

D = y².y + y².(-4) + 2.y + 2.(-4) + (-2y²).y + (-2y²).(-2) + 1.y + 1.(-2)

D = y³ - 4y² + 2y - 8 - 2y³ + 4y² + y - 2

D = (y³ - 2y³) + (-4y² + 4y²) + (2y + y) + (-8 - 2)

D = -y³ + 3y - 10 (1)

Thay y = -2/3 vào (1), ta có

D = -y³ + 3y - 10

D = -(-2/3)³ + 3.(-2/3) - 10

D = 8/27 - 2 - 10

D = -316/27

Vậy: ...

\(\text{Vì }a^2-b^2=c^2-d^2=1\Leftrightarrow\hept{\begin{cases}a^2=b^2+1\left(1\right)\\d^2=c^2-1\left(2\right)\end{cases}}\)

\(\text{Ta có: }a^2d^2-a^2d^2=0\)

\(\Rightarrow a^2.\left(c^2-1\right)-d^2.\left(b^2+1\right)=0\)

\(\Rightarrow a^2c^2-b^2d^2-a^2-d^2=0\)

\(\Rightarrow a^2c^2-b^2d^2=a^2+d^2\)

Vậy \(a^2c^2-b^2d^2=a^2+d^2\)

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

<=>(2x - 1). (3 - x)  = ( 1-x ) . (x - 2) - (x -2 ) . (x + 3)

<=>(2x - 1). (3 - x)  =  (x - 2) (1-x-x-3)

<=>(2x - 1). (3 - x)  =  (x - 2) (-2x-2)

\(\Leftrightarrow6x-2x^2-x+x=-2x^2-2x+4x+4\)

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

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\)

\(a^2+ab+b^2=\)\(a^2+2.a.\frac{b}{2}+\frac{b^2}{4}+\frac{3b^2}{4}=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge0\)

Xét trường hợp : \(a^2+ab+b^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(a+\frac{b}{2}\right)^2=0\\\frac{3b^2}{4}=0\end{cases}\Leftrightarrow}a=b=0\)vô lí vì a khác b

=> \(a^2+ab+b^2>0\)

\(a^2+ab+b^2=\left(a+\frac{1}{2}b\right)^2+\frac{3}{4}b^2>0\)

(với a\(\ne\)b)

Học tốt!!!!!!!!!!!!!

\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}\)

a,\(ĐKXĐ:x\ne0;x\ne3;x\ne1\)

\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}=\left(\frac{9}{x\left(x-3\right)}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3\left(1-x\right)}\)

\(=\left(\frac{9+\left(x-2\right)\left(x-3\right)-x.x}{x\left(x-3\right)}\right).\frac{x}{3\left(1-x\right)}=\frac{9+x^2-5x+6-x^2}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}\)

\(=\frac{-5x+15}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=\frac{-5\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=-\frac{5}{3\left(1-x\right)}\)

b, \(x=\frac{1}{2}\)

\(\Rightarrow P=-\frac{5}{3\left(1-\frac{1}{2}\right)}=-\frac{5}{3.\frac{1}{2}}=-5:\frac{3}{2}=-\frac{10}{3}\)

c, Để \(P\in z\)thì \(3\left(1-x\right)\inƯ\left(5\right)=\left(-5;-1;1;5\right)\)

\(3\left(1-x\right)=-5\Rightarrow1-x=-\frac{5}{3}\Rightarrow x=\frac{8}{3}\)

\(3\left(1-x\right)=-1\Rightarrow1-x=-\frac{1}{3}\Rightarrow x=\frac{4}{3}\)

\(3\left(1-x\right)=1\Rightarrow1-x=\frac{1}{3}\Rightarrow x=\frac{2}{3}\)

\(3\left(1-x\right)=5\Rightarrow1-x=\frac{5}{3}\Rightarrow x=-\frac{2}{3}\)

A = ( x - 2 )² - ( 2x + 1 )² 

A = x² - 4x + 4 - 4x² + 4x + 1 

A = - 3x² + 5 

B = ( x - 2y )² - ( x - 2y ) . ( 2y + x ) 

B = x² - 4xy + 4y² - ( 2xy + x² - 4y² - 2xy ) 

B = x² - 4xy + 4y² - 2xy - x² + 4y² + 2xy 

B = 8y² - 4xy