\(B=-2x^2+4x-5\)

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

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

\(=-2\left[\left(x-1\right)^2+\frac{3}{2}\right]\)

\(=-2\left[\left(x-1\right)^2\right]-3\le3< 0\forall x\)

\(B=-2x^2+4x-5\)

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

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

\(B=-2\left[\left(x-1\right)^2+\frac{3}{2}\right]\)

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

Mà \(\hept{\begin{cases}-2\left(x-1\right)^2\le0\forall x\\-3< 0\end{cases}\Rightarrow B< 0\forall x}\)

a) \(2x^2+4x+2\)

\(=2\left(x^2+2x+1\right)\)

\(=2\left(x^2+x+x+1\right)\)

\(=2\left[x\left(x+1\right)+\left(x+1\right)\right]\)

\(=2\left(x+1\right)^2\)

b) \(9\left(2x-5\right)-6y\left(5-2x\right)\)

\(=9\left(2x-5\right)+6y\left(2x-5\right)\)

\(=\left(9+6y\right)\left(2x-5\right)\)

\(=3\left(3+2y\right)\left(2x-5\right)\)

\(A=\left(2x-1\right)^2-\left(5+x\right)\left(5-x\right)+4x\)

\(=4x^2-4x+1-\left(25-x^2\right)+4x\)

\(=4x^2-4x+1-25+x^2+4x\)

\(=5x^2-24\)

Thay x = -2 vào bt ,ta được: \(5.\left(-2\right)^2-24=-4\)

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

\(=3x^4-15x^3+6x^2\)

tìm x biết 2x+5 chia hết cho 3x-1

\(=1-4x-x\left(x^2-4\right)\)

\(=1-4x-x^3-4x\)

\(=1-x^3-8x\)

\(5x\left(x-2y\right)+2\left(2y-x\right)^2\)

\(=5x\left(x-2y\right)-2\left(x-2y\right)^2\)

\(=\left(x-2y\right)\left[5x-2\left(x-2y\right)\right]\)

\(=\left(x-2y\right)\left(5x-2x+4y\right)\)

\(=\left(x-2y\right)\left(3x+4y\right)\)

Câu trên nhầm nhé là 4( x - 2y ) ( 2x -y )

\(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(4-y\right)^2-\left(4-y\right)^3\)

\(=\left(4-y\right)^2\left[7x-\left(4-y\right)\right]\)

\(NếuA+B=C\)

\(B=C-A\)

\(A=C-B\)

\(NếuA.B=C\)

\(B=\frac{C}{A}\)

\(A=\frac{C}{B}\)

bn làm đúg r đó

Ta có: x²y + xy - 2x² - 3x + 4 = 0

=> x²(y - 2) + x(y - 2) - (x + 1) = -5

=> (x² + x)(y - 2) - (x + 1) = -5

=> x(x + 1)(y - 2) - (x + 1) = -5

=> (x - 1)[x(y - 2) - 1] = -5

=>  x - 1; x(y - 2) - 1 \(\in\)Ư(-5) = {1; -1; 5; -5}

Với : x - 1 = 1           => x = 2

      x(y - 2)  - 1 = -5 => x(y - 2) = -4   => y - 2 = -2      => y = 0

x - 1 = -1                => x = 0

x(y - 2) - 1 = 5  => x(y - 2) = 6       (ktm vì x = 0)

x - 1 = 5                => x = 6

x(y - 2) - 1 = -1      => x(y - 2) = 0                 => y - 2 = 0          => y = 2

x - 1 = -5              => x = -4

x(y - 2) - 1 = 1    => x(y - 2) = 2              => y - 2 = -1/2             => y = 3/2

Vậy ...