Выделите квадрат двучлена из квадратного трехчлена: 1) 2x²-3x+7 2) -3x²+4x-1 3) 5x²-3x 4)...

Решите задачу:

$Ax^2+Bx+C=A(x^2+ \frac{B}{A} x+ \frac{C}{A}) = \\\ =A(x^2+2\cdot \frac{B}{2A} x+(\frac{B}{2A})^2-(\frac{B}{2A})^2+ \frac{C}{A}) = \\\ =A((x+\frac{B}{2A})^2-\frac{B^2}{4A^2}+ \frac{4AC}{4A^2}) = A((x+\frac{B}{2A})^2+ \frac{4AC-B^2}{4A^2}) = \\\ =A(x+\frac{B}{2A})^2+ \frac{4AC-B^2}{4A}$

$2x^2-3x+7= 2(x^2- \frac{3}{2} x+ \frac{7}{2} )= 2(x^2-2\cdot \frac{3}{4} x+ ( \frac{3}{4} )^2-( \frac{3}{4} )^2+\frac{7}{2} )= \\\ =2((x- \frac{3}{4} )^2- \frac{9}{16} +\frac{56}{16} )= 2((x- \frac{3}{4} )^2+\frac{65}{16} )=2(x- \frac{3}{4} )^2+\frac{47}{8}$

$-3x^2+4x-1=-3(x^2- \frac{4}{3} x+ \frac{1}{3} )= \\\ =-3(x^2-2\cdot \frac{2}{3} x+ ( \frac{2}{3} )^2-( \frac{2}{3} )^2+\frac{1}{3} )= -3((x- \frac{2}{3} )^2- \frac{4}{9} +\frac{3}{9} )= \\\ =-3((x- \frac{2}{3} )^2- \frac{1}{9} )= -3(x- \frac{2}{3} )^2+ \frac{1}{3}$

$5x^2-3x=5(x^2- \frac{3}{5} x)=5(x^2-2\cdot \frac{3}{10} x+(\frac{3}{10})^2-(\frac{3}{10})^2)= \\\ =5((x-\frac{3}{10})^2-\frac{9}{100})=5(x-\frac{3}{10})^2-\frac{9}{20}$

$-4x^2+8x=-4(x^2-2x)=-4(x^2-2\cdot1\cdot x+1^2-1^2)= \\\ =-4((x-1)^2-1)=-4(x^2-1)^2+4$