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a) số lẻ wa
b)(x - 1)3 - (x + 3) . (x2 - 3x +9) + 3 . (x + 2) . (x - 2) = 2
\(VT=3x-40\)
\(\Leftrightarrow3x-40=2\)
\(\Leftrightarrow3x=42\)
\(\Leftrightarrow x=14\)
a) (a+b)3- (a-b)3- 2ab
=a3+3a2b+3ab2+b3-(a3-3a2b+3ab2-b3)-2ab
=a3+3a2b+3ab2+b3-a3+3a2b-3ab2+b3-2ab
=2b3+6a2b-2ab
b) (x-2). (x2+2x+4) - x.(x2-1)+x+5
=x3-8-x3+x+x+5
=2x-3
a) 3(x - 1)2 - 3x(x - 5) = 3(x2 - 2x + 1) - 3x2 + 15x = 3x2 - 6x + 3 - 3x2 + 15x = 9x + 3 = 21 => x = (21 - 3) : 9 = 18 : 9 = 2
b) 3(x + 2)2 + (2x - 1)2 - 7(x + 3)(x - 3) = 3(x2 + 4x + 4) + 4x2 - 4x + 1 - 7(x2 - 9) = 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63
= 8x + 76 = 36 => x = (36 - 76) : 8 = -40 : 8 = -5
a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)
B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
a) \(A=\left(x^2-\frac{1}{2}x\right)^2+\frac{3}{4}\left(x+\frac{2}{3}\right)^2+\frac{2}{3}>0\)
Ko biết xét khoảng:v
a ) VP = \(\left(x+a\right).\left(x+b\right)=x^2+bx+ax+ab\)
VT = \(x^2+\left(a+b\right).x+ab=x^2+ax+bx+ab\)
\(\Rightarrow VT=VP\)
b ) VP : \(\left(x+a\right).\left(x+b\right)\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\) ( Vế đầu áp dụng luôn ở câu a )
\(=x^2.x+x^2.c+bx.x+bx.c+ax.x+ax.c+ab.x+ab.c\)
\(=x^3+cx^2+bx^2+cbx+ax^2+cax+abx+abc\)
\(=x^3+\left(cx^2+bx^2+ax^2\right)+\left(cbx+cax+abx\right)+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+ac+bc\right).x+abc\)
Vậy \(\left(x+a\right).\left(x+b\right).\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)
a) VP =\(\left(x+a\right)\left(x+b\right)=x^2+bx+\text{ax+ab}\)
\(VT=x^2+\left(a+b\right).x+ab=x^2+ax+bx+ab\\ =>VT=VP\)
b) VP : \(\left(x+a\right).\left(x+b\right).\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\)( Vế đầu áp dụng luôn ở câu a )
\(=x^2.x+x^2.c+bx.x+bx.c+\text{ax}.x+\text{ax}.c+ab.c+ab.c\\ =x^3+cx^2+bx^2-cbx+\text{ax}^2+ca.x+ab.x+abc\\ \)
\(=x^3+\left(cx^2+bx^2+\text{ax}^2\right)-\left(cbx+c\text{ax}+abx\right)+abc\\ =x^3-\left(a+b+c\right)x^2+\left(ab+ac+bc\right).x+abc\)
Vậy \(\left(x+a\right)\left(x-b\right)\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)
\(\left(x+a\right)\left(x+b\right)=x\left(x+b\right)+a\left(x+b\right)\)
\(=x^2+xb+ax+ab\)
\(=x^2+\left(a+b\right)x+ab\)
Áp dụng :
a/ \(\left(x+5\right)\left(x+2\right)=x^2+\left(5+2\right).x+5.2=x^2+7x+10\)
b/ \(\left(x+8\right)\left(x-3\right)=x^2+\left(8-3\right)x+8.\left(-3\right)=x^2+5x-24\)
c/ \(\left(x-7\right)\left(x-4\right)=x^2+\left[\left(-7\right)+\left(-3\right)\right]x+\left(-7\right)\left(-3\right)=x^2-10x+21\)
d/ \(\left(x-9\right)\left(x+1\right)=x^2+\left(-9+1\right)x+\left(-9\right).1=x^2-8x-9\)
phân tách (x + a)( x+ b) và x^2 +( a + b )x + ab để biết được cách nó ghép