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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. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
\(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left(x^2+bx+ax+ab\right)\left(x+c\right)\)
\(=x^3+bx^2+ax^2+abx+cx^2+bcx+acx+abc\)
\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+bcx+cax\right)+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
\(\Rightarrowđpcm\)
Ta có: (x+a)(x+b)(x+c) = x3 + (a+b+c)x2 +(ab+bc+ca)x + abc
VT = (x2+ax+bx+ab)(x+c)
= x3 + ax2 + bx2 + abx + cx2 + cax + bcx + abc (1)
VP = x3 + (a+b+c)x2 +(ab+bc+ca)x + abc
= x3 + ax2 + bx2 + abx + cx2 + cax + bcx + abc (2)
Từ (1) và (2), suy ra:
(x+a)(x+b)(x+c) = x3 + (a+b+c)x2 +(ab+bc+ca)x + abc
TC:a+b+cd=2p=>b+c=2p-a
=>(b+c)2=(2p-a)2
=>b2+2bc+c2=4p2-4pa+a2
=>b2+2bc+c2-a2=4p2-4pa
=>2bc+b2+c2-a2=4p(p-a) ĐPCM
Ta có : \(a+b+cd=2p\Rightarrow b+c=2p-a\)
\(\Rightarrow\left(b+c\right)^2=\left(2p-a\right)^2\)
\(\Rightarrow b^2+2bc+c^2=4p^2-4pa+a^2\)
\(\Rightarrow b^2+2bc+c^2-a^2=4p^2-4pa\)
\(\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
\(\RightarrowĐPCM\)
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 \(\left(x+a\right)\left(x-b\right)\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)
a: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3x^2y+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
a/ Chứng minh:
\(\left(x+a\right)\left(x+b\right)\)
\(=x^2+bx+ax+ab\)
\(=x^2+\left(ax+bx\right)+ab\)
\(=x^2+x\left(a+b\right)+ab=VP\) (đpcm)
b/ Chứng minh:
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)
\(=x^3+cx^2+ax^2+acx+bx^2+bcx+abx+abc\)
\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+bcx+acx\right)+abc\)
\(=x^3+x^2\left(a+b+c\right)+x\left(ab+bc+ac\right)+abc=VP\) (đpcm)