Chúng minh các hằng đẳng thức
a) \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
b) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
Chúng minh các hằng đẳng thức
a) \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
b) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
a) Biến đổi vế trái ta có:
\(\left(x+a\right)\left(x+b\right)\)
= \(x^2+xb+xa+ab\)
= \(x^2+\left(a+b\right)x+ab=VP\)
Vậy đẳng thức đc CM
b) Biến đổi VT ta có:
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
= \(\left(x^2+xa+xb+ab\right)\left(x+c\right)\)
= \(x^3+x^2a+x^2b+x^2c+xab+xac+xbc+abc\)
= \(x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)= VP
Vậy đẳng thức đc CM
Chứng minh đẳng thức:
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
Giúp bài này nha
Chứng minh hằng đẳng thức :
\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)
\(=ab+bc+ca-x^2\)
Biết \(2x=a+b+c\)
\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\) là Vế Phải
\(ab+bc+ca-x^2\)là vế trái .
Biến đổi VP ta có :
\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)
\(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ab\)
\(=3x^2-2x\left(a+b+c\right)+\left(ab+bc+ca\right)\)
Thay \(a+b+c\)là \(2x\)ta được :
\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)= VP
\(=-x^2+ab+bc+ca=VT\)
=> đpcm
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chứng minh rằng
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
chứng minh các hằng đẳng thức
a. (x+a).(x+b) = \(x^2+\left(a+b\right).x+a.b\)
b. (x+a).(x+b).(x+c)= \(x^3+\left(a+b+c\right).x^2+\left(ab+bc+ca\right).x+abc\)
c. (a+b+c).\(\left(a^2+b^2+c^2-ab-bc-ca\right)=a^3+b^3+c^3-3abc\)
a)(x+a)(x+b)
=x(x+b)+a(x+b)
=x2+xb+ax+ab
=x2+(a+b).x+a.b
Vậy (x+a)(x+b)=x2+(a+b).x+a.b
b)(x+a)(x+b)(x+c)
=x(x+b)(x+c)+a(x+b)(x+c)
=(x2+xb)(x+c)+(ax+ab)(x+c)
=x2(x+c)+xb(x+c)+ax(x+c)+ab(x+c)
=x3+x2.c+x2.b+xbc+ax2+axc+abx+abc
=x3+(a+b+c).x2+(ab+bc+ca).x+abc
Vậy (x+a)(x+b)(x+c)=x3+(a+b+c).x2+(ab+bc+ca).x+abc
c)(a+b+c)(a2+b2+c2-ab-bc-ca)
=a(a2+b2+c2-ab-bc-ca)+b(a2+b2+c2-ab-bc-ca)+c(a2+b2+c2-ab-bc-ca)
=a3+ab2+ac2-a2.b-abc-a2.c+ba2+b3+bc2-ab2-b2.c-bca+ca2+cb2+c3-cab-bc2-c2.a
=a3+b3+c3 -abc-bca-cab
=a3+b3+c3 -3abc
Vậy (a+b+c)(a2+b2+c2-ab-bc-ca)=a3+b3+c3 -3abc
ngu như con hà cày
ta có:
(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}=\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+x\right)\left(y+z\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\frac{9}{4\left(xy+yz+zx\right)}=\frac{9}{4}\)
Cho a,b,c khác 0 và cho x,y,z tùy ý. Chứng minh rằng: \(\frac{bc\left(a-x\right)\left(a-y\right)\left(a-z\right)}{\left(a-b\right)\left(a-c\right)}+\frac{ca\left(b-x\right)\left(b-y\right)\left(b-z\right)}{\left(b-c\right)\left(b-a\right)}+\frac{ab\left(c-x\right)\left(c-y\right)\left(c-z\right)}{\left(c-a\right)\left(c-b\right)}=abc-xyz\)
Rút gọn biểu thức:
a) \(A=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ca}{\left(b-c\right)\left(b-a\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
b) \(B=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+\dfrac{1}{x^3}}\)
Chứng minh đẳng thức sau :
a) \(x^2+y^2=\left(x+y\right)^2-2xy\)
b)\(\left(a+b\right)^2-\left(a-b\right)\cdot\left(a+b\right)=2b\left(a+b\right)\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=ab\)
a) \(x^2+y^2=x^2+y^2+2xy-2xy=\left(x+y\right)^2-2xy\)
b) \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)=\left(a+b\right)^2-\left(a^2-b^2\right)=a^2+2ab+b^2-a^2+b^2\)
\(=2ab+2b^2=2b\left(a+b\right)\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=2b.2a=4ab\)
a: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
b: \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-a+b\right)\)
\(=2b\left(a+b\right)\)
c: \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\)