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1) \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=ab\left(a+b\right)-b^2c-bc^2+a^2c-ac^2\)
\(=ab\left(a+b\right)-c\left(b^2-a^2\right)-c^2\left(a+b\right)\)
\(=ab\left(a+b\right)-c\left(a+b\right)\left(a-b\right)-c^2\left(a+b\right)\)
\(=\left(a+b\right)\left(ab-ac+bc-c^2\right)\)
\(=\left(a+b\right)\left[a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a+b\right)\left(b-c\right)\left(a+c\right)\)
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+2ab+b^2-2ab\right)+6a^2b^2\)
\(=\left(a^2+2ab+b^2-3ab\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=\left(a+b\right)^2-3ab+3ab\times\left(-2ab\right)+6a^2b^2\)
\(=-3ab-6a^2b^2+6a^2b^2\)
= - 3ab
\(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=\left(x^3-6x^2y+9xy^2\right)+\left(y^3-6xy^2+9x^2y\right)\)
\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)=x\left(x-3y\right)^2+y\left(y-3x\right)^2\)
b/
\(\left(a+b\right)^3+\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3\)
\(=2a^3+6ab^2=2a\left(a^2+3b^2\right)\)
c/
\(\left(a+b\right)^3-\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)
\(=6a^2b+2b^3=2b\left(b^2+3a^2\right)\)
d/
\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-\left(3a^2b+3ab^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
e/
\(a^3-b^3=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)\)
\(a)\)\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d+a+b-c-d}{a-b+c-d+a-b-c+d}=\frac{2\left(a+b\right)}{2\left(a-b\right)}=\frac{a+b}{a-b}\) \(\left(1\right)\)
Lại có :
\(\frac{a+b+c+d}{a-b+c-d}=\frac{a+b-c-d}{a-b-c+d}=\frac{a+b+c+d-a-b+c+d}{a-b+c-d-a+b+c-d}=\frac{2\left(c+d\right)}{2\left(c-d\right)}=\frac{c+d}{c-d}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)\(\Leftrightarrow\)\(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\) \(\left(3\right)\)
Lại có :
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\) \(\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\) suy ra \(\frac{a}{c}=\frac{b}{d}\) ( đpcm )
Chúc bạn học tốt ~
\(b)\)\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\) ( vì \(a+b+c=0\) )
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
Vậy ...
Chúc bạn học tốt ~
Chứng minh rằng:
a, a3+b3= (a+b)3-3ab(a+b)
b, a3-b3= (a-b)3+3ab(a-b)
làm kiểu như vở bài tập nha
a) Biến đổi vế phải ta có::
\(\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=a^3+b^3=VT\)
=>đpcm
b) Biến đổi vế phải ta có:
\(\left(a-b\right)^3+3ab\left(a-b\right)=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2=a^3-b^3=VT\)
=>đpcm
2) b)
Do \(a+b+c=9\Rightarrow\left(a+b+c\right)^2=81\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=81\)
\(\Rightarrow2\left(ab+bc+ac\right)=81-141=-60\)
\(ab+bc+ac=-60:2=-30\)
a, B=x^3 + 3xy +y^3 = x^3 +3xy(x+y)+y^3 (vì x+y=1)
= (x+y)^3
= 1^3 =1
b, (a+b+c)^2 =a^2 +b^2 +c^2 +2ab +2bc +2ac
9^2 = 141 +2(ab+bc+ac)
-60 = 2(ab+bc+ac)
ab+ac+bc=-30
Vậy M=-30
c, N =(x+y)^3 -3(x+y)(x^2+y^2) +2(x^3+y^3)
= x^3 + 3x^2 .y + 3xy^2 + -3(x^3+xy^2 +x^2 .y+y^3)+ 2x^3 +2y^3
= x^3 +3x^2 .y + 3xy^2 - 3x^3 -3xy^2 -3x^2 .y -3y^3 +2x^3 +2y^3
= 0
Vậy N=0 .Chúc bạn học tốt.
Ta có: a^3+b^3 = (a+b)(a^2-ab+b^2)
= a^3-a^2b+ab^2+a^2b-ab^2+b^3
= a^3-3a^2b+2a^2b+3ab^2-2ab^2+3a^2b-2a^2b-3ab^2+2ab^2+b^3
= (a^3+3a^2b+3ab^2+b^3)-(3a^2b+3ab^2)+(2a^2b-2a^2b)+(2ab^2-2ab^2)
= (a+b)^3-3ab(a+b) (đpcm)
a3 + b3 = ( a + b ) 3 - 3ab( a + b )
a3 + b3 =a^3+3a^2b+3ab^2-3a^b-3ab^2
a3 + b3 =a^2+b^2(đpcm)