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Giả sử a3 + b3 + c3 = 3abc, ta có :
a3 + b3 + c3 - 3abc = 0
Đưa về hằng đẳng thức mở rộng a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
<=> (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = 0
Mà a + b + c = 0
=> 0.(a2 + b2 + c2 - ab - bc - ca) = 0 (đúng)
Vậy , với a + b + c = 0 thì
a3 + b3 + c3 = 3abc
1. \(a^3+b^3+c^3-3abc\)
\(=a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2+c^3-3abc\)
\(=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab.\left(a+b+c\right)\)
\(=\left(a+b+c\right).\left[\left(a+b\right)^2-c.\left(a+b\right)+c^2\right]-3ab.\left(a+b+c\right)\)
\(=\left(a+b+c\right).\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right).\left(a^2+b^2+c^2-bc-ab-ca\right)\)
Mà \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right).\left(a^2+b^2+c^2-bc-ab-ca\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\RightarrowĐpcm.\)
2. Dễ rồi.
3.
\(A=2.\left(x-y\right).\left(x^2+xy+y^2\right)-3.\left(x^2+2xy+y^2\right)\)
\(A=4.\left(x^2+xy+y^2\right)-3x^2-6xy-3y^2\)
\(A=4x^2+4xy+4y^2-3x^2-6xy-3y^2\)
\(A=x^2-2xy+y^2\)
\(A=\left(x-y\right)^2\)
Thay \(x-y=2\) vào ta có:
\(A=\left(x-y\right)^2\)\(=2^2=4\)
4. \(A=x^2-3x+5\)
\(A=x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(A=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(\Rightarrow x-\dfrac{3}{2}=0\)
\(\Rightarrow x=\dfrac{-3}{2}\)
\(\Rightarrow Min_A=\dfrac{11}{4}\Leftrightarrow x=\dfrac{-3}{2}\)
\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(B=4x^2-4x+1+x^2+4x+4\)
\(B=5x^2+5\)
Ta có: \(5x^2\ge0\)
\(\Rightarrow5x^2+5\ge0\)
\(\Rightarrow Min_B=5\Leftrightarrow x=0\)
bạn phải tách từng câu ra. chứ kiểu này k ai trả lời cho đâu
2)
a)x2-y2=(x+y).(x-y)=(87+13).(87-13)=100.74=7400
b)x3-3x2+3x-1=(x-1)3=(101-1)3=1003=1000000
c)x3+9x2+27x+27=(x+3)3=(97+3)3=1003=1000000
4)
a)x2-6x+10=x2-6x+9+1=(x-3)2+1>=1>0 voi moi x
b)4x-x2-5= -(x2-4x+5)= -(x2-4x+4+1)= -(x-2)2 - 1<0 voi moi x
\(1)\)
\(a)\)\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100\left(100+1\right)}{2}\)
\(A=5050\)
\(b)\)\(B=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^8+1\right).....\left(2^{64}+1\right)+1\)
\(............\)
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1\)
\(B=2^{128}\)
Chúc bạn học tốt ~
\(1)\)
\(c)\)\(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(C=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-2\left(a+b\right)^2\)
\(C=2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2\)
\(C=2c^2\)
\(2)\)
\(a)\)\(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(VP=a^3+3a^2b+3ab^2+b^3-3ab\left(a+b\right)\)
\(VP=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)\)
\(VP=a^3+b^3=VT\) ( đpcm )
\(b)\)\(VT=a^3+b^3+c^3-3abc\)
\(VT=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(VT=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(VT=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(VT=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\) ( đpcm )
Từ đó suy ra :
\(i)\)\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)\(\Rightarrow\)\(a+b+c=0\)
Hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
Chúc bạn học tốt ~
a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\) *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\) cho nhanh*
b) \(a^3-b^3+c^3+3abc=\left(a^3-3a^2b+3ab^2-b^2\right)+3a^2b-3ab^2+c^3+3abc\)
\(=\left(a-b\right)^3+c^3+\left(3a^2b-3ab^2+3abc\right)\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)
c) \(a^3-b^3-c^3-3abc=\left[a^3-3a^2b+3ab^2-b^3\right]+3a^2b-3ab^2-c^3-3abc\)
\(=\left[\left(a-b\right)^3-c^3\right]+3ab\left(a-b-c\right)=\left(a-b-c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b-c\right)\)
\(=\left(a-b-c\right)\left[a^2-2ab+b^2+ac-bc+c^2+3ab\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
a,(x+1)2
b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}
c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}
a) \(x^3+2x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)=0\)
\(TH1:x+1=0\Leftrightarrow x=-1\)
\(TH2:x^2+x+1=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)
Mà \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)nên loại TH2
Vậy x = 1
Câu a), x = -1 nha, kết luận nhầm
b) \(x^3-4x^2+12x-27=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+9\right)-4x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-7x+9\right)=0\)
\(TH1:x-3=0\Leftrightarrow x=3\)
\(TH2:x^2-7x+9=0\)
\(\cdot\Delta=\left(-7\right)^2-4.9=13\)
Vậy pt của TH2 có 2 nghiệm phân biệt
\(x_1=\frac{7+\sqrt{13}}{2}\);\(x_2=\frac{7-\sqrt{13}}{2}\)
Do a+b+c=0 nên a+b=-c => -(a+b)=c; thay vào ta có:
\(a^3+b^3-\left(a+b\right)^3=a^3+b^3-\left(a^3+3a^2b+3ab^2+b^3\right)\)
\(=-3a^2b-3ab^2=-\left(3ab\left(a+b\right)\right)\)
\(=-\left(-3abc\right)=3abc\)
Từ trên ta có: \(\left(x-3\right)^3+\left(2x-3\right)^3=\left(3\left(x-2\right)\right)^3=\left(3x-6\right)^3\)
\(=\left(x-3+2x-3\right)^3\)
Coi x-3 là a; 2x-3 là b thì 3x- 6 là c;
Mà a+b =c nên : \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)
\(=>3ab\left(a+b\right)=0=>3abc=0\)
\(=>\left\{{}\begin{matrix}x-3=0\\2x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
CHÚC BẠN HỌC TỐT......
Click tai đây để xem lời giải
sai oy bạn ơi , cái này là a+b+c=0 nhé
An Trịnh Hữu à mk nhìn nhầm
cho mình hỏi phần tìm x, phần CM mình lm đc rồi
cho mình hỏi phần tìm x, phần CM mình lm đc rồi
ak nham
cảm ơn bạn !