Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
có a+b+c = 0
=> a^2+b^2+c^2+2(ab+bc+ac) = 0
mà a^2+b^2+c^2 = 2
=> ab+bc+ac = -1
=> a^2b^2+b^2c^2+a^2c^2 + 2ab^2c+2a^2bc+2abc^2 = 1
=>a^2b^2+b^2c^2+a^2c^2 + 2abc(b+a+c) = 1
=>a^2b^2+b^2c^2+a^2c^2 = 1
Ta bình phương cái a^2+b^2+c^2 lên
đc là
a^4+b^4+c^4 + 2a^2b^2+2a^2c^2+2b^2c^2=4
=> a^4+b^4+c^4 + 2(a^2b^2+a^2c^2+b^2c^2) = 4
mà ở trên là a^2b^2+b^2c^2+a^2c^2 = 1
=> a^4+b^4+c^4 +1 =4
a^4+b^4+c^4 = 3 :3
nhớ thank nha :3
Ta có: a + b + c = 0
=> ( a + b + c )2 = 0
=> a2 + b2 + c2 + 2ab +2ac+ 2bc = 0
=> 2 + 2( ab + ac + bc ) = 0
=> 2( ab + ac +bc ) = - 2
=> ab + ac + bc = -1
=> ( ab + ac + bc )2 = 1
=> a2b2 + a2c2 + b2c2 + 2a2bc + 2ab2c + 2abc2 = 1
=> a2b2 + a2c2 + b2c2 + 2abc( a + b + c ) = 1
=> a2b2 + a2c2 + b2c2 + 2abc x 0 = 1
=> a2b2 + a2c2 + b2c2 = 1 ( * )
Ta có: a2 + b2 + c2 = 2
=> ( a2 + b2 + c2 )2 = 22
=> a4 + b4 + c4 + 2a2b2 + 2a2c2 + 2b2c2 = 4
=> a4 + b4 + c4 + 2( a2b2 + a2c2 + b2c2 ) = 4
Từ ( * ) => a4 + b4 + c4 + 2 x 1 = 4
=> a4 + b4 + c4 = 4 - 2 = 2
~~~~
Phần còn lại tương tự, cậu tự làm nhóe :3 Chúc cậu học tốt ~~
=a, a(b2+c2)+b(a2+c2)+c(a2+b2)+2abc
= ab2+ac2+ba2+bc2+ca2+cb2+2abc
= c2(a+b)+ab(a+b)+c(a2+b2+2ab)
= c2(a+b)+ab(a+b)+c(a+b)2
= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)
= (a+b)(c2+ab+ca+cb)
= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
=(a+b)(a+c)(b+c)
b, a(b-c)3+b(c-a)3+c(a-b)3
= a(b-c)3-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)2(a-b)-3b(b-c)(a-b)2-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(b-c+a-b)-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(a-c)-b(a-b)3+c(a-b)3
= (b-c)3(a-b)-3b(b-c)(a-b)(a-c)-(a-b)3(b-c)
= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)
=(b-c)(a-b)(b2-2bc+c2-3ab+3bc-a2+2ab-b2)
= (b-c)(a-b)(c2-a2+bc-ab)
= (b-c)(a-b)\(\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
= (b-c)(a-b)(c-a)(c+a+b)
c, a2b2(a-b)+b2c2(b-c)+c2a2(c-a)
= a2b2(a-b)-b2c2\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c2a2(c-a)
= a2b2(a-b)-b2c2(a-b)-b2c2(c-a)+c2a2(c-a)
= b2(a-b)(a2-c2)+c2(c-a)(a2-b2)
= b2(a-b)(a-c)(a+c)-c2(a-c)(a-b)(a+b)
= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)
= (a-c)(a-b)(b2a+b2c-c2a-c2b)
= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)
= (a-c)(a-b)(b-c)(ab+ac+bc)
d, a4(b-c)+b4(c-a)+c4(a-b)
= a4(b-c)-b4[(b-c)+(a-b)]+c4(a-b)
= (b-c)(a4-b4)+(a-b)(c4-b4)
= (b-c)(a2-b2)(a2+b2)+(a-b)(c2-b2)(c2+b2)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b2+c2)
= (b-c)(a-b)(a3+ab2+ba2+b3-bc2-b3-cb2-c3)
= (b-c)(a-b)(a3+ab2+ba2-bc2-c3-cb2)
= (b-c)(a-b)(a3-c3)+b2(a-c)+b(a2-c2)
= (b-c)(a-b)(a-c)(a2+ac+c2)+b2(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a2+ac+c2+b2+ab+ac)
= (a-b)(b-c)(c-a)(a2+b2+c2+ab+bc+ca)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow ab+bc+ca=-\frac{1}{2}\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Lại có:\(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Rightarrow a^4+b^4+c^4+\frac{1}{2}=1\)
\(\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)
Bài 3:
a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)
\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)
\(\Leftrightarrow x^3-27-x^3+16x=21\)
=>16x=48
hay x=3
b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\)
=>-2x=4-8=-4
hay x=2
Ta có \(\left(a^2+b^2+c^2\right)^2=4\Rightarrow a^4+b^4+c^4=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
Mà \(\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)=2\)
=> \(ab+bc+ca=-1\Rightarrow\left(ab+bc+ca\right)^2=1\)
=> \(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\Rightarrow a^2b^2+b^2c^2+c^2a^2=1\)
=> \(a^4+b^4+c^4=4-2=2\)
^.^
Theo bài ra ta có : a2 + b2 + c2 = 2 .
Do đó : ( a2 + b2 + c2 )2 = 22 .
⇒ a4 + b4 + c4 = 4 .
Vậy a4 + b4 + c4 = 4 .
\(Mik\)\(làm\)\(rồi\)\(nhưng\)\(k\)\(biết\)\(đúng\)\(k???\)\(Admin\)\(cho\)\(ý\)\(kiến\)\(nha!!!\)
\(Ta\)\(có:\)\(a^2+b^2+c^2=10\Rightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)\)
\(=10^2=100=a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)\)
\(\Rightarrow a^4+b^4+c^4=100-\left(2\left(a^2b^2+a^2c^2+b^2c^2\right)\right)\)
\(Ta\)\(có:\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)\)
\(0=10+2\left(ab+ac+bc\right)\Rightarrow2\left(ab+ac+bc\right)=-10\)
\(\Rightarrow ab+ac+bc=-5\)
\(\left(ab+ac+bc\right)^2=a^2b^2+a^2c^2+b^2c^2+2\left(a^2bc+ab^2c+abc^2\right)\)
\(\left(-5\right)^2=25=a^2b^2+a^2c^2+b^2c^2+2\left(a^2bc+ab^2c+abc^2\right)\)
\(25=a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)\)
\(25=a^2b^2+b^2c^2+a^2c^2+2abc.0\Rightarrow a^2b^2+a^2c^2+b^2c^2=25\)
\(Vậy\)\(a^4+b^4+c^4=100-\left(2.25\right)=100-50=50\)
\(Mong\)\(ONLINE\)\(MATH\)\(ủng\)\(hộ\)