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\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)
a, \(x^3+y^3+z^3=3xyz\Rightarrow x^3+y^3+z^3-3xyz=0\)( 1 )
Nhận xét : \(\left(x+y\right)^3=x^3+y^3+3x^2y+3xy^2\Rightarrow x^3+y^3=\left(x+y\right)^3-3x^2-3xy^2\)
Thay vào ( 1 ) ta có :
\(\left(x+y\right)^3+c^3-3x^2y-3xy^2-3xyz\)
\(=\left(z+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(z+y+z\right)\left(z^2+2xy+y^2-xz-yz+z^2\right)-3xyz\left(z+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(z^2+x^2+y^2-xy-yz-xz\right)\)
Vì theo đầu bài ta có: \(x+y+z=0\)nên ta có ( DPCM ) ..... học cho tốt nhé!
\(a)x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2+z^3-3xyz=0\)
\(\) \(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(\right.\) \(\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(\right.\) \(x^2+2xy+y^2-xz-yz+z^2-3xy)=0\)
Mà \(x+y+z=0\)
\(\Rightarrow0=0\left(đpcm)\right.\)
\(b)\left(x^2y^2+y^2z^2+x^2z^2+2\left.x^2yz+2xy^2z+2xyz^2\right)\right.=x^2y^2+y^2z^2+x^2z^2\)
\(\Leftrightarrow2\left(\right.\) \(x^2yz+xy^2z+xyz^2)=0\)
\(\Leftrightarrow2\left(x+y+z\right)\left(xyz\right)=0\)
Mà \(x+y+z=0\)
\(\Rightarrow0=0\left(đpcm\right)\)
\(c)\) Ta có:\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(\right.\) \(x^2yz+xy^2z+xyz^2)=0\)
\(\Rightarrow2\left(\right.\) \(xy+yz+xz^{})=-\left(\right.\) \(x^2+y^2+z^2)\)
\(\Rightarrow4\left(\right.\) \(xy+yz+xz)^2=\) \(x^4+y^4+z^4+2\left(\right.\) \(x^2y^2+y^2z^2+x^2z^2)\left(1\right)\)
Mà ta có: \(\left(xy+yz+xz\right)^2=x^2y^2+y^2z^2+x^2z^2\) (theo câu b)
\(\Leftrightarrow2\left(xy+yz+xz\right)^2=2\left(\right.\) \(x^2y^2+y^2z^2+x^2z^2)\left(2\right)\)
\(\left(1\right)-\left(2\right)\Leftrightarrow2\left(xy+yz+xz\right)^2=x^4+y^4+z^4\left(đpcm\right)\)
2.Câu hỏi của Lãnh Hàn Thần - Toán lớp 8 - Học toán với OnlineMath
a) Có a +b +c=0
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(ab+bc+ac\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left[a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right]\)\(\Rightarrow\)\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
1, mk nhớ k lầm thì mk đã từng làm cho bn rồi ,kq=1/2
2,Dễ CM \(x^2+y^2+z^2\ge xy+yz+xz\) ,dấu "=" xảy ra <=>x=y=z
\(=>\left(x+y+z\right)^2\ge\left(xy+yz+xz\right)+2\left(xy+yz+xz\right)=3\left(xy+yz+xz\right)\)
\(=>9\ge3\left(xy+yz+xz\right)=>xy+yz+xz\le\frac{9}{3}=3\)
=>GTLN của xy+yz+xz=3
3)x3+y3+z3=3xyz
<=>x3+y3+z3-3xyz=0
<=>(x+y+z)(x2+y2+z2-xy-yz-xz)=0
<=>x+y+z=0 hoặc x2+y2+z2-xy-yz-xz=0
(+)x+y+z=0 thì x+y=-z;y+z=-x;x+z=-y
thế vô P =-1
(+)x2+y2+z2-xy-yz-xz=0
TH này thì x=y=z
thế vô P=2
Ta có:
\(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow a^2+2ab+b^2=c^2\)
\(\Leftrightarrow a^2+b^2-c^2=-2ab\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2=4a^2b^2\)
\(\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=\left(a^2+b^2+c^2\right)^2\)
PS: Lỡ tay ghi a, b, c rồi nên dùng a, b, c luôn nha.
2/ a+b+c=0 suy ra (a+b+c)2=0
-> a2+b2+c2+2ab+2ac+2bc=0
Mà ta có a2+b2+c2=14 nên thu được ab+ac+bc = -7
->(ab+ac+bc)2 = (-7)2 -> a2b2+a2c2+b2c2+2abc(a+b+c)=49
->a2b2+a2c2+b2c2=49
Lại có (a2+b2+c2)2=a4+b4+c4+2a2b2+2a2c2+2b2c2=142
Suy ra a4+b4+c4+2.49=196
Ta thu được a4+b4+c4=98
sai vi chung minh cau 1 lech sang cau 2