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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 + b + c = 0
=> a + b = -c
=> (a + b)2 = (-c)2
=> a2 + b2 + 2ab = c2
=> a2 + b2 - c2 = -2ab
=> (a2 + b2 - c2)2 = (-2ab)2
=> a4 + b4 + c4 + 2a2b2 - 2a2c2 - 2b2c2 = 4a2b2
=> a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2a2c2
Khi đó a2 + b2 + c2 = 14
<=> (a2 + b2 + c2)2 = 142
=> a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2a2c2 = 196
=> a4 + b4 + c4 + a4 + b4 + c4 = 196 (Vì a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2a2c2)
=> 2(a4 + b4 + c4) = 196
=> a4 + b4 + c4 = 98
\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)
1. Cần sửa lại thành \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
Ta có : \(a^2+b^2+c^2-3=2\left(a+b+c\right)\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\) \(\Leftrightarrow a=b=c=1\)
2. Cần sửa lại thành : \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
Ta có : \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\) \(\Leftrightarrow a=b=c\)
3. Ta có : \(a+b+c=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=\frac{-\left(a^2+b^2+c^2\right)}{2}=-\frac{1}{2}\)\(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Lại có : \(1=\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2+b^2+c^2\right)=1-2.\frac{1}{4}=\frac{1}{2}\)
tài năng toán học hoàng lê bảo ngọc,tui công nhận bn 3 lần/ngày
a) Ta có : x(x + 4)(x - 4) - (x2 + 1)(x2 - 1)
= x(x2 - 16) - (x4 - 1)
= x3 - 16x - x4 + 1
= x(x2 - 16 - x3) + 1
\(a,x.\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x\left(x^2-16\right)-x^4+1=x^3-16x=x^4+1\)
\(a^2+b^2+c^2=1\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=1\Leftrightarrow0-2\left(ab+bc+ca\right)=1\Leftrightarrow ab+bc+ca=-\frac{1}{2}\)
\(M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)
\(=1-2\left(\frac{1}{4}-2abc\left(a+b+c\right)\right)=1-\frac{1}{2}+4abc.0=\frac{1}{2}\)
dùng thước đo và so sánh BH và HC nếu ab = ac thì có thể suy ra HB = HC không
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Rightarrow2ac+2bc+2ab=-14\)
\(\Rightarrow ac+ab+bc=-7\)
\(\left(ac+bc+ab\right)^2=49\)
\(a^2c^2+b^2c^2+a^2b^2+2abc^2+2ab^2c+2a^2bc=49\)
\(\Rightarrow a^2c^2+b^2c^2+a^2b^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2c^2+b^2c^2+a^2b^2=49\)
Có \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)^2=196\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)
\(\Rightarrow a^4+b^4+c^4=196-2.49=196-98=98\)