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![](https://rs.olm.vn/images/avt/0.png?1311)
(x+2)(x+5)(x+3)(x+4)-24=(X^2+7x+10)(x^2+7x+12)-24. Đặt x^2+7x+10=t đa thức đã cho trở thành t*(t+2)-24=t^2+2t-24=t^2+6t-4t-24=t*(t+6)-4(t+6). =(t+6)(t-4).
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow ab+bc+ca=\frac{-1}{2}\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)( 1 )
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
Mà theo ( 1 ) nên có \(a^2+b^4+c^4=\frac{1}{2}\)
P/S:Hướng lm là như vầy nhé !
Cho a + b + c = 0 và a2 + b2 +c2= 1 Tính giá trị của biểu thức M = a4+b4+c4 Giúp mk vs nha!!
Tham khảo
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có (2x-4)^4 >= 0 khi x = 2
/4-2x />= 0 khi x = 2
Vậy min A = 1986 khi x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{\left(2^4+4\right).\left(6^4+4\right).\left(10^4+4\right).\left(14^4+4\right)}{\left(4^4+4\right).\left(8^4+4\right).\left(12^4+4\right).\left(16^4+4\right)}\)
\(=\frac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right).\left(9^2+1\right).\left(11^2+1\right).\left(13^2+1\right).\left(15^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right).\left(9^2+1\right).\left(11^2+1\right).\left(13^2+1\right).\left(15^2+1\right).\left(17^2+1\right)}\)
\(=\frac{1^2+1}{17^2+1}=\frac{1}{145}\)
\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)\left(14^4+4\right)}{\left(4^4+4\right)\left(8^4+4^4\right)\left(12^4+4\right)\left(16^4+4\right)}\)
\(=\frac{4\left(2^4+6^4+10^4+14^4\right)}{4\left(4^4+8^4+12^4+16^4\right)}\)
\(=\frac{4.76848}{4.90624}\)
\(=\frac{307392}{362496}=\frac{1601}{1888}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
<=> \(\left(a+b\right)^3-3ab\left(a+b\right)+3\left(a+b\right)^2-6ab+4\left(a+b\right)+4=0\)
<=> \(\left[\left(a+b\right)^3+2\left(a+b\right)^2\right]-3ab\left(a+b+2\right)+\left(a+b\right)^2+4\left(a+b\right)+4=0\)
<=> \(\left(a+b\right)^2\left(a+b+2\right)-3ab\left(a+b+2\right)+\left(a+b+2\right)^2=0\)
<=> \(\left(a+b+2\right)\left(\left(a+b\right)^2-3ab+a+b+2\right)=0\)
<=> \(\left(a+b+2\right)\left(a^2+b^2-ab+a+b+2\right)=0\)(1)
Có: \(a^2+b^2-ab+a+b+2=\frac{1}{2}\left[\left(a-b\right)^2+\left(a+1\right)^2+\left(b+1\right)^2\right]+1>0\)
=> (1) <=> a + b + 2 = 0 <=> a + b = -2
Thế vào tìm M .
Cố gắng học tốt giúp đỡ mọi người nhiều hơn nhé! :))))
Số 4 (^v^)