Tính giá trị P = \(a^2-b^2-c^2-2bc-20a\) , biết a+b+c=10
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\(a^2-b^2-c^2-2bc-2ac-2ab\)
\(=a^2-b^2-c^2-2\left(bc+ac+bc\right)\)
\(=\left(a-b-c\right)^2=10^2=100\)
Sửa đề tí nha
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\(a^2-b^2-c^2-2bc-2ac-2ab\)
\(=>a^2-b^2-c^2-2\left(bc+ac+ab\right)\)
\(=\left(a+b+c\right)^2\)
\(=10^2=100\)
Ủng hộ mik nha thanks nhiều
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\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)
-Ta có hằng đẳng thức: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(P=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}+2bc+2ca+2ab\)
\(=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}=\dfrac{2\left(b^3c^3+c^3a^3+a^3b^3\right)}{a^2b^2c^2}=\dfrac{2.\left(ab+bc+ca\right)\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=\dfrac{2.0.\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=0\)
-C/m hằng đẳng thức trên:
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=a^2-b^2-c^2-2bc-20a\)
\(=a^2-\left(b+c\right)^2-20a\)
\(=a^2-\left(a-10\right)^2-20a\)(vì a + b + c = 10)
\(=a^2-a^2+20a-100-20a\)
\(=-100\)