Câu hỏi của Vũ Đăng Thành

Question

tính giá trị của BT $a^4+b^{4^{ }}+c^4+\dfrac{1}{4}$ biết a+b+c = 0 và $a^2+b^2+c^2=1$

Nguyễn Lê Phước Thịnh · Accepted Answer

Ta có: a+b+c=0nên $\left(a+b+c\right)^2=0$$\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0$$\Leftrightarrow2ab+2ac+2bc=-1$$\Leftrightarrow ab+ac+bc=\dfrac{-1}{2}$$\Leftrightarrow\left(ab+ac+bc\right)^2=\dfrac{1}{4}$$\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}$$\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}$$\Leftrightarrow a^2b^2+a^2c^2+b^2c^2=\dfrac{1}{4}$Ta có: $a^2+b^2+c^2=1$$\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1$$\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1$$\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=1$$\Leftrightarrow a^4+b^4+c^4+2\cdot\dfrac{1}{4}=1$$\Leftrightarrow a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}$$\Leftrightarrow a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}$Vậy: $a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}$Câu trả lời: Ta có: a+b+c=0nên $\left(a+b+c\right)^2=0$$\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0$$\Leftrightarrow2ab+2ac+2bc=-1$$\Leftrightarrow ab+ac+bc=\dfrac{-1}{2}$$\Leftrightarrow\left(ab+ac+bc\right)^2=\dfrac{1}{4}$$\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}$$\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}$$\Leftrightarrow a^2b^2+a^2c^2+b^2c^2=\dfrac{1}{4}$Ta có: $a^2+b^2+c^2=1$$\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1$$\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1$$\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=1$$\Leftrightarrow a^4+b^4+c^4+2\cdot\dfrac{1}{4}=1$$\Leftrightarrow a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}$$\Leftrightarrow a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}$Vậy: $a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}$

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