Question

Cảm ơn. Mik đang cần gấp

Answer 1

Câu 1:

Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

Thay \(a+b+c=0\) vào biểu thức ta được:

\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow a^3+b^3=3abc\left(đpcm\right)\)

Vậy \(a^3+b^3=3abc\) khi \(a+b+c=0\)

Answer 2

Câu 3:

\(\text{a) }x^2+x+1\\ =x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{4}\right)^2\right]+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ \text{Ta có : }\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ \text{ Vậy biểu thức luôn nhận giá trị dương}\text{ }\forall x\\ \)

\(\text{b) }2x^2+2x+1\\ =2x^2+2x+\dfrac{1}{2}+\dfrac{1}{2}\\ =2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}\\ =2\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{1}{2}\\ =2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\\ \text{Ta có: }2\left(x+\dfrac{1}{2}\right)^2\forall x\\ 2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\forall x\\ \text{Vậy giá trị của biểu thức luôn nhận giá trị dương }\forall x\\ \)

Answer 3

\(\text{d) }x^2+4y^2+z^2-2x-6z+8y+15\\ =x^2+4y^2+z^2-2x-6z+8y+1+4+9+1\\ =\left(x^2-2x+1\right)+\left[\left(2y\right)^2+8y+4\right]+\left(z^2-6z+9\right)+1\\ =\left(x^2-2x+1^2\right)+\left[\left(2y\right)^2+2\cdot2y\cdot2+2^2\right]+\left(z^2-2\cdot z\cdot3+3^2\right)+1\\ =\left(x-1\right)^2+\left(2y+2\right)^2+\left(z+3\right)^2+1\\ \text{Ta có : }\left(x-1\right)^2\ge0\forall x\\ \left(2y+2\right)^2\ge0\forall x\\ \left(z+3\right)^2\ge0\forall x\\ \left(x-1\right)^2+\left(2y+2\right)^2+\left(z+3\right)^2\ge0\forall x\\ \left(x-1\right)^2+\left(2y+2\right)^2+\left(z+3\right)^2+1\ge1\forall x\\ \)

Vậy biểu thức luôn nhận giá trị dương \(\forall x\)