Câu hỏi của trần thảo mai

Question

bài 4: cho a,b là các số thực khác 0 thỏa mãn: a3+8b3+1=6ab

tính giá trị biểu thức: a+2b

Trần Tuấn Hoàng · Accepted Answer

$a^3+8b^3+1=6ab$$\Rightarrow\left(a+2b\right)^3-6a^2b-12ab^2+1-6ab=0$$\Rightarrow\left(a+2b\right)^3+1-6ab\left(a+2b+1\right)=0$$\Rightarrow\left(a+2b+1\right)\left[\left(a+2b\right)^2-\left(a+2b\right)+1\right]-6ab\left(a+2b+1\right)=0$$\Rightarrow\left(a+2b+1\right)\left(a^2+4ab+4b^2-a-2b+1-6ab\right)=0$$\Rightarrow\left(a+2b+1\right)\left(a^2-2ab+4b^2-a-2b+1\right)=0$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\a^2-2ab+4b^2-a-2b+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a^2-2a\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b\right)+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a^2-2a+1-1\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b+\dfrac{1}{4}-\dfrac{1}{4}\right)+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a-1\right)^2-\dfrac{1}{2}+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2-\dfrac{1}{2}+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a-1\right)^2+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\a=1;b=\dfrac{1}{2}\end{matrix}\right.$*$a+2b+1=0\Rightarrow a+2b=-1$*$a=1;b=\dfrac{1}{2}\Rightarrow a+2b=1+2.\dfrac{1}{2}=2$ Câu trả lời: $a^3+8b^3+1=6ab$$\Rightarrow\left(a+2b\right)^3-6a^2b-12ab^2+1-6ab=0$$\Rightarrow\left(a+2b\right)^3+1-6ab\left(a+2b+1\right)=0$$\Rightarrow\left(a+2b+1\right)\left[\left(a+2b\right)^2-\left(a+2b\right)+1\right]-6ab\left(a+2b+1\right)=0$$\Rightarrow\left(a+2b+1\right)\left(a^2+4ab+4b^2-a-2b+1-6ab\right)=0$$\Rightarrow\left(a+2b+1\right)\left(a^2-2ab+4b^2-a-2b+1\right)=0$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\a^2-2ab+4b^2-a-2b+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a^2-2a\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b\right)+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a^2-2a+1-1\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b+\dfrac{1}{4}-\dfrac{1}{4}\right)+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a-1\right)^2-\dfrac{1}{2}+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2-\dfrac{1}{2}+1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\dfrac{1}{2}\left(a-1\right)^2+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}a+2b+1=0\a=1;b=\dfrac{1}{2}\end{matrix}\right.$*$a+2b+1=0\Rightarrow a+2b=-1$*$a=1;b=\dfrac{1}{2}\Rightarrow a+2b=1+2.\dfrac{1}{2}=2$

trần thảo mai · Answer

.Câu trả lời: .

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP