Câu hỏi của Lê Hoàng Thảo Nguyên

Question

Tìm x,biết

a) (2x+1)^2 - 4(x+2)^2 = 12

b) (x^3)^2 - ( x-4 ).(x-8) = 1

c) ( x-3) . ( x^2 +3x +9 ) + x. (x-3) . (3-x) = 1

d) ( x+1)^3 - ( x - 1 )^3 - 6. (x-1) ^2 = -19

( Dấu "^" là dấu mũ ạ , e ko...

Trần Thanh Phương · Accepted Answer

a) $\left(2x+1\right)^2-4\left(x+2\right)^2=12$

$\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12$

$\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0$

$\Leftrightarrow-12x-27=0$

$\Leftrightarrow x=\frac{-9}{4}$

b) xem lại đề

c) $\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1$

$\Leftrightarrow x^3-27-x\left(x-3\right)^2=1$

$\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0$

$\Leftrightarrow x^3-28-x^3+6x^2-9x=0$

$\Leftrightarrow6x^2-9x-28=0$

$\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0$

$\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0$

$\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2$

$\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.$

$\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}$

d) $\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19$

$\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0$

$\Leftrightarrow12x+15=0$

$\Leftrightarrow x=\frac{-5}{4}$Câu trả lời: a) $\left(2x+1\right)^2-4\left(x+2\right)^2=12$

$\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12$

$\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0$

$\Leftrightarrow-12x-27=0$

$\Leftrightarrow x=\frac{-9}{4}$

b) xem lại đề

c) $\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1$

$\Leftrightarrow x^3-27-x\left(x-3\right)^2=1$

$\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0$

$\Leftrightarrow x^3-28-x^3+6x^2-9x=0$

$\Leftrightarrow6x^2-9x-28=0$

$\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0$

$\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0$

$\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2$

$\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.$

$\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}$

d) $\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19$

$\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0$

$\Leftrightarrow12x+15=0$

$\Leftrightarrow x=\frac{-5}{4}$

Trần Thanh Phương · Answer

Theo giả thiết:

$\left(a+b+c\right)^2=3\left(ab+bc+ca\right)$

$\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca$

$\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0$

$\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0$

$\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0$

$\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0$

Dễ thấy $VT\ge0\forall a;b;c$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}a-b=0\b-c=0\c-a=0\end{matrix}\right.$$\Leftrightarrow a=b=c$(đpcm)Câu trả lời: Theo giả thiết: