Câu hỏi của lê thị hương giang

Question

Tìm x,biết:

$\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42$

lê thị hương giang · Accepted Answer

$\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42$

$\Rightarrow x^3+9x^2+27x+27-\left[x\left(9x^2+6x+1\right)\right]+8x^2+1-3x^2=42$

$\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^2+1-3x^2=42$

$\Rightarrow-8x^3+8x^2+26x+28=42$Câu trả lời: $\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42$

$\Rightarrow x^3+9x^2+27x+27-\left[x\left(9x^2+6x+1\right)\right]+8x^2+1-3x^2=42$

$\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^2+1-3x^2=42$

$\Rightarrow-8x^3+8x^2+26x+28=42$

lê thị hương giang · Answer

$\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42$

$\Rightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42$

$\Rightarrow x^3+9x^2+27x+28-9x^3-6x^2-x+8x^3-3x^2=42$

$\Rightarrow26x+28=42$

$\Rightarrow26x=14\Rightarrow x=\dfrac{7}{13}$Câu trả lời: $\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42$

$\Rightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42$

$\Rightarrow x^3+9x^2+27x+28-9x^3-6x^2-x+8x^3-3x^2=42$

$\Rightarrow26x+28=42$

$\Rightarrow26x=14\Rightarrow x=\dfrac{7}{13}$

Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)