(2x + y)² - (2x - 3)² = (2x + y - 2x + 3)(2x + y + 2x - 3) = (y + 3)(4x + y - 3)

\(\left(2x+y\right)^2-4x^2+12x-9\)

\(=\left(2x+y\right)^2-\left(2x-3\right)^2\)

\(=\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)

\(=\left(y+3\right)\left(4x+y-3\right)\)

\(4\cdot2-12x+9=x+7\)

\(\Leftrightarrow8-12x+9=x+7\)

\(\Leftrightarrow-12x-x=7-8-9\)

\(\Leftrightarrow-13x=-10\)

\(\Leftrightarrow x=\dfrac{10}{13}\)

Vậy \(x=\dfrac{10}{13}\)

Ta có: \(4x^2-12x+9=x+7\)

\(\Leftrightarrow4x^2-13x+2=0\)

\(\text{Δ}=\left(-13\right)^2-4\cdot4\cdot2=137\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{13-\sqrt{137}}{8}\\x_2=\dfrac{13+\sqrt{137}}{8}\end{matrix}\right.\)

=(4x2-12x+9)=[(2x)2-2.2x.3+32]=-(2x-3)2

Đặt \(C=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=\left|2x-1\right|+\left|2x-3\right|\)

\(=\left|2x-1\right|+\left|3-2x\right|\)

\(\ge\left|\left(2x-1\right)+\left(3-2x\right)\right|=\left|2\right|=2\)

Vậy \(C_{min}=2\)

#)Giải :

\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=\left|2x-1\right|+\left|2x-3\right|\)

\(=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|=2\)

Dấu ''='' xảy ra khi x = 1

\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=|2x-1|+|2x-3|\)

\(=|2x-1|+|3-2x|\ge|2x-1+3-2x|=2\)

Dấu"=" xảy ra \(\Leftrightarrow\left(2x-1\right)\left(3-2x\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

Chỉ là góp ý:V

a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)

\(x^2+2x+1=\left(x+1\right)^2\)

b) \(x^2+y^2+2xy=\left(x+y\right)^2\)

c) \(9x^2+12x+4=\left(3x+2\right)^2\)

d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)

a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2  

b) Ta có  Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3

Xét phương trình hoành độ giao điểm 4 x 2   =   12 x   –   9   ↔   4 x 2   −   12 x   +   9 có ∆’ = 0 nên phương trình có nghiệm kép hay đường thẳng tiếp xúc parabol tại một điểm.

Đáp án: B

a:

ĐKXĐ: \(x^2+3x>=0\)

=>x(x+3)>=0

=>\(\left[{}\begin{matrix}x>=0\\x< =-3\end{matrix}\right.\)

 \(\sqrt{16}-\sqrt{x^2+3x}=0\)

=>\(\sqrt{x^2+3x}=\sqrt{16}\)

=>x^2+3x=16

=>x^2+3x-16=0

\(\text{Δ}=3^2-4\cdot1\cdot\left(-16\right)=9+64=73>0\)

Do đó: Phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{73}}{2}\\x_2=\dfrac{-3+\sqrt{73}}{2}\end{matrix}\right.\)

b:

ĐKXĐ: \(x\in R\)

 \(3x-1-\sqrt{4x^2-12x+9}=0\)

=>\(\sqrt{\left(2x-3\right)^2}=3x-1\)

=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(2x-3\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(x+2\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{4}{5}\left(nhận\right)\end{matrix}\right.\)

c:

ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+8>=0\\2x^2-10x+11>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\\\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=4\end{matrix}\right.\)

 \(\sqrt{2x^2-10x+11}=\sqrt{x^2-6x+8}\)

\(\Leftrightarrow2x^2-10x+11=x^2-6x+8\)

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>x=3(loại) hoặc x=1(nhận)