\(x^2+4x+5=2\sqrt{2x+3}\)

\(ĐK:x\ge-\dfrac{3}{2}\)

\(pt\Leftrightarrow(2x+3-2\sqrt{2x+3}+1)+x^2+2x+1=0\)

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

Vì \(\left(\sqrt{2x+3}-1\right)^2\ge0;-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}(\sqrt{2x+3}-1)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}=1\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)}\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy, pt có nghiệm duy nhất là x=-1

\(4A-3x^2+7-6x=x^2+3A-4x-3\)

\(\Rightarrow4A-3A=\left(x^2+3x^2\right)-\left(4x-6x\right)-\left(3+7\right)\)

\(\Rightarrow A=4x^2-\left(-2x\right)-10\)

\(\Rightarrow A=4x^2+2x-10\)

\(A=16x^2-y^2-16x^2+8x=8x-y^2\\ A=8\cdot3-\left(-1\right)^2=24-1=23\\ B=64x^3-80x-64x^3-1=-80x-1\\ B=-80\cdot\dfrac{1}{5}-1=-16-1=-17\)

Bài 1: 

a: \(49-4x^2=\left(7-2x\right)\left(7+2x\right)\)

b: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

c: \(x^2+18xy+81y^2=\left(x+9y\right)^2\)

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)

b: \(B=\left(x+2\right)^2-\left(2x-1\right)^2\)

\(=x^2+4x+4-4x^2+4x-1\)

\(=-3x^2+8x+3\)

\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x}=\dfrac{x^2+2x+6}{x\left(x-3\right)}\)  đkxđ: x khác 3 , x khác 0

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-3\right)}+\dfrac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}-\dfrac{x^2+2x+6}{x\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+2x}{....}+\dfrac{x^2-3x-2x+6}{.....}-\dfrac{x^2+2x+6}{...}=0\)

\(\Leftrightarrow x^2+2x+x^2-3x-2x+6-x^2-2x-6=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

đk : x khác -3 ; 3

\(\Rightarrow\left(x+3\right)^2-4x^2=\left(x-3\right)^2\Leftrightarrow-3x^2+6x+9=x^2-6x+9\)

\(\Leftrightarrow4x^2-12x=0\Leftrightarrow4x\left(x-3\right)=0\Leftrightarrow x=3\left(ktm\right);x=0\)

\(\dfrac{x+3}{x-3}-\dfrac{4x^2}{x^2-9}=\dfrac{x-3}{x+3}\) đkxđ: x khác 3 ; x khác -3

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+6x+9}{...}-\dfrac{4x^2}{.....}-\dfrac{x^2-6x+9}{....}=0\)

\(\Leftrightarrow x^2+6x+9-4x^2-x^2+6x-9=0\)

\(\Leftrightarrow-4x^2+12x=0\)

\(\Leftrightarrow-4x\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=3\left(kotm\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt có S thuộc \(\left\{0\right\}\)