Bài 6:

1) \(A=\left(x-2\right)^2-5\)

Ta có \(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2-5\ge-5\forall x\)

Dấu "=" xảy ra khi \(x=2\)

Vậy \(A_{min}=-5\) khi \(x=2\)

2) \(B=\left|x+\dfrac{1}{2}\right|+2\)

Ta có \(\left|x+\dfrac{1}{2}\right|\ge0\forall x\Rightarrow\left|x+\dfrac{1}{2}\right|+2\ge2\forall x\)

Dấu "=" xáy ra khi \(x=-\dfrac{1}{2}\)

Vậy \(B_{min}=2\) khi \(x=-\dfrac{1}{2}\)

a. \(\sqrt{1-4x+4x^2}+5=x-2\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\)

\(\Leftrightarrow\left|1-2x\right|-x=-7\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2x-x=-7\\2x-1-x=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\x=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-6\end{matrix}\right.\)

b. ĐKXĐ: \(x\ge-3\)
\(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\)

\(\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}-6\sqrt{3+x}=4\)

\(\Leftrightarrow4\sqrt{3+x}=4\) \(\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\) ( thỏa mãn đk )

neither she nor i went out last night, mình nhầm 

neither..nor

neither she or i went out last night

\(y=f\left(x\right)=-3x+1\)

Ta có với mọi \(x_1,x_2\in R\) 

\(x_1>x_2\Leftrightarrow-3x_1< -3x_2\Leftrightarrow-3x_1+1< -3x_1+1\Leftrightarrow f\left(x_1\right)< f\left(x_2\right)\)

Vì \(x_1>x_2\) mà \(f\left(x_1\right)< f\left(x_2\right)\) nên \(y=f\left(x\right)=-3x+1\) nghịch biến trên \(R\)

7B

Ta có  \(\dfrac{5}{x^2+4x+4}=\dfrac{5}{\left(x+2\right)^2}\) 

Để biểu thức có nghĩa thì \(\left(x+2\right)^2\ne0\Leftrightarrow x+2\ne0\Leftrightarrow x\ne-2\)

a. \(7\left(2x-0,5\right)-3\left(x+4\right)=4-5\left(x-0,7\right)\)

\(\Rightarrow14x-3,5-3x-12=4-5x+3,5\)

\(\Rightarrow14x-3x+5x=4+3,5+3,5+12\)

\(\Rightarrow16x=23\)

\(\Rightarrow x=\dfrac{23}{16}\)

Vậy \(S=\left\{\dfrac{23}{16}\right\}\)

b. \(5x^3-2x^2-7x=0\)

\(\Rightarrow x\left(5x^2-2x-7\right)=0\)

\(\Rightarrow x\left(x-\dfrac{7}{5}\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{7}{5}=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)

Vậy \(S=\left\{0;\dfrac{7}{5};-1\right\}\)