a/ \(\left|1-2x\right|>7\Leftrightarrow\left[{}\begin{matrix}1-2x=7\\1-2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x< -6\\2x< 8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -3\\x< 4\end{matrix}\right.\)

b/ \(\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\) ( vì -5<0)

\(\Leftrightarrow x>3\)

sao ko trả lời câu c

(\(x-3\))² + (2y - 1)² = 0

          (\(x\) - 3)² ≥ 0 ∀ \(x\)

        (2y - 1)² ≥ 0 ∀ y

⇔ (\(x\) - 3)² + (2y - 1)²= 0

⇔ \(\left\{{}\begin{matrix}x-3=0\\3y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{3}\end{matrix}\right.\)

(4\(x-3\))⁴ + (y + 2)² ≤ 0

(4\(x\) - 3)⁴ ≥ 0 ∀ \(x\)

(y + 2)² ≥ 0 ∀ y

⇔(4\(x\) - 3)⁴   + (y+2)² ≥ 0

⇔ (4\(x\) - 3)⁴ + (y + 2)² ≤ 0 ⇔

⇔\(\left\{{}\begin{matrix}4x-3=0\\y+2=0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-2\end{matrix}\right.\)

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

 đó chính là -4 minh khong muon giai ra ta lau lam ban

rút 4 ra ngoài nhan bạn  4(2(x+1/x)^2+(x^2+1/x^2)^2-(x^2+1/x^2)(x+1/x)^2=(x+4)^2 

mik xét cái này cho dễ nhìn nhan 

2(x+1/x)^2-(x^2+1/x^2)(x+1/x)^2

= (x+1/x)^2(2-x^2-1/x^2)

= -(x+1/x)^2(x^2-2+1/x^2)

= -(x+1/x)^2(x-1/x)^2=-(x^2-1/x^2)^2

thế ở trên ta có 

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

4(-x^4+2-1/x^4+x^4+2+1/x^4)=x^2+8x+16

4.4=x^2+8x+16 

suy ra x^2+8x=0 

x(x+8)=0

suy ra x=0 hoặc x=-8 

mak nhìn để bài thì x=0 ko được nên x=-8

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{x}{x\left(\sqrt{x+4}+2\right)}=\lim\limits_{x\rightarrow0^+}\dfrac{1}{\sqrt{x+4}+2}=\dfrac{1}{4}\)

\(f\left(0\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(mx^2+2m+\dfrac{1}{4}\right)=2m+\dfrac{1}{4}\)

Hàm liên tục tại x=0 khi: \(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)=f\left(0\right)\)

\(\Leftrightarrow2m+\dfrac{1}{4}=\dfrac{1}{4}\Leftrightarrow m=0\)