Điều kiện x khác 0

     \(\left(5x^4-3x^3\right):2x^3=\frac{1}{2}\) 

\(\Rightarrow\frac{5}{2}x-\frac{3}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{5}{2}x=2\Rightarrow x=\frac{4}{5}\)

Đặt \(log_2x=t\Rightarrow t\ge4\)

Phương trình trở thành: \(\sqrt{t^2-2t-3}=m\left(t-3\right)\)

\(\Leftrightarrow\sqrt{\left(t+1\right)\left(t-3\right)}=m\left(t-3\right)\)

\(\Leftrightarrow\sqrt{t+1}=m\sqrt{t-3}\)

\(\Leftrightarrow m=\sqrt{\dfrac{t+1}{t-3}}\)

Hàm \(f\left(t\right)=\sqrt{\dfrac{t+1}{t-3}}\) nghịch biến khi \(t\ge4\)

\(\lim\limits_{t\rightarrow+\infty}\sqrt{\dfrac{t+1}{t-3}}=1\) ; \(f\left(4\right)=\sqrt{5}\)

\(\Rightarrow1< f\left(t\right)\le\sqrt{5}\Rightarrow1< m\le\sqrt{5}\)

Đáp án D

a)\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\)

     \(4x^2-11x-3-\left(4x^2-29x+7\right)=15\)

     \(4x^2-11x-3-4x^2+29x-7=15\)

      \(18x-10=15\)

       \(x=\frac{25}{18}\)

b)\(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

     \(\left(x+1\right)\left(3x-5-3x+1\right)=x-4\)

       \(\left(x+1\right).\left(-4\right)-x+4=0\)

        \(-4x-4-x+4=0\)

          \(x=0\)

a) \(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x+1\right)=15\)

\(\left(4x+1\right)\left(x-3-x+7\right)=15\)

\(\left(4x+1\right)4=15\)

\(4x+1=15:4\)

\(4x+1=3,75\)

\(x=\frac{11}{16}\)

b) \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\left(x+1\right)\left(3x-5-3x+1\right)=x-4\)

\(\left(x+1\right).-4=x-4\)

\(-4x-4-x+4=0\)

\(-5x=0\)

\(x=0\)

k nhé

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

\(\left(3x+4\right)^2\left(2y+\frac{3}{7}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}3x=-4\\2y=-\frac{3}{7}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-4}{3}\\y=\frac{-3}{14}\end{cases}}\)