nhân vào rồi chi cho x,xong đặt là ra

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

ĐKXD phức tạp nên ko tìm ngay

Đặt \(x^2=t>0\Rightarrow\sqrt{12-\dfrac{3}{t}}+\sqrt{4t-\dfrac{3}{t}}=4t\)

Đặt \(\sqrt{4t-\dfrac{3}{t}}=a\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{t}=4t-a^2\\3=4t^2-ta^2\end{matrix}\right.\)

\(\Rightarrow\sqrt{4\left(4t^2-ta^2\right)-\left(4t-a^2\right)}+a=4t\)

\(\Rightarrow\sqrt{16t^2-4ta^2-4t+a^2}=4t-a\)

\(\Rightarrow16t^2-4ta^2-4t+a^2=\left(4t-a\right)^2\)

\(\Rightarrow16t^2-4ta^2-4t+a^2=16t^2-8ta+a^2\)

\(\Rightarrow4ta^2-8ta+4t=0\)

\(\Rightarrow4t\left(a-1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=0\left(loại\right)\\a=1\end{matrix}\right.\)

\(\Rightarrow\sqrt{4t-\dfrac{3}{t}}=1\Rightarrow4t^2-t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{3}{4}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Thử lại 2 nghiệm vào pt ban đầu đều thỏa mãn

Tham khảo:

Giải phương trình: \(\sqrt{12-\dfrac{3}{x^2}}+\sqrt{4x^2-\dfrac{3}{x^2}}=4x^2\) - Hoc24

ĐKXĐ: x\(\ne-2\)

Ta co 

\(x^2+\frac{4x^2}{\left(x+2\right)^2}=12\)

=> \(x^2-2.x.\frac{2x}{x+2}+\frac{4x^2}{\left(x+2\right)^2}\)\(+2.x.\frac{2x}{x+2}\)=12

=> \(\left(x-\frac{2x}{x+2}\right)^2+\frac{4x^2}{x+2}-12=0\)

=>\(\frac{x^4}{\left(x+2\right)^2}+\frac{4x^2}{x+2}-12=0\)(1)

Đặt \(\frac{x^2}{x+2}=y\)

(1)<=>y²+4y-12=0

   <=>(y+6)(y-2)=0

Đến đây dễ rồi bạn tự làm tiếp nhé

a, \(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

Vậy..............

b, \(\Leftrightarrow\left(x+\frac{2x}{x-2}\right)^2-2x.\frac{2x}{x-2}=12\)

\(\Leftrightarrow\left(\frac{x^2}{x-2}\right)^2-\frac{4x^2}{x-2}=12\)

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

Đến đây đơn giản rồi nhé

\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow-5x^2-9x+2=0\)
\(\Leftrightarrow-5x^2-10x+x+2=0\)
\(\Leftrightarrow-5x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(-5x+1\right)\left(x+2\right)=0 \)
\(\Leftrightarrow\left(-5x+1\right)=0\) Hoặc \(x+2=0\)
\(\Leftrightarrow x=\frac{1}{5}\)Hoặc \(x=-2\)
 

(x+2)(3-4x)=x²+4x+4

<=>(x+2)(3-4x)=(x+2)²

<=>(x+2)(3-4x)-(x+2)²=0

<=>(x+2)(3-4x-x-2)=0

<=>(x+2)(1-5x)=0

<=>x+2=0 hoặc 1-5x=0

<=>x=-2 hoặc x=1/5