\(\dfrac{3}{5}x-\dfrac{25}{4}=\dfrac{1}{2}:x+\dfrac{5}{4}\)

=>\(\dfrac{3}{5}x-\dfrac{25}{4}-\dfrac{1}{2x}+\dfrac{5}{4}\)

=>\(\dfrac{3}{5}x-\dfrac{1}{2x}=\dfrac{30}{4}\)

=>\(\dfrac{3x}{5}-\dfrac{1}{2x}=\dfrac{30}{4}\)

=>\(\dfrac{6x^2-5}{10x}=\dfrac{30}{4}\)

=>\(6x^2-5=10x\cdot\dfrac{30}{4}=5x\cdot15=75x\)

=>\(6x^2-75x-5=0\)

=>\(x=\dfrac{75\pm\sqrt{5745}}{12}\)

1) \(\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\)

\(\Leftrightarrow25x^2-1=25x^2-7x+15\)

\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)

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

\(\Leftrightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Leftrightarrow5x=0\Leftrightarrow x=0\)

   x(x+4)(4-x)+(x-5)(x²+5x+25)=3

   (x²+4x)(4-x)+x³+5x²+25x-5x²-25x-125=3

   4x²-x³+16x-4x²+x³+5x²+25x-5x²-25x=3+125

   (4x²-4x²)-(x³-x³)+(16x+25x-25x)+(5x²-5x²)=128

   16x=128

   x=128:16

   x=8

Ko thấy j hết á bạn

1)

\(3\left(x-2\right)+4\left(x-1\right)=25\)

\(3x-6+4x-4=25\)

\(7x-10=25\\ 7x=35\\ x=5\)

2)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)

\(\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(5x-3-x+1\right)=0\)

\(\left(x-2\right)\left(4x-2\right)=0\)

\(=>\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

3)

\(\left(x-2\right)^2=4\left(x-1\right)^2\)

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

\(x^2-4x+4=4x^2-8x+4\)

\(x^2-4x+4-4x^2+8x-4=0\)

\(-3x^2+4x=0\)

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

\(=>\left[{}\begin{matrix}x=0\\-3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(3\left(x-2\right)+4\left(x-1\right)=25\) 

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\) 

\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)

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

\(\left(x-2\right)^2=4\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-2\right)^2-4\left(x-1\right)^2=0\) 

\(\Leftrightarrow\left[\left(x-2\right)-2\left(x-1\right)\right]\left[\left(x-2\right)+2\left(x-1\right)\right]=0\) 

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-4=0\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

a) \(\left(x-3\right)\left(4-5x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\4-5x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{4}{5}\end{array}\right.\)

b) \(\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=5\\x-1=-5\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-4\end{array}\right.\)

a)(x-3)(4-5x)\(\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x-3=0\\4-5x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{4}{5}\end{array}\right.\)

Vậy x=3 và \(\frac{4}{5}\)

b) \(\left(x-1\right)^2=25\Rightarrow\begin{cases}x-1=5\\x-1=-5\end{cases}\)

\(\Rightarrow\begin{cases}x=6\\x=-4\end{cases}\)

Vậy x=-4 và 6

\(a.\)

\(\left(x-3\right)\left(4-5x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\4-5x=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=0,8\end{array}\right.\)

Vậy :       \(x\in\left\{0,8;3\right\}\)

\(b.\)

\(\left(x-1\right)^2=25\)

\(\Leftrightarrow\left(x-1\right)^2=\left(\pm5\right)^2\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=-5\\x-1=5\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=6\end{array}\right.\)

Vậy :        \(x\in\left\{-4;6\right\}\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)