b: \(\Leftrightarrow\dfrac{x-2}{A}=\dfrac{\left(5x-1\right)\left(x-2\right)}{x^2\left(5x-1\right)+3\left(5x-1\right)}=\dfrac{x-2}{x^2+3}\)

hay \(A=x^2+3\)

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

1)-2x – (x – 17) = 34 – (-x + 25)

-2x-x-_17

17x + 3. ( -16x – 37) = 2x + 43 - 4x

<=>17x-48x-111=-2x+43

<=>-29x=154

<=> \(x=-\frac{154}{29}\)

-3. (2x + 5) -16 < -4. (3 – 2x)

\(\Leftrightarrow-6x-31< -12+8x.\)

\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)

a: Ta có: \(6-4x=5(x+3)+3\)

\(\Leftrightarrow6-4x-5x-12-3=0\)

\(\Leftrightarrow-9x=9\)

hay x=-1

b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)

\(\Leftrightarrow15x+45-30=10x-30+5x+25\)

\(\Leftrightarrow15=-5\left(loại\right)\)

c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)

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

d: Ta có: \(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\)

\(\Leftrightarrow2+x-2=x^2+2x\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

giải phần còn lại giúp mình được ko?

giúp mik với

tìm số ngyên x :

a) -2x-(x-17)=34-(-x+25)
-2x - x +17 =34 + x -25
-3x + 17 =9 +x
-3x + 17 - 9 -x=0
-4x + 8 = 0
-4x = -8
x =2 

mình cần gấp

Đề bài yêu cầu gì?

3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)

\(=x^3+125-x^3+8x^2-16x+16x\)

\(=8x^2+125\)