1/ Ta có: \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)

\(\Rightarrow5x^2-15x=5x^2-11x+2-5\)

\(\Rightarrow-15x=-11x-3\)

\(\Rightarrow-15x+11x=-3\)

\(\Rightarrow-4x=-3\Rightarrow x=\frac{3}{4}\).

2/ Ta có: \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)

\(\Rightarrow2x^2-5x+2-2x^2+x+21-3=0\)

\(\Rightarrow-4x+20=0\Rightarrow-4x=-20\Rightarrow x=5\).

Chúc bn hc tốt! 

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

<=> \(5x^2-15x=\left(5x^2-x-10x+2\right)-5\)

<=> \(5x^2-15x=5x^2-11x-3\)

<=> \(5x^2-15x-5x^2+11x+3=0\)

<=> \(-4x+3=0\)

<=> \(x=\frac{3}{4}\)

2/ \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)

<=> \(2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)

<=> \(2x^2-5x+2-2x^2+x+21=3\)

<=> \(-4x+21=3\)

<=> \(4x=17\)

<=> \(x=\frac{17}{4}\)

Cám ơn hai bạn nhiều

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

đề là j vậy bạn 

Đề là nhân đa thức với đa thức

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

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

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

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

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

a; \(\dfrac{1}{2}x\) + \(\dfrac{5}{2}\) = \(\dfrac{7}{2}\)\(x\) - \(\dfrac{3}{4}\)

     \(\dfrac{3}{4}\) + \(\dfrac{5}{2}\) = \(\dfrac{7}{2}\)\(x\) - \(\dfrac{1}{2}\)\(x\) 

    \(\dfrac{3}{4}\) + \(\dfrac{10}{4}\) = 3\(x\)

    3\(x\)          = \(\dfrac{13}{4}\)

     \(x\)          = \(\dfrac{13}{4}\) : 3

     \(x=\dfrac{13}{12}\) 

Vậy \(x=\dfrac{13}{12}\)