Ta có : 

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

\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)

\(\Leftrightarrow\)\(x^2-4x+15=-7\)

\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)

\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)

Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)

\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)

Vậy không có giá trị nào của x thoã mãn đề bài 

Chúc bạn học tốt ~ 

\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)

<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)

Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)

Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)

Vây x= 2 hoặc 3; 4

a,x- 2/3= -1/2

x= -1/2 + 2/3

x= 1/6

b,20%.x - x +1/5 =3/4

1/5.x - x +1/5 = 3/4

1/5.x-x =3/4 - 1/5

x.(1/5 - 1)=11/20

x.-4/5 =11/20

x=11/20 : -4/5=-11/16

a)  \(x-\frac{2}{3}=-\frac{1}{2}\)

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

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

b)  \(20\%x-x+\frac{1}{5}=\frac{3}{4}\)

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

\(\Leftrightarrow-\frac{4}{5}x=\frac{11}{5}\)

\(\Leftrightarrow x=\frac{11}{5}:\left(-\frac{4}{5}\right)\)

\(\Leftrightarrow x=-\frac{11}{4}\)

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)

\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(c,\\ \left(2x+1\right)^3=9.81\\ \left(2x+1\right)^3=3^2.3^4\\ \left(2x+1\right)^3=3^6\\ \left(2x+1\right)^3=\left(3^2\right)^3=9^3\\ Vậy:2x+1=9\\ 2x=9-1\\ 2x=8\\ x=\dfrac{8}{2}\\ x=4\)

a ) 5x=2y

b) 7x = 3y

c ) 3+x /7+y= 3/7

=> (3+x).7= (7+y).3

21 + 7x = 21 + 3y

=> 7x = 3y

=> x/y = 3/7 => x= 20:(7+3).3=6 

=> y = 20 - 6 = 14

Vậy (x;y) thuộc {(6;14)}