a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)

=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)

=>\(6x-\dfrac{39}{4}=1\)

=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)

=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)

b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)

=>\(2x-6=3x+6-x+1\)

=>2x-6=2x+7

=>-6=7(vô lý)

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

=>\(x^2+3x+x^2-2x=2x^2-2x\)

=>3x-2x=-2x

=>3x=0

=>x=0

d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)

=>\(3x^2-3x-2x-4-2x=x^2-x\)

=>\(3x^2-7x-4-x^2+x=0\)

=>\(2x^2-6x-4=0\)

=>\(x^2-3x-2=0\)

=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)

 20 . 2^x + 1 = 10.4^2 + 1 

 20 . 2^x + 1 = 10 . 16 + 1

20 . 2^x + 1 = 161

20 . 2^x = 161 - 1

20 . 2^x = 160

2^x = 8

2^x = 2^3

=> x = 3

Ban co giai dc phan 2 ko ?

( 4 - x : 2 )^3 - 1 = 2 . ( 2^3 - 5 : 2^0 )

( 4 - x : 2 )^3 - 1 = 2 . ( 8 - 5 : 1 )

( 4 - x : 2 )^3 - 1 = 2 . 3

( 4 - x : 2 )^3 - 1 = 6

( 4 - x : 2 )^3  = 7

=> ko tìm đc x

`#3107.101107`

\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)

`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`

`<=> x^3 - 25x - x^3 - 8 = 5`

`<=> -25x - 8 = 5`

`<=> -25x = 13`

`<=> x = -13/25`

Vậy, `x = -13/25`

_____

\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)

`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`

`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`

`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`

`<=> 12x - 4 = -19`

`<=> 12x = -15`

`<=> x = -15/12 = -5/4`

Vậy, `x = -5/4.`

________

`@` Sử dụng các hđt:

`1)` `A^2 + B^2 = (A - B)(A + B)`

`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`

`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`

`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`

`5)` `(A - B)^2 = A^2 - 2AB + B^2.`

a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)

=>\(x\left(x^2-25\right)-x^3-8=5\)

=>\(x^3-25x-x^3-8=5\)

=>-25x=13

=>\(x=-\dfrac{13}{25}\)

b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)

=>\(6x^2+2-6x^2+12x-6=-19\)

=>12x-4=-19

=>12x=-15

=>x=-5/4

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

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

\(\frac{7}{6}x=\frac{3}{20}\Rightarrow x=\frac{9}{70}\)

b) \(-5^{\frac{1}{2}x+1}=\frac{3}{4}-\frac{7}{6}\)

\(-5^{\frac{1}{2}x}.\left(-5\right)=-\frac{5}{12}\)

\(-5^{\frac{1}{2}x}=\frac{1}{12}\)

mà -5^1/2x mang giá trị âm

1/12 có giá trị dương

=> không tìm được x

c) \(\frac{2x-2}{3}=\frac{7x+3}{2-1}\)

\(\frac{2x-2}{3}=7x+3=\frac{21x+9}{3}\)

=> 2x - 2 = 21x + 9

=> 2x - 21x = 9 + 2

-19x = 11

x = -11/19

phần d bn lm như phần a nha

Ta có\(\frac{2^x+2^{x+1}+2^{x+2}}{7}=\frac{3^x+3^{x+1}+3^{x+2}}{13}\)

\(\Rightarrow\frac{2^x\left(1+2+2^2\right)}{7}=\frac{3^x\left(1+3+3^2\right)}{13}\)

\(\Rightarrow\frac{2^x\left(1+2+4\right)}{7}=\frac{3^x\left(1+3+9\right)}{13}\)

\(\Rightarrow\frac{2^x.7}{7}=\frac{3^x.13}{13}\)

\(\Rightarrow2^x=3^x\)

\(\Rightarrow x=0\)