Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

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

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

`a) (x^2 - 4x  +5 ) - (x^2 - 2x + 1)=3`

`=> x^2 - 4x + 5 - x^2 + 2x - 1 = 3`

`=> (x^2 - x^2) - (4x - 2x) + (5-1) = 3`

`=> -2x + 4 =3`

`=> -2x = 3-4`

`=> -2x = -1`

`=>x = -1 : -2`

`=> x = 1/2`

Vậy `x=1/2`

ko phải "-" đâu mà là nhân đấy

giữa 2 ngoặc đó

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a: |4x-1|=1

=>\(\left[\begin{array}{l}4x-1=1\\ 4x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=2\\ 4x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=0\end{array}\right.\)

Thay x=1/2 vào A(x), ta được:

\(A\left(\frac12\right)=\left(\frac12\right)^4-4\cdot\left(\frac12\right)^3+2\cdot\left(\frac12\right)^2-5\cdot\frac12+6\)

\(=\frac{1}{16}-4\cdot\frac18+2\cdot\frac14-\frac52+6=\frac{1}{16}-\frac12+\frac12-\frac52+6\)

\(=\frac{1}{16}-\frac{40}{16}+\frac{96}{16}=\frac{97-40}{16}=\frac{57}{16}\)

Thay x=0 vào A(x), ta được:

\(A\left(0\right)=0^4-4\cdot0^3+2\cdot0^2-5\cdot0+6=6\)

b: \(A\left(x\right)-B\left(x\right)=3x^2-x-3x^3-x^2+x^4-2x^2+6\)

=>A(x)-B(x)=\(x^4-3x^3+\left(3x^2-x^2-2x^2\right)-x+6\)

=>A(x)-B(x)=\(x^4-3x^3-x+6\)

=>\(B\left(x\right)=A\left(x\right)-\left(x^4-3x^3-x+6\right)\)

=>\(B\left(x\right)=x^4-4x^3+2x^2-5x+6-x^4+3x^3+x-6=-x^3+2x^2-4x\)

c: Đặt B(x)=0

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

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

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

mà \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3>0\forall x\)

nên x=0

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

a) \(\left(x-3\right)^2-4=0\)

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

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

\(\left(x-3\right)^2=\pm4\)

\(\left(x-3\right)^2=\pm2^2\)

\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

\(4x^2+12x+9-4x^2+1=22\)

\(12x+10=22\)

\(12x=22-10\)

\(12x=12\)

\(x=1\)

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

\(16x^2-9-16x^2+40x-25=16\)

\(-34+40x=16\)

\(40x=16+34\)

\(40x=50\)

\(x=\frac{50}{40}=\frac{5}{4}\)

d) \(x^3-9x^2+27x-27=-8\)

\(x^3-9x^2+27x-27+8=0\)

\(x^3-9x^2+27x-19=0\)

\(\left(x^2-8x+19\right)\left(x-1\right)=0\)

Vì \(\left(x^2-8x+19\right)>0\) nên:

\(x-1=0\)

\(x=1\)

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

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

\(3x+1=2\)

\(3x=2-1\)

\(3x=1\)

\(x=\frac{1}{3}\)