1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

1) `2x(3x-1)-(2x+1)(x-3)`

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

+) Lỗi nhỏ: Sai ở chỗ: \(\left|x-2+4-3x\right|=\left|-2x-2\right|\)

+) Lỗi lớn: Dấu bằng xảy ra:  \(\hept{\begin{cases}\left(x-2\right)\left(4-3x\right)\ge0\\\left(-2x+2\right)\left(2x-3\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{3}{2}\le x\le1\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le1\)( làm tắt )

Nhưng mà thử vào chọn x= 1=>  A = 3 > 1. Nên bài này sai. 

Làm lại nhé!

A = | x - 2 | + | 2 x - 3  | + | 3  x - 4 |

 = | x - 2 | + | 2 x - 3  | + 3 | x - 4/3 |

= | x -2 | + | x - 4/3 | + | 2x -3 | +2 | x - 4/3 |

= ( | 2 - x | + | x - 4/3 | ) + ( | 3 - 2x  | + | 2x - 8/3 | )

\(\ge\)| 2 -x + x - 4/3 | + | 3 - 2x + 2x -8/3 |

= 2/3 + 1/3 = 1

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(2-x\right)\left(x-\frac{4}{3}\right)\ge0\\\left(3-2x\right)\left(2x-\frac{8}{3}\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{4}{3}\le x\le\frac{3}{2}\end{cases}}\Leftrightarrow\frac{4}{3}\le x\le\frac{3}{2}\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

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

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

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

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

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

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

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

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

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

C

ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-1;\dfrac{-3}{2};-2\right\}\)

Ta có: \(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)

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

\(\Leftrightarrow\dfrac{8x+12-4x-2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2-6x-12}{\left(2x+2\right)\left(2x+4\right)}\)

\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}\)

\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}=0\)

\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{4x+10}{\left(2x+2\right)\left(2x+4\right)}=0\)

\(\Leftrightarrow\left(4x+10\right)\left(\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{1}{\left(2x+2\right)\left(2x+4\right)}\right)=0\)

\(\Leftrightarrow2\left(2x+5\right)\left(\dfrac{\left(2x+2\right)\left(2x+4\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}+\dfrac{\left(2x+1\right)\left(2x+3\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(4x^2+8x+4x+8+4x^2+6x+2x+6\right)=0\)(Vì \(\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left(2x+5\right)\left(8x^2+20x+14\right)=0\)

mà \(8x^2+20x+14>0\forall x\)

nên 2x+5=0

\(\Leftrightarrow2x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{2}\)

Vậy: \(S=\left\{-\dfrac{5}{2}\right\}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\\dfrac{1}{2x+1}-\dfrac{1}{2x+3}-\dfrac{1}{2x+4}+\dfrac{1}{2x+2}=0\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow\dfrac{2x+3-2x-1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{2x+4-2x-2}{\left(2x+4\right)\left(2x+2\right)}=0\)

\(\Leftrightarrow\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{2}{\left(2x+4\right)\left(2x+2\right)}=0\)

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

......

(Vô lí)

 \(24x-4\left(2x-\frac{3}{4}\right)-4\left(3+\frac{2x}{2}\right)=36-3\left(x-\frac{3}{2}\right)-3\left(3-\frac{2x}{3}\right)\)

Đề như này đúng không bạn

1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42

a: Sửa đề: (5-2x)(5+2x)+2x(x+3)=4-2x^2

=>25-4x^2+2x^2+6x=4-2x^2

=>6x+25=4

=>6x=-21

=>x=-7/2

b: (3x-2)(-2x)+5x^2=-x(x-3)

=>-6x^2+4x+5x^2=-x^2+3x

=>4x=3x

=>x=0

c: =>7-(4x^2-9)=x^2+8x+16

=>7-4x^2+9-x^2-8x-16=0

=>-5x^2-8x=0

=>5x^2+8x=0

=>x(5x+8)=0

=>x=0 hoặc x=-8/5