=>|x-3|-|x-2|=5

TH1: x<2

Pt sẽ là 3-x-(2-x)=5

=>3-x-2+x=5

=>1=5(loại)

Th2: 2<=x<3

Pt sẽ là 3-x-(x-2)=5

=>3-x-x+2=5

=>5-2x=5

=>x=0(loại)

Th3: x>=3

Pt sẽ là x-3-(x-2)=5

=>x-3-x+2=5

=>-1=5(loại)

`TH_1`: \(x\ge3\)

Khi đó:

\(\left|x-3\right|-\left|2-x\right|=5\\ \Leftrightarrow x-3-\left(x-2\right)=5\\ \Leftrightarrow x-3-x+2=5\\ \Leftrightarrow-1=5\left(loại\right)\)

`TH_2`: \(x< 2\)

Khi đó:

\(\left|x-3\right|-\left|2-x\right|=5\\ \Leftrightarrow3-x-\left(2-x\right)=5\\ \Leftrightarrow3-x-2+x=5\\ \Leftrightarrow1=5\left(loại\right)\)

`TH_3`: \(2\le x< 3\)

Khi đó:

\(\left|x-3\right|-\left|2-x\right|=5\\ \Leftrightarrow3-x-\left(x-2\right)=5\\ \Leftrightarrow3-x-x+2=5\\ \Leftrightarrow5-2x=5\\ \Leftrightarrow x=0\left(loại\right)\)

Vậy PT trên vô nghiệm

số bé là:

   (1995-5)/(200-1)=10

số lớn là:

    10+1995=2005

           Đáp số: 10 và 2005

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`

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

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

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

\(=4x^2+12x+9+4x^2-12x+9-\left(4x^2-9\right)\)

\(=8x^2+18-4x^2+9=4x^2+27\)

d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)

\(=x^3-1-8x^3-27=-7x^3-28\)

e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)

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

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

=2

ngược dấu hai chỗ điều kiện rồi bạn

a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)

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

\(=8x^2+2x-3\)

Lời giải:

a.

$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$

$=4x^2-12x+9+4x^2-4x-15$

$=24-8x$
b.

$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$

c.

$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$

$=6x^2-24x-6x^2-32x+10$

$=8x-10$

d.

$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$

$=x^2-9-x^2+10x-25=10x-34$

e.

$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$

$=-3x^2y+3xy^2=3xy(y-x)$

a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)

\(=4x^2-12x+9+2x-4x^2+15-6x\)

\(=-16x+24\)

b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)

\(=6x^2-24x+30x-6x^2-10+2x\)

\(=8x-10\)