Lời giải:

$3^{n+2}-2^{n+2}+3^n-2^n=9.3^n-4.2^n+3^n-2^n$

$=(9.3^n+3^n)-(4.2^n+2^n)=10.3^n-5.2^n$

$=10.3^n-10.2^{n-1}=10(3^n-2^{n-1})\vdots 10$ với mọi $n\in\mathbb{N}^*$

`|2x+1|-3=x+4`

`<=>|2x+1|=x+4+3=x+7(x>=-7)`

`**2x+1=x+7`

`<=>x=7-1=6(tm)`

`**2x+1=-x-7`

`<=>3x=-6`

`<=>x=-2(tm)`

`|3x-5|=1-3x(x<=1/3)`

`**3x-5=1-3x`

`<=>6x=6`

`<=>x=1(l)`

`**3x-5=3x-1`

`<=>-5=-1` vô lý

`|2x+2|+|x-1|=10`

Nếu `x>=1`

`pt<=>2x+2+x-1=10`

`<=>3x+1=10`

`<=>3x=9`

`<=>x=3(tm)`

Nếu `x<=-1`

`pt<=>-2x-2+1-x=10`

`<=>-1-3x=10`

`<=>-11=3x`

`<=>x=-11/3(tm)`

Nếu `-1<=x<=1`

`pt<=>2x+2+1-x=10`

`<=>x+3=10`

`<=>x=7(l)`

Vậy `S={3,-11/3}`

d) 

+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\)  (Nhận)

+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)

+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)

+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)

  Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\) 

M= x²  + 2xy - 3x - 3y + y² + 10

= x² + 2xy + y² - 3(x + y) + 10

= (x + y)² - 3(x + y) + 10

= 4 - 6 + 10 ( do x + y = 2 )

= 8

Vậy M = 8

đề là j nhỉ :>?

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

\(=\left(2x^2+6x-x-3\right)-\left(3x^2-4x-6x+8\right)+5x\)

\(=\left(2x^2+5x-3\right)-\left(3x^2-10x+8\right)+5x\)

\(=2x^2+5x-3-3x^2+10x-8+5x\)

\(=x^2+20x-11\)

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

\(=10x^3-15x^2+5x-2x\left(x^2-2x+x-2\right)\)

\(=10x^3-15x^2+5x-2x^3+4x^2-2x^2+4x\)

\(=8x^3-13x^2+9x\)

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

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

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

\(C=\left(3x+2\right)\left(x-1\right)-2x\left(x+3\right)-2x+1\\ C=3x^2-x-2-2x^2-6x-2x+1\\ C=x^2-9x-1\)

chs game ít thôi bn

a, 3x(2x+1)+4<2x(3x-1)-6

\(\Leftrightarrow\)6x²+3x+4<6x²-2x-6

\(\Leftrightarrow\)6x²+3x-6x²+2x < -6-4

\(\Leftrightarrow\)5x<-10

\(\Leftrightarrow\)x<-2

Vậy bpt có nghiệm x<-2

b,(2x-3)² < (2x+5)(2x-5)

\(\Leftrightarrow\)4x²-6x+9 < 4x²-25

\(\Leftrightarrow\)4x²-6x-4x² < -25-9

\(\Leftrightarrow\)-6x < -34

\(\Leftrightarrow\)x > \(\frac{17}{3}\)

Vậy bpt có nghiệm x > \(\frac{17}{3}\)

a)3x(2x+1)+4<2x(3x-1)-6
⇔\(6x^2+3x+4< 6x^2-2x-6\)
⇔\(6x^2-6x^2+3x+2x< -6-4\)
⇔5x < -10
⇔x > -2
Vậy bpt có tập nghiệm {x/x > -2}

b) \(\left(2x-2\right)^2< \left(2x+5\right)\left(2X-5\right)\)
⇔\(\left(2x\right)^2-2.2x.x+2^2< \left(2x\right)^2-5^2\)
⇔\(4x^2-8x+4< 4x^2-25\)
⇔\(-8x< -29\)
⇔\(x>\frac{29}{8}\)
Vậy bpt có tập nghiệm {x/\(x>\frac{29}{8}\)}

Tôi học lớp 3 ok

jmkhgfda????

tui học lớp 4 

\(\left|2x-1\right|-3=11\)

=> \(\left|2x-1\right|=11+3\)

=> \(\left|2x-1\right|=14\)

=> \(\orbr{\begin{cases}2x-1=14\\2x-1=-14\end{cases}}\)

=> \(\orbr{\begin{cases}2x=15\\2x=-13\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{15}{2}\\x=-\frac{13}{2}\end{cases}}\)

\(\left|-2x+3\right|-3=x\)

=> \(\left|-2x+3\right|=x+3\) (Đk: \(x+3\ge0\) => \(x\ge-3\))

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

=> \(\orbr{\begin{cases}-3x=0\\-x=-6\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=6\end{cases}}\) (tm)