Bài 10:

a: Để A là phân số thì n+2<>0

hay n<>-2

b: Khi n=0 thì A=3/2

Khi n=2 thì A=3/(2+2)=3/4

Khi n=-7 thì A=3/(-7+2)=-3/5

Bài 9:

1)9/x = -35/105               2) 12/5 = 32/x                   3)x/2 = 32/x                            x = 9. (-35)/105              x.12/5 = x.32/x                    2x.x/2 = 2x.32/x        

        x = -3                              x.12/5=32                         xx = 2.32

                                                        x= 32:12/5                x^2 = 2.32

                                                         x = 40/3                   x^2 = 64

                                                                                         x = 8

4) x-2/4 = x-1/5

      5(x-2) = 4(x-1)
       5x - 10 = 4x - 4
        5x - 4x = 10 - 4
         x = 6   

  Bài 10:Cho biểu thức 

VP đâu bạn?

Lời giải:

1. $(x+2)-2=0$

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

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

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

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

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

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

\(\Leftrightarrow x=\dfrac{19}{5}\)

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

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

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

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)

Ko thấy j hết á bạn

1)

\(3\left(x-2\right)+4\left(x-1\right)=25\)

\(3x-6+4x-4=25\)

\(7x-10=25\\ 7x=35\\ x=5\)

2)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)

\(\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)

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

\(\left(x-2\right)\left(4x-2\right)=0\)

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

3)

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

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

\(x^2-4x+4=4x^2-8x+4\)

\(x^2-4x+4-4x^2+8x-4=0\)

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

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

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

1. x(x + 1) - x² + 1 = 0

<=> x(x + 1) - (x² - 1) = 0

<=> x(x + 1) - (x + 1)(x - 1) = 0

<=> (x - x + 1)(x + 1) = 0

<=> x + 1 = 0\

<=> x = -1

2. 4x(x - 2) - 6 + 3x = 0

<=> 4x(x - 2) - (3x - 6) = 0

<=> 4x(x - 2) - 3(x - 2) = 0

<=> (4x - 3)(x - 2) = 0

<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)

3. x(x + 2) - 3(x + 2) = 0

<=> (x - 3)(x + 2) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)