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đúng vậy

a)(x-1)(x²+x+1)-x(x+2)(x-2)=5

=>x³-1-4x-x³=5

=>x³-x³+4x-1=5

=>4x-1=5

=>4x=6

=>x=3/2

b)(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=15

=>x³-6x²+12x-8-x³+27+6x²+12x+6=15

=>(x³-x³)-(-6x²+6x²)+(12x+12x)-8+27+6=15

=>24x+25=15

=>24x=-10

=>x=-5/12

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

=>6x²+12x+6-2x³-6x²-6x-2+2x³-2=1

=>(6x²-6x²)+(12x-6x)-(-2x³+2x³)+6-2-2=1

=>6x+2=1

=>6x=-1

=>x=-1/6

x²-4x+7 = 0 ⇔ x² -4x + 4 + 3 = 0 

⇔ (x-2)²+3=0 ⇔ (x-2)²=-3 (vô lí)

Vậy pt vô nghiệm

*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm

Ta có: \(x^2-4x+7=0\)

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

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

mà \(\left(x-2\right)^2+3\ge3>0\forall x\)

nên \(x\in\varnothing\)(đpcm)

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

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

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

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)

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

=>x+3-6x+12=-5

=>-5x+15=-5

=>-5x=-20

=>x=4(nhận)

e: ĐKXĐ: x<>-2

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)

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

=>\(2x^2-4x+8-2x^2-16=5x+10\)

=>5x+10=-4x-8

=>9x=-18

=>x=-2(loại)

f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)

\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)

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

=>\(\dfrac{\left(x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

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

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

=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)

=>8x=-10

=>x=-5/4(nhận)

1x2= 2       1x2x3=6             1x2x3x4=24               1x2x3x4x5=120            1x2x3x4x5x6=720                   1x2x3x4x5x6x7=5040 

1x2x3x4x5x6x7x8=40320                 1x2x3x4x5x6x7x8x9=362880           1x2x3x4x5x6x7x8x9x10=3628800

1 x 2 = 2

1 x 2 x 3 = 6

1 x 2 x 3 x 4 = 24

1 x 2 x 3 x 4 x 5 = 120

1 x 2 x 3 x 4 x 5 x 6 = 720

1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040

1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320

1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362880

1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800

1 . 2 = 2

1 .2 .3 = 6

1 .2 .3 .4 = 24

1 .2 .3 .4 .5 = 120

1 .2 .3 .4 .5 .6 = 720

1 .2 .3 .4 .5 .6 .7 = 5040

1 .2 .2 .4 .5 .6 .7 .8 = 40320

1 .2 .3 .4 .5 .6 .7 .8 .9 =362880

1 .2 .3 .4 .5 .6 .7 .8 .9 .10 = 3628800

          hok tốt

1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)

ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)

<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)

<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)

<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)

<=> \(\frac{3x+10}{x^2+2x-3}=0\)

<=> \(3x+10=0\)

<=> \(x=-\frac{10}{3}\)

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

Suy ra: \(\left(x+1\right)^2\cdot\left(x^2-x+1\right)-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=2\left(x+2\right)^2\)

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

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

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

\(\Leftrightarrow2x^2\left(x+1\right)-2\left(x+2\right)^2=0\)