x+1/x^2+x+1-x-1/x^2-x+1=2(x+2)^2/x^6-1
1/x-1-x^3-x/x^2+1(x/x^2-2x+1-1/x^2-1)
[2/(x+1)^3.(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
(x/x^2-36-x-6/x^2+6x):2x-6/x^2+6x+x/6-x
giúp mik với ;-; mik cần gấp
1. (x-6)^2 = 2(x-6)
2. 2(x-3)^2 = (x-3)(x+5)
3. 4(x-3)=2x-5(2x+3)
4. x2 +4 -2 (x-1) = (x-2)^2
5. x-3/5 = 6 - 1-2x/3
6. x+2 = 6-5x/2
7. x+2/5 - x+3 = x-2/2
8. 2x-5/x-4 = 2x+1/x+2
9. X+3/x-3 - x-1/x+3 = x2 + 4x + 6/x2 -9
10. 3x-3/x2-9 -1/x-3 = x+1/x+3
11. X+1/x-1 - x-1/x+1 = 4/x2 -1
Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!
Tìm x, biết
a/ (x-1)(x^2+x+1)-x(x+2)(x-2)=5
b/ (x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=15
c/6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1
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.
Giải các phương trình sau
d) \(\dfrac{1}{x-2}\)-\(\dfrac{6}{x+3}\)=\(\dfrac{5}{6-x^2-x}\)
e) \(\dfrac{2}{x+2}\)-\(\dfrac{2x^2+16}{x^3+8}\)=\(\dfrac{5}{x^2-2x+4}\)
f) \(\dfrac{x+1}{x^2+x+1}\)-\(\dfrac{x-1}{x^2-x+1}\)=\(\dfrac{2\left(x+2\right)^2}{x^6-1}\)
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)
1 x 2 = ?
1 x 2 x 3 = ?
1 x 2 x 3 x 4 = ?
1 x 2 x 3 x 4 x 5 = ?
1 x 2 x 3 x 4 x 5 x 6 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = ?
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.giải phương trình :
1)1 + 2/x-1 + 1/x+3=x^2+2x-7/x^2+2x-3
2)x/x^2+5x+6=2/x^2+3x+2 (x=3)
3)1/x^2+9x+20 - 1/x^2+8x+12=x^2-2x-33/x^2+8x+15 (x=-5,7)
4)x+5/3x-6 - 1/2=2x-3/2x-4 (x=25/7)
5)x-1/x^3+1 + 2x+3/x^2-x+1=2x+4/x+1 - 2(x=0)
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
1) (3x-2)/3-2=(4x+1)/42) (x-3)/4+(2x-1)/3=(2-x)/63) 1/2 (x+1)+1/4 (x+3)=3-1/3 (x+2)4) (x+4)/5-x+4=x/3-(x-2)/25) (4-5x)/6=2(-x+1)/2 6) (-(x-3))/2-2=5(x+2)/4 7)2(2x+1)/5-(6+x)/3=(5-4x)/158) (7-3x)/2-(5+x)/5=1 9)(x-1)/2+3(x+1)/8=(11-5x)/310)(3+5x)/5-3=(9x-3)/4