( x^3 + 1 ) : (x^2 - x + 1 )
Những câu hỏi liên quan
a, (3x+2)2 - (3x-2)2 =5x+38 b, 3(x-2)2 +9(x-1) =3(x2+x-3)
c, (x+3)3 -(x-3)2 -(x-3)2 =6x+18 d, (x-1)3-x(x+1)2=5x(2-x)-11(x+2)
e, (x+1)(x2-x+1)-2x=x(x-1)(x+1) f, (x-2)3+(3x-1)(3x+1)=(x+1)3
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
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1. x+1/x-1 - x-1/x+1 = 4/x2-1
2. x+2/x-2 + x/x+2 = 2
3. 2x/x-1 - 1/x+2 = 2
4. x/x2-25 - 1-x/x-5 = 1/x+5
5. x-1/x+2 - 9/x2-4 = -3/x-2
6. 3x-3/x2-9 - 1/x-3 = x+1/x+3
Xem chi tiết
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
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a) x^3 + 1 (x + 1)(x^2 - x + 1) x^9 + x^7 - 3x^2 - 3 x^7(x^2 + 1) - 3(x^2 + 1) (x^2 + 1)(x^7 - 3).Điều kiện của x để giá trị của biểu thức Q xác định là x neq -1, x^7 neq 3, x neq -3, x neq 4.b) Q left[frac{x^7 -3}{x^3 + 1}.frac{(x - 1)(x + 1)(x^2 - x + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - frac{2(x + 6)}{x^2 + 1}right].frac{(2x + 1)^2}{(x + 3)(4 - x)} left[frac{x^7 - 3}{x^3 + 1}.frac{(x - 1)(x^3 + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - frac{2(x + 6)}{x^2 + 1}right].frac{(2x + 1)^2}{(x + 3)(4 - x)}
Đọc tiếp
a) \(x^3 + 1 = (x + 1)(x^2 - x + 1)\)
\(x^9 + x^7 - 3x^2 - 3 = x^7(x^2 + 1) - 3(x^2 + 1) = (x^2 + 1)(x^7 - 3)\).
Điều kiện của x để giá trị của biểu thức Q xác định là \(x \neq -1, x^7 \neq 3, x \neq -3, x \neq 4\).
b) \(Q = \left[\frac{x^7 -3}{x^3 + 1}.\frac{(x - 1)(x + 1)(x^2 - x + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - \frac{2(x + 6)}{x^2 + 1}\right].\frac{(2x + 1)^2}{(x + 3)(4 - x)}\)
\(= \left[\frac{x^7 - 3}{x^3 + 1}.\frac{(x - 1)(x^3 + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - \frac{2(x + 6)}{x^2 + 1}\right].\frac{(2x + 1)^2}{(x + 3)(4 - x)}\)
. Tìm x biết rằng:
a)(x + 1)3 – (x + 2)(x – 1)2 – 3(x – 3)(x + 3) = 5
b)(x + 1)3 + (x – 1)3 = (x + 2)3 + (x – 2)3
c) (x + 1)3 - (x - 1)3 - 6(x - 1)2 = -10
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
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(X-1)^3 = (1-x)^2
Giải phương trình:
A) 1-x/x+1 +3 = 2x+3/x+1
B) (x+2)^2/2x-3 -1 = x^2-10/2x-3
C) 5x-2/2-2x + 2x-1/2 = 1 + x^2+x-3/x-1
D) 5-2x/3 - (x-1)(x+1)/1-3x = (x+2)(1-3x)/9x-3
E) x-3/x-2 + x-2/x-4 = -1
F) 1 + x/3-x = 5/(x+2)(3-x) + 2/x+2
G) x+1/x-1 - x-1/x+1 = 3x( 1 - x-1/x+1 )
H) 1-6x/x-2 + 9x-4/x+2 = x(3x-2)+1/x^2-4
I) 3x-1/x-1 - 2x+5/x+3 + 4/x^2+2x-3 = 1
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
Trả lời:
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)\(\left(đkxđ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow4+2x=2x+3\)
\(\Leftrightarrow2x-2x=3-4\)
\(\Leftrightarrow0x=-1\)(không thỏa mãn)
Vậy \(S=\varnothing\)
b, \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\)\(\left(đkxđ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{\left(x+2\right)^2-\left(2x-3\right)}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow x^2+2x+7=x^2-10\)
\(\Leftrightarrow x^2+2x-x^2=-10-7\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(tm)
Vậy \(S=\left\{\frac{-17}{2}\right\}\)
c, \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)\(\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\frac{2-5x}{2x-2}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)
\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)
\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)}+\frac{2\left(x^2+x-3\right)}{2\left(x-1\right)}\)
\(\Rightarrow2-5x+2x^2-3x+1=2x-2+2x^2+2x-6\)
\(\Leftrightarrow2x^2-8x+3=2x^2+4x-8\)
\(\Leftrightarrow2x^2-8x-2x^2-4x=-8-3\)
\(\Leftrightarrow-12x=-13\)
\(\Leftrightarrow x=\frac{13}{12}\)(tm)
Vậy \(S=\left\{\frac{13}{12}\right\}\)
Xem thêm câu trả lời
giải phương trình:
a) 2/x+1 - 1/x-3= 3x-11/x^2-2x-3
b) 3/x-2 +1/x=-2/x.(x-2)
c) x-3/x+3 - 2/x-3=3x+1/9-x^2
d) 2/x+1 - 1/x-2=3x-5/x^2-x-2
e) x-2/x+2 + 3/x-2=x^2-11/x^2-4
f) x+3/x+1 + x-2/x=2
g) x+5/x-5 - x-5/x+5=20/x^2-25
h) x+4/x+1 + x/x-1=2x^2/x^2-1
i) x+1/x-1 - 1/x+1=x^2+2/x^2-1
Giải các phương trình sau:1) 2 1 5 x 2) 2 1 5 x x 3) 3 1 2 x x 4) 3 2 2 x x5) 2 1 5 x x 6) 3 2 x x7) 2 3 2 1 x x 8) 2 1 4 1 0 x x 29) 2 5 4 3 1 1 23 2 3 1x xx x x x 10) 1 7 3 23 3 9x x xx x x 11) 5 296 2 1 3 116 4 4x xx x x 12) 2 412 1 2 1 2 1 2 1x xx x x x 13) 2 1 2 22 2xx x x x 14) 22 42 6 2 2 2 3
Đọc tiếp
Giải các phương trình sau:
1) 2 1 5 x 2) 2 1 5 x x
3) 3 1 2 x x 4) 3 2 2 x x
5) 2 1 5 x x 6) 3 2 x x
7) 2 3 2 1 x x 8) 2 1 4 1 0 x x 2
9) 2 5 4 3 1 1 2
3 2 3 1
x x
x x x x
10) 1 7 3 2
3 3 9
x x x
x x x
11) 5 296 2 1 3 1
16 4 4
x x
x x x
12)
2 4
1
2 1 2 1 2 1 2 1
x x
x x x x
13) 2 1 2 2
2 2
x
x x x x
14) 22 4
2 6 2 2 2 3
Tìm x biết : 6(x+2)(x-3)-3(x-2)^2-3(x-1)(x+1)=1
3(x+2)^2+(2x-1)^2-7(x+3)(x-3)=36
(x-1)(x^2+x+1)+x(x+2)(2-x)=5
(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2
Tìm x biết : 6(x+2)(x-3)-3(x-2)^2-3(x-1)(x+1)=1
3(x+2)^2+(2x-1)^2-7(x+3)(x-3)=36
(x-1)(x^2+x+1)+x(x+2)(2-x)=5
(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2
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!!!!!
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