Tính \(Q=\frac{x^6-6x^5+12^4-8x^3+2015}{x^6-8x^3-12x^4-6x+2015}\) với \(x^2-2x-1=0.\)
Tính \(Q=\frac{x^6-6x^5+12^4-8x^3+2015}{x^6-8x^3-12x^4-6x+2015}\) với \(x^2-2x-1=0.\)
Biết \(x^2-2x-1=0\). Tính biểu thức \(\dfrac{x^6-6x^5+12x^4-8x^3+2015}{x^6-8x^3-12x^2+6x+2015}\)
Ta có : \(x^2-2x-1=0
\)
\(\Leftrightarrow \)\((x-1)^2=2\)
\(\Leftrightarrow
\)\(\left[\begin{array}{}
x-1=\sqrt{2}\\
x-1=-\sqrt{2}
\end{array} \right.\)
Đặt P = \(\dfrac{x^6-6x^5+12x^4-8x^3+2015}{x^6-8x^3-12x^2+6x+2015}\)
=\(\dfrac{(x^6-2x^5-x^4)-(4x^5-8x^4-4x^3)+(5x^4-10x^3-5x^2)-(2x^3-4x^2-2x)+(x^2-2x-1)+2016}
{(x^6-2x^5-x^4)+(2x^5-4x^4-2x^3)+(5x^4-10x^3-5x^2)+(4x^3-8x^2-4x)+(x^2-2x-1)+12x+2016}\)
=\(\dfrac{x^4(x^2-2x-1)-4x^3(x^2-2x-1)+5x^2(x^2-2x-1)-2x(x^2-2x-1)+(x^2-2x-1)+2016}
{x^4(x^2-2x-1)+2x^3(x^2-2x-1)+5x^2(x^2-2x-1)+4x(x^2-2x-1)+(x^2-2x-1)+12x+2016}\)
=\(\dfrac{2016}{12x + 2016}\)
=\(\dfrac{2016}{12(x+1)+2004}\)
=\(\dfrac{168}{x+1+167}\)
=\(\left[\begin{array}{}
\dfrac{168}{\sqrt{2}+167}\\
\dfrac{168}{-\sqrt{2}+167}
\end{array} \right.\)
Chú thích: Hình như mẫu là \(-6x\) chứ không phải \(6x
\) bạn ạ. Hay là mình phân tích sai thì cho mình xin lỗi nhé.
Giải các phương trình sau:
\(\frac{3}{4x-20}-\frac{15}{2x^2-50}+\frac{7}{6x+30}=0\)
\(\frac{8x^2}{3-12x^2}+\frac{1+8x}{4+8x}=\frac{-2x}{3-6x}\)
\(\frac{1}{x^2-2x+1}+\frac{1}{x^2+2x=1}=\frac{2}{x^2-1}\)
\(\frac{1}{x^2+1}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{4}{5}\)
Thực hiên cá phép tính sau
a) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
b) \(\frac{12x}{5y^3}-\frac{15y^4}{8x^3}\)
c) \(\frac{x^2-4}{3x+12}\) \(\frac{x+4}{2x-4}\)
a)\(\Rightarrow\frac{3}{2.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}\)
\(\Rightarrow\frac{3x-x+6}{2x.\left(x+3\right)}\)
\(\Rightarrow\frac{2x+6}{2x.\left(x+3\right)}=\frac{2.\left(x+3\right)}{2x.\left(x+3\right)}=\frac{2}{2x}=\frac{1}{x}\)
b
=\(\frac{96x^4-75y^7}{40x^3y^3}\)
c, phan tich ra:
=\(\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}=\frac{x+2}{6}\)
=
a) \(\frac{1+8x}{8x+4}=\frac{2x}{6x-3}-\frac{8x^2}{3-12x^2}\)
b)(x-2)(x-3)<(x-4)2-2(x+3)
a) ĐKXĐ: \(x\notin\left\{\frac{1}{2};\frac{-1}{2}\right\}\)
Ta có: \(\frac{1+8x}{8x+4}=\frac{2x}{6x-3}-\frac{8x^2}{3-12x^2}\)
\(\Leftrightarrow\frac{8x+1}{4\left(2x+1\right)}=\frac{2x}{3\left(2x-1\right)}+\frac{8x^2}{3\left(4x^2-1\right)}\)
\(\Leftrightarrow\frac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}=\frac{2x\cdot4\cdot\left(2x+1\right)}{12\left(2x+1\right)\left(2x-1\right)}+\frac{32x^2}{12\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(3\left(8x+1\right)\left(2x-1\right)=8x\left(2x+1\right)+32x^2\)
\(\Leftrightarrow3\left(16x^2-8x+2x-1\right)=16x^2+8x+32x^2\)
\(\Leftrightarrow3\left(16x^2-6x-1\right)=48x^2+8x\)
\(\Leftrightarrow48x^2-18x-3-48x^2-8x=0\)
\(\Leftrightarrow-26x-3=0\)
\(\Leftrightarrow-26x=3\)
hay \(x=-\frac{3}{26}\)
Vậy: \(S=\left\{-\frac{3}{26}\right\}\)
b) Ta có: \(\left(x-2\right)\left(x-3\right)< \left(x-4\right)^2-2\left(x+3\right)\)
\(\Leftrightarrow x^2-5x+6< x^2-8x+16-2x-6\)
\(\Leftrightarrow x^2-5x+6< x^2-10x+10\)
\(\Leftrightarrow x^2-5x+6-x^2+10x-10< 0\)
\(\Leftrightarrow5x-4< 0\)
\(\Leftrightarrow5x< 4\)
hay \(x< \frac{4}{5}\)
Vậy: S={x|\(x< \frac{4}{5}\)}
Tìm x biết
1) 8x ^ 3 - 12x ^ 2 + 6x - 1 = 0
2) x ^ 3 - 6x ^ 2 + 12x - 8 = 27
3) x ^ 2 - 8x + 16 = 5 * (4 - x) ^ 3
4) (2 - x) ^ 3 = 6x(x - 2)
5) (x + 1) ^ 3 - (x - 1) ^ 3 - 6 * (x - 1) ^ 2 = - 10
6) (3 - x) ^ 3 - (x + 3) ^ 3 = 36x ^ 2 - 54x
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) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)
\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6
\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6
\(\Leftrightarrow\) 11x = 9
\(\Leftrightarrow\) x = 0,8
Vậy S = {0,8}
2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12
\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)
\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x
\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20
\(\Leftrightarrow\) 17x = 7
\(\Leftrightarrow\) x = \(\frac{7}{17}\)
Vậy S = {\(\frac{7}{17}\)}
3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15
\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)
\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3
\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3
\(\Leftrightarrow\) 4x = 2
\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)
Vậy S = {\(\frac{1}{2}\)}
4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12
\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)
\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24
\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24
\(\Leftrightarrow\) 5x = -58
\(\Leftrightarrow\) x = -11,6
Vậy S = {-11,6}
5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)
\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x
\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10
\(\Leftrightarrow\) -8x = -1
\(\Leftrightarrow\) x = \(\frac{1}{8}\)
Vậy S = {\(\frac{1}{8}\)}
6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12
\(\Leftrightarrow\frac{3.\left(12x+1\right)}{12}=\frac{4.\left(9x+1\right)}{12}-\frac{3-5x}{12}\)
\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x
\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3
\(\Leftrightarrow\) -5x = -2
\(\Leftrightarrow x=\frac{2}{5}\)
7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)
\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0
\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4
\(\Leftrightarrow\) -3x = -18
\(\Leftrightarrow\) x = 6
Vậy S = {6}
8) x\(^2\) - x - 6 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0
\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0
\(\Leftrightarrow\) (x - 3).(x + 2) = 0
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -2
Vậy S = {3; -2}
Giúp em giải 2 phương trình này với?
a)\(8x^3-12x^2+6x+1=29\)
b)\(\frac{x+3}{x+2}+\frac{x+6}{x+5}=\frac{x+5}{x+4}+\frac{x+4}{x+3}\)
cau a: 8x^3 -12x^2 + 6x + 1 =29
<=>8x^3 - 12x^2 + 6x - 28 =0
<=>(8x^3 - 16x^2)+(4x^2 - 8x)+(14x-28)=0
<=>8x^2 ( x-2) + 4x(x-2) + 14(x-2)=0
<=>(x-2)(8x^2 + 4x +14)=0
<=>8x^2 +4x +14 =0 <=> 8(x^2 +1/2 x +7/4)=0<=>(x^2 +2* x*1/4 + 1/16) +27/16 =0 <=>(x+ 1/4)^2=-27/16 (0xay ra) (loai)
=>(x-2)(8x^2 +4x+14)=0 <=> x-2=0 <=>x=2
Vay tap nghiem phuong trinh S={2}
b) x+3/x+2 +x+6/x+5=x+5/x+4 + x+4/x+3
=(x+3/x+2 -1) +(x+6/x+5 - 1)=(x+5/x+4 -1) +(x+4/x+3 -1)
=1/x+2 +1/x+5 -1/x+4 -1/x+3 =0
x=1000
Giải pt sau
a.(2x+3)(x-5)=4x2+6x
b.x/2x-6 - x/2x+2 = 2x/(x+1)(x-3)
c.giải bpt sau : 12x+1/12 ≤ 9x+1/3 - 8x+1/4