Giúp mình làm với ạ,mình đang gấp lắm.Cảm ơn ạ
a)x3 + 2x2 + 4x + 8=0
b)(2x-1)2 -3/2= 3x
c) x3 - 2x2 - 24x + 8 = 0
d) |3x-2| + |2/3 - x| = 4/3
Bài1: giải các pt sau:
a, 3-4x+24+6x= x+27+3x
b, 5-(6-x)=4(3-2x)
c, x-(x+1)/3 = (2x+1)/5
d,(2x-1)/3 - (5x+2)/7 = x+13
Bài 2:
a, (x-1)(3x+1)=0
b, (x-5)(7-x)=0
c, ( x-1)(x+5)(-3x+8)=0
d, x(x^2 - 1 )=0
Giúp mình 2 bài này với , mình đang cần gấp , CẢM ƠN M.N ạ><
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
Tìm x, biết:
a) 2(5x-8)-3(4x-5) = 4(3x-4) + 11;
b) 2 x ( 6 x - 2 x 2 ) + 3 x 2 ( x - 4 ) = 8;
c) 2 ( x 3 - 1 ) - 2 x 2 ( x + 2 x 4 ) + ( 4 x 5 + 4 ) x = 6;
d)(2x)2(4x-2)-(x3 -8x2) = 15.
a) x = 2 7 b) x = 2.
c) x = 2 d) x = 1.
Tìm x biết:
a) 27x3-54x2+36x-8=0
b) x3+15x2+75x+125=0
c) x3-18x2+108x-216=0
d) (x-1)3-x.(x2+3x)=2
e) (x+1)3+(x-2)3-2x2.(x-1,5)=3
Giải chi tiết giúp mình nha.Cảm ơn nhiều
Bài 1 : giải phương trình
a) (8x + 3)(2x - 1) = (2x - 1)2
b) (x - 5)2 - 36 = 0
c) (4x - 3)2 - 4(x + 3)2
d) x3 - 3x -2 = 0
e) x3 + 2x2 - 4x - 8 = 0
tìm x, biết:
a) 9x2+36=0
b) 3(x+4)-x2-4x=0
c) x(2x-1)-(x-2)(2x+1)=0
d) (2x-3)2-4x2=00
e)1 phần 3.x2-3x=0
f) x3-x2-x+1=0
ráng giúp mình nha
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Cứu với ạ
Làm tính chia
1) (x3 – 3x2 + x – 3) : (x – 3) 2) (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3)
3) (x – y – z)5 : (x – y – z)3 4) (x2 + 2x + x2 – 4) : (x + 2)
5) (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) | 6) (2x3 – 5x2 + 6x – 15):(2x – 5) |
a) \(x^3-x^2+3x-3>0\)
\(\Leftrightarrow x^2\left(x-1\right)+3\left(x-1\right)>0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\)
Mà: \(x^2+3>0\forall x\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
b) \(x^3+x^2+9x+9< 0\)
\(\Leftrightarrow x^2\left(x+1\right)+9\left(x+1\right)< 0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+1\right)< 0\)
Mà: \(x^2+9>0\forall x\)
\(\Leftrightarrow x+1< 0\)
\(\Leftrightarrow x< -1\)
d) \(4x^3-14x^2+6x-21< 0\)
\(\Leftrightarrow2x^2\left(2x-7\right)+3\left(2x-7\right)< 0\)
\(\Leftrightarrow\left(2x^2+3\right)\left(2x-7\right)< 0\)
Mà: \(2x^2+3>0\forall x\)
\(\Leftrightarrow2x-7< 0\)
\(\Leftrightarrow2x< 7\)
\(\Leftrightarrow x< \dfrac{7}{2}\)
d) \(x^2\left(2x^2+3\right)+2x^2>-3\)
\(\Leftrightarrow2x^4+3x^2+2x^2+3>0\)
\(\Leftrightarrow2x^4+5x^2+3>0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x^2+3\right)>0\)
Mà:
\(x^2+1>0\forall x\)
\(2x^2+3>0\forall x\)
\(\Rightarrow x\in R\)
a: =>x^2(x-1)+3(x-1)>0
=>(x-1)(x^2+3)>0
=>x-1>0
=>x>1
b: =>x^2(x+1)+9(x+1)<0
=>(x+1)(x^2+9)<0
=>x+1<0
=>x<-1
c: 4x^3-14x^2+6x-21<0
=>2x^2(2x-7)+3(2x-7)<0
=>2x-7<0
=>x<7/2
d: =>x^2(2x^2+3)+2x^2+3>0
=>(2x^2+3)(x^2+1)>0(luôn đúng)
Giúp mình với ạ ;-;
a) (x-1)(2x^2-8)=0
b)3x^2-8x+5=0
c)(7x-1).2x-7x+1=0
d)(4x+2)(x-1)=1phần2x(x-1)
e)5x-2 phần 3 = 5-3x phần 2
f) 2x-1 phần x-1 + 1= 1 phần x-1
g) 1 phần x-2 + 3= x-3 phần 2-x
(x-1)(2x^2-8)=0
\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)
3x^2-8x+5=0
áp dụng công thức bậc 2 ta có:
\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)
\(\Rightarrow x=\dfrac{5}{3};x=1\)
(7x-1).2x-7x+1=0
\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)
d: \(\Leftrightarrow\left(4x+2\right)\left(x-1\right)-\dfrac{1}{2}x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+2-\dfrac{1}{2}x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-\dfrac{7}{2}x+2\right)=0\)
=>x=1 hoặc x=4/7
e: \(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)
=>10x-4=15-9x
=>19x=19
hay x=1
f: \(\Leftrightarrow\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)
=>2x-1+x-1=1
=>3x-2=1
hay x=1(loại)
g: =>1+3x-6=3-x
=>3x-5-3+x=0
=>4x-8=0
=>x=2(loại)
1. Tìm x,y:
a) (x+2)2 + (x-3)2 = 2x ( x+ 7)
b) x3- 3x2 + 3x - 126 = 0
c) x2 + y2 - 2x + 4y + 5 = 0
d) 2x2 - 2xy + y2 + 4x + 4 = 0
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)