{[(2x+14:8-3]:2}-1=0
giúp mình giai nhanh nha
(2x-1)^3-8x+4=0
Giúp mình nhanh với ạ!
(2x - 1)³ - 8x + 4 = 0
(2x - 1)³ - 4x(2x - 1) = 0
(2x - 1)[(2x - 1)² - 4x] = 0
(2x - 1)[(2x - 1)(2x - 1) - 4x] = 0
(2x - 1)[2x(2x - 1) - 1.(2x - 1) - 4x] = 0
(2x - 1)(4x² - 2x - 2x + 1 - 4x) = 0
(2x - 1)(4x² + 1) = 0
⇒ 2x - 1 = 0 hoặc 4x² + 1 = 0
*) 2x - 1 = 0
2x = 1
x = 1/2
*) 4x² + 1 = 0
4x² = -1 (vô lý vì 4x² ≥ 0 với mọi x)
Vậy x = 1/2
`(2x-1)^3-8x+4=0`
`-> (2x-1)^3 - 4(2x-1)=0`
`-> (2x-1)^2*(2x-1)-4(2x-1)=0`
`-> (2x-1)* [(2x-1)^2-4]=0`
`->` TH1: `2x-1=0`
`-> 2x=1`
`-> x=1/2`
TH2: `(2x-1)^2-4=0`
`-> (2x-1)^2=4`
`-> (2x-1)^2=(+-2)^2`
`->`\(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={1/2; 3/2; -1/2}`.
a) x^2 - 25 - 3 *(x - 5) =0
b) (3x + 1 )^2 = (2x - 5 )
c) 2x^2 - 7x + 6 = 0
giúp mình vs
*đừng giải tắt nha*
mình cần gấp
a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)
c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)
a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)
b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương
c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)
\(\left(3x+1\right)^2=\left(2x-5\right)^2\)
\(\left[{}\begin{matrix}3x+1=2x-5\\3x+1=5-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\5x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{5}\end{matrix}\right.\)
a) 1-(5/3/8+x-7/5/24)=16/2/3=0
Giúp mình nha mình đg cần gấp
\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)
Ta có: \(1-\left(\dfrac{5}{\dfrac{3}{8}}+\dfrac{x-7}{\dfrac{5}{24}}\right)-\dfrac{16}{\dfrac{2}{3}}=0\)
\(\Leftrightarrow\dfrac{40}{3}+\dfrac{24\left(x-7\right)}{5}=23\)
\(\Leftrightarrow\left(x-7\right)\cdot\dfrac{24}{5}=\dfrac{29}{3}\)
\(\Leftrightarrow x-7=\dfrac{145}{72}\)
hay \(x=\dfrac{649}{72}\)
a) (x\(^2\)+ 3x - 3)\(^2\) - 12x\(^2\) - 36x + 36 = 0
b) x\(^3\) + 1 + 3(x\(^2\) - x +1) = 0
c) x\(^3\) - 2x\(^2\) + 4x - 8 = 0
Giúp mình với ạ
Giai các bpt sau
a,(x+2)(x-3)>0
b,(x-1)(x+4)≤0
giúp mk vs ạ ai nhanh mk tick nka
a: (x+2)(x-3)>0
nên x+2;x-3 cùng dấu
=>x>3 hoặc x<-2
b: (x-1)(x+4)<=0
nên x-1 và x+4 khác dấu
=>-4<=x<=1
Cho A= x^3 -2x+n B=x-2 Tìm n để A chia hết cho B
(x-3)^2-2x+6=0
x^2-5x+6=0
Giúp mình nhanh vs ạ
\(A=x^3-2x+n\)
\(B=n-2\)
\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)
Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)
Để \(A\text{⋮}B\)
⇒ \(n+4=0\)
⇒ \(n=-4\)
Tìm x
a) ( 2x + 1 )2- 4x2 + 2x2 - 2 = 0
b) ( x - 2 ) . ( x + 2 ) - ( x + 3 )2 - 2x - 5 = 0
Giúp mình với ;-;
a. (2x + 1)2 - 4x2 + 2x2 - 2 = 0
<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x2 - 1) = 0
<=> (4x + 1) + 2x2 - 2 = 0
<=> 4x + 1 + 2x2 - 2 = 0
<=> 2x2 + 4x - 2 + 1 = 0
<=> 2x2 + 4x - 1 = 0
<=> 2x2 + 4x = 1
<=> 2x(x + 2) = 1
Vì 1 chỉ có tích là 1 . 1 nên:
<=> \(\left[{}\begin{matrix}2x=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)
Tìm x biết
a,(x+4)(3x-5)=0
b, x2-2x+10x-20=0
c, 2x2+7x+3=0
GIÚP MÌNH VỚI NHA
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
=>\(x\in\left\{0;1;-1\right\}\)