tìm x biết: (4x-12)(x^3+64)=0
(3x-12)(x^2-4)=0
(x+3)^3:3-1=-10
(3x-1)^3-2=-66
m.n giúp mik vs tối mik phải nộp r ak
Những câu hỏi liên quan
x mũ 3 - x mũ 2 +3x - 3 > 0
x mũ 3 + x mũ 2 + 9x + 9 < 0
4x mũ 3 - 14x mũ 2+ 6x - 21<0
men nào giúp mik vs
tối nay mik phải đi hok r
bài 1 tìm x
a)3x(x-3)+4x-12=0
b)(x+1)(x^2-x+1)-x^3+2x-=17
c)(x-3)(x+5)+(x-1)^2-6x^4y^2:3x^2y^2=15x
giúp mik vs nhanh ak cảm ơn nhìu!
a: \(3x\left(x-3\right)+4x-12=0\)
=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)
=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(3x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)
\(\Leftrightarrow x^3+1-x^3+2x=17\)
=>2x+1=17
=>2x=17-1=16
=>\(x=\dfrac{16}{2}=8\)
c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)
=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)
=>\(15x=-14\)
=>\(x=-\dfrac{14}{15}\)
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2 . ( x³ -1)-2x²(x+2x⁴) +(4x⁵+4)x=6. 3. (X²-4x+16)(x+4)-x(x+1)(x+2)+3x²= 0 4 . ( 8x +2 ) (1-3x) + ( 6x-1)(4x-10) =-50 Đề bài là tìm x nha mn Nhanh giúp mik vs
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
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Giúp mik với đang cần gấp
Bài 2: Tìm x biết:
a) x.(x + 1) – x² + 2 = 0
b) 2.(3x + 2) – (2x + 12) = 0
c) 2x³(2x – 3) – x²(4x² – 6x + 2) = 0
d) (3x + 2)(x – 1) – 3(x + 1)(x – 2) = 4
tìm x biết:
a)2/3x-3/2x=5/12
b)2/5+3/5.(3x-3,7)=-53/10
c)7/9:(2+3/4x)+5/9=23/27
d)-2/3.x+1/5=3/10
e)|x|-3/4=5/3
f)|2x-1/3|+5/6=1
giúp mik vs mik đang cần gấp bn nào giải mik cũng tick nha
giải hết giúp mik các bn nhé
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
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b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
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c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)
\(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)
\(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)
2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)
2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{21}{8}\)
\(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2
\(\dfrac{3}{4}\)\(x\) = \(\dfrac{5}{8}\)
\(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{5}{6}\)
Vậy \(x=\dfrac{5}{6}\)
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Bài 1: Tìm x biết a) x^3 - 4x^2 - x + 4= 0 b) x^3 - 3x^2 + 3x + 1=0 c) x^3 + 3x^2 - 4x - 12=0 d) (x-2)^2 - 4x +8 =0
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
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Tìm x,biết
b) |3/4x-5|-2/3=|-1/4|
a) (3x-1) (-1/4 - 5x) (-2/7x + 3)=0
c)|-2/5 - 3x| - |-7/9 + 2x|=0
d) |-3/7x - 1| + |5+2/3x|=0
e)|3x-1|+|9x^2 - 1|=0
f) x+2/2018 + x+4/2016 = x+6/2014 - 1
Mọi người giải giúp e với ạ chiều e phải nộp r
ngu thế à bạn
Tìm x
Y) x^2-x-6=0
Z) 3x² –5x–8=0
J) 25x^2-4=0
R) 2(x+3)-x^2-3x=0
U. x³–3x² –x+3=0
Giúp mik vs mình cần gấp
y) \(x^2-x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;3\right\}\) là nghiệm của pt.
z) \(3x^2-5x-8=0\\ \Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{8}{3};-1\right\}\) là nghiệm của pt.
j) \(25x^2-4=0\\ \Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{2}{5};\dfrac{-2}{5}\right\}\) là nghiệm của pt.
r) \(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{-3;2\right\}\) là nghiệm của pt.
u) \(x^3-3x^2-x+3=0\\ \Leftrightarrow x^2\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{-1;1;3\right\}\) là nghiệm của pt.
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Xem thêm câu trả lời
Tìm x
Y) x^2-x-6=0
Z) 3x² –5x–8=0
J) 25x^2-4=0
R) 2(x+3)-x^2-3x=0
U. x³–3x² –x+3=0
Giúp mik vs mình cần gấp
y: Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
z: Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)
j: Ta có: \(25x^2-4=0\)
\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
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r: Ta có: \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
u: Ta có: \(x^3-3x^2-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)
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