tìm x biết:
a) ( x - 5 ) . x = 0
b) ( x - 2 ) . ( 1 - x ) = 0
c) ( x + 1 ) . ( x2 + 4 ) = 0
d) ( 2x - 4 ) .' ( 9 - 3x ) = 0
m.n làm nhanh cho mình, mình đang cầm gấp! cảm ơn m.n!
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) 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
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Tìm x, biết:
a) 7x2 - 28 = 0
b) \(\dfrac{2}{3}\)x(x2 - 4) = 0
c) 2x(3x - 5) - (5 - 3x) = 0
d) (2x - 1)2 - 25 = 0
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
a,7x2 - 28 = 0
=> 7x2 = 28 => x2 = 4 => x = 2
b,2/3x(x2 - 4) = 0
=>2/3x(x - 2)(x + 2) = 0
=> x ∈ {0 ; 2 ; -2}
c,2x(3x - 5) - (5 - 3x) = 0
= 2x(3x - 5) + (3x - 5)
= (3x - 5)(2x + 1) = 0
=> x ∈ { 5/3 ; -1/2}
d, (2x - 1)2 - 25 = 0
=> (2x - 4)(2x - 6) = 0
=> x ∈ {2 ;3}
a,7x2 - 28 = 0
=> 7x2 = 28 => x2 = 4 => x = 2
b,2/3x(x2 - 4) = 0
=>2/3x(x - 2)(x + 2) = 0
=> x ∈ {0 ; 2 ; -2}
c,2x(3x - 5) - (5 - 3x) = 0
= 2x(3x - 5) + (3x - 5)
= (3x - 5)(2x + 1) = 0
=> x ∈ { 5/3 ; -1/2}
d, (2x - 1)2 - 25 = 0
=> (2x - 4)(2x - 6) = 0
=> x ∈ {2 ;3}
Tìm x biết:
a, 16x² – 9(x + 1)²= 0
b, x2 (x – 1) – 4x2 + 8x – 4 = 0
c, x(2x – 3) – 2(3 – 2x) = 0
d, (x – 3)(x² + 3x + 9) – x(x + 2)(x – 2) = 1
e, 4x² + 4x – 6 = 2
f, 2x² + 7x + 3 = 0
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Tìm tập hợp giá trị của x biết:
a,(x-1)(x-2)>0
b,2x-3<0
c,(2x-4)(9-3x)>0
d,2x/3-3/4<0
Bài 5. Tìm x, biết:
a) x (2x - 7) + 4x -14 = 0
b) x3 - 9x = 0
c) 4x2 -1 - 2(2x -1)2 = 0
d) (x3 - x2 ) - 4x2 + 8x - 4 = 0
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Giúp mình 4 câu này với :<
Bài 1: Tìm x
a) 2x(x - 3) + x - 3 = 0
b) x2 (x - 6) - x2 + 36 = 0
c) 5 - 6x + x2 = 0
d) x(x + 5) + x(x + 15) = 0
Cảm ơn các bạn nhaaaa
a: Ta có: \(2x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
Bài 9: Tìm x, biết:
a)|-2x+1,5|=1/4
b)3/2-|1 1/4+3x|=1/4
c)|4x-1| - |3x-1/2|=0
d)|x-1|-2x=1/2
Giúp mình với mình đang cần gấp
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)
b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)
Tìm x biết:
a) 2x2 - 4 = 0
b) (x + 1)2 = 4
c) (2x - 1)2 - 9 = 0
d) x2 - x = 0
a: =>2x^2=4
=>x^2=2
=>\(x=\pm\sqrt{2}\)
b: =>(x+1)^2-4=0
=>(x+1+2)(x+1-2)=0
=>(x+3)(x-1)=0
=>x=1 hoặc x=-3
c: =>(2x-1)^2-3^2=0
=>(2x-1-3)(2x-1+3)=0
=>(2x-4)(2x+2)=0
=>x=2 hoặc x=-1
d: x^2-x=0
=>x(x-1)=0
=>x=0 hoặc x=1