Tìm x biết
|2x| - |-2.5| =|-7.5| với x> 0
|x| = |x +1|
|x+2| +(x -1) =0
|x+1|+(x+4) =3x
[\(\dfrac{1}{2}\)x-3].[\(\dfrac{2}{3}\)x+\(\dfrac{1}{2}\)]=0
|2x|-|-2.5|=|-7.5
|1-3x|=x-7 [ dieu kien : x-7>,=0]
Giải:
a) \(\left(\dfrac{1}{x}-3\right)\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\\dfrac{2}{3}x+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3\\\dfrac{2}{3}x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
b) \(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Leftrightarrow\left|2x\right|-2,5=7,5\)
\(\Leftrightarrow\left|2x\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vây ...
c) \(x-7\ge0\Leftrightarrow x\ge7\)
\(\left|1-3x\right|=x-7\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=x-7\\1-3x=7-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x-x=-7-1\\-3x+x=7-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x=-8\\-2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy ...
Tìm x biết a) x(x-25)=0 b)2x(x-4)-x(2x-1)=-28 c)x^2 -5x=0 d)(x-2)^2-(x+1)(x+3)=-7 e)(3x+5).(4-3x)=0 f)x^2-1/4=0
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
Tìm x,biết:
a) (x-3)^2-4=0
b) x^2-9=0
c) x(x-2x)-y^2-8=0
d) 2x(x-1)-2x^2+x-5=0
e) x(x-3)-(x+1)(x-2)=0
f) x(3x-1)-3x^2-7x=0
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 = 4
<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy S = { 5 ; 1 }
b) x2 - 9 = 0
<=> x2 = 9
<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy S = { 3 ; -3 }
c) x( x - 2x ) - x2 - 8 = 0
<=> x2 - 2x2 - x2 - 8 = 0
<=> -2x2 - 8 = 0
<=> -2x2 = 8
<=> x2 = -4 ( vô lí )
<=> x = \(\varnothing\)
Vậy S = { \(\varnothing\)}
d) 2x( x - 1 ) - 2x2 + x - 5 = 0
<=> 2x2 - 2x - 2x2 + x - 5 = 0
<=> -x - 5 = 0
<=> -x = 5
<=> x = -5
Vậy S = { -5 }
e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0
<=> x2 - 3x - ( x2 - x - 2 ) = 0
<=> x2 - 3x - x2 + x + 2 = 0
<=> - 2x + 2 = 0
<=> -2x = -2
<=> x = 1
Vậy S = { 1 }
f) x( 3x - 1 ) - 3x2 - 7x = 0
<=> 3x2 - x - 3x2 - 7x = 0
<=> -8x = 0
<=> x = 0
Vậy S = { 0 }
d) 2x(x - 1) - 2x2 + x - 5 = 0
=> 2x2 - 2x - 2x2 + x - 5 = 0
=> -x = 5
=> x = -5
e) x(x - 3) - (x + 1)(x - 2) = 0
=> x2 - 3x - (x2 - x - 2) = 0
=> x2 - 3x - x2 + x + 2 = 0
=> -2x = - 2
=> x = 1
f) x(3x - 1) - 3x2 - 7x = 0
=> 3x2 - x - 3x2 - 7x = 0
=> -8x = 0
=> x = 0
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: 1) (x+1)(3-x) / (4-2x) lớn hơn hoặc =0
2) (x+3)(3x-2) / (2x+1)(4-3x) >0
1,X=-1 hoặc 3
2,Tìm x sao cho (x+3) và (3x-2) ko bằng 0
Cần lắm nhờ mấy bạn
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
Giúp mình với
a)\(x^2+x-x^2+2=0\)\(\Rightarrow x+2=0\)\(\Rightarrow x=-2\)
b)\(2\left(3x+2\right)-2\left(x+6\right)=0\)
\(\Rightarrow2\left(3x+2-x-6\right)=0\)
\(\Rightarrow2\left(2x-4\right)=0\)
\(\Rightarrow2x-4=0\Rightarrow x=2\)
c)\(4x^4-6x^3-4x^4+6x^3-2x^2=0\)
\(\Rightarrow-2x^2=0\Rightarrow x=0\)
d)\(\left(3x^2-x-2\right)-3\left(x^2-x-2\right)=4\)
\(\Rightarrow3x^2-x-2-3x^2+3x+6=4\)
\(\Rightarrow2x+4=4\Rightarrow2x=0\Rightarrow x=0\)
Tìm x, biết:
a) 3x(x - 1) + x - 1 = 0;
b) (x - 2)( x 2 + 2x + 7) + 2( x 2 - 4) - 5(x - 2) = 0;
c) ( 2 x - 1 ) 2 - 25 = 0;
d) x 3 + 27 + (x + 3)(x - 9) = 0.
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
BT2: Tìm x 2, 3x(x-4)+2x-8=0 3, 4x(x-3)+x^2-9=0 4, x(x-1)-x^2+3x=0 5, x(2x-1)-2x^2+5x=16
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
Bài 2: Tìm x, biết: a) (x + 2)^2 – 2(x + 2)(x – 5) = 0. b) 2x^2 + 3x – 5 = 0. c) x + 2 ^2 x 2 + 2x^3 = 0. d) (3x-1)^2-4(x+5)^2=0
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)