tìm x biết:
a, 49x^2 -4=0
b, (x+3)^2 - (x+2)(x-2)=11
c,(2x +1)^2 - (x-3)^2 - 3(x+5)(x-5) =5
d, (3x +1)(3x-1)=8
Giai giùm mình với
Tìm x, biết:
a)9x2–6x–3 = 0
b) (2x + 1)2–4(x + 2)2= 9
c)3(x –1)2–3x(x –5) = 21
d) (x + 3)2–(x –4)(x + 8) = 1
giúp mk với ạ
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
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) 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}
a: =>(x-5)(x+5)+(x-5)(3x-15)=0
=>(x-5)(x+5+3x-15)=0
=>(x-5)(4x-10)=0
=>x=5 hoặc x=5/2
c: =>x^3-3x^2+2x^2-6x-8x+24=0
=>(x-3)(x^2+2x-8)=0
=>(x-3)(x+4)(x-2)=0
=>\(x\in\left\{3;-4;2\right\}\)
Tìm x:
a)2x3-18x=0
b)(3x-2).(2x+1)-6x.(x+2)=11
c)(x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
c. (x - 1)3 - (x + 2)(x2 - 2x + 4) = 3(1 - x2)
<=> (x3 - 3x2 + 3x - 1) - (x3 - 2x2 + 4x + 2x2 - 4x + 8) = 3 - 3x2
<=> x3 - 3x2 + 3x - 1 - x3 + 2x2 - 4x - 2x2 + 4x - 8 = 3 - 3x2
<=> x3 - x3 - 3x2 + 2x2 - 2x2 + 3x2 + 3x - 4x + 4x = 3 + 1 + 8
<=> 3x = 12
<=> x = 4
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
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x biết:
a) (3x + 5)2 - 4x2 = 0
b) 25x4 - (4x - 3)2 = 0
c) (3x + 7)2 - (2x - 3)2 = 0
d) (4x - 1)2 - (5 - 3x)2 = 0
a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
. Tìm x, biết:
a) 6x.(x – 5) + 3x.(7 – 2x) = 18 b) 2x.(3x + 1) + (4 – 2x).3x = 7 c) 0,5x.(0,4 – 4x) + (2x + 5).x = -6,5 | d) (x + 3)(x + 2) – (x - 2)(x + 5) = 6 e) 3(2x - 1)(3x - 1) – (2x - 3)(9x - 1) = 0 |
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}