Tìm x
a, x. [x-1]=0
b, [x+1] . [x-2]=0
c, x. [ x-2]. [x-3]=0
Dấu chấm là nhân nha các bạn
Giúp mình vs. Thank
Những câu hỏi liên quan
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: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
Đúng 0
Bình luận (0)
1.tìm x,y biết
a, x.(y-3)≥0
b, (2.x-1).(y-1)≤0
c,(x-1).(2.k+1)≥0
2. tìm x,y ϵ Z biết
a, x(x+3)=0
b,(x-2).(5-x)=0
c,(x-1).(x^2+1)=0
d, x.y+3.x-7.y=21
e,x.y+3.x-2y=11
GIẢI GIÚP MÌNH VỚI, MÌNH ĐANG CẦN GẤP LẮM Ạ!!!!!
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
Đúng 0
Bình luận (0)
Tìm x
a) 4x(x-2) + x-2 = 0
b) (3x-1)^2 - 9 = 0
c) x^3 - 8 + (x-2)(x+1) = 0
`a)4x(x-2)+x-2=0`
`<=>(x-2)(4x+1)=0`
`<=>[(x-2=0),(4x+1=0):}`
`<=>[(x=2),(x=-1/4):}`
Vậy `S={2;-1/4}.`
`b)(3x-1)^3-9=0`
`<=>(3x-1-3)(3x-1+3)=0`
`<=>(3x-4)(3x+2)=0`
`<=>[(3x-4=0),(3x+2=0):}`
`<=>[(x=4/3),(x=-2/3):}`
Vậy `S={4/3;-2/3}.`
`c)x^3-8+(x-2)(x+1)=0`
`<=>(x-2)(x^2+2x+4)+(x-2)(x+1)=0`
`<=>(x-2)(x^2+3x+5)=0`
Mà `x^2+3x+5=(x+3/2)^2+11/4>=11/4>0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
Đúng 2
Bình luận (0)
a) Ta có: \(4x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b)Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Đúng 0
Bình luận (0)
Tìm x
a) 4x(x-2) + x - 2 = 0
b) (3x-1)^2 - 9 = 0
c) x^3 - 8 + (x-2)(x+1) = 0
a, \(4x\left(x-2\right)+x-2=0\Leftrightarrow\left(4x+1\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{1}{4};x=2\)
b, \(\left(3x-1\right)^2-9=0\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\Leftrightarrow x=\dfrac{4}{3};x=-\dfrac{2}{3}\)
c, \(x^3-8+\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\ne0\right)=0\Leftrightarrow x=2\)
Đúng 0
Bình luận (0)
a) Ta có: \(4x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b) Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
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
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Tìm x
a,(7x+4)^2-(7x+4)(7x-4)=0
b, 5( x + 3 )( x - 3 ) + ( 2x + 3 )^2+(x-6)^=10
c, (x + 1)^3 + (x – 2)^3 – 2x^2 (x – 1,5) = 3
d,( x + 2)(x^2 – 2x + 4)(x – 2)(x^2 + 2x + 4) = – 65
e, 4x^2 + 4x – 5 = 2
f,16x^2 – 9(x + 1)^2 = 0
Các bạn giúp mình vs mai mình phải nộp rùii
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
TÌM X
a. 3.(x^2-x+2)-x.(2+3x)=0
b. (x-1)^2 + (x-1)(x+2)=0
c. 2x^3 +3x^2+2x+3=0
d. 2x^2+x=6
\(a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\\ \Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+x-6=0\\ \Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
Đúng 4
Bình luận (0)
Giải các phương trình sau:
a) 3x - 12 = 0
b) (x-2)(2 x+3)=0
c) x+2/x-2- 6/x+2= x^2/x^2-4
Các bạn ơi giúp mình!!!!!
a \(\Leftrightarrow3x=12\Leftrightarrow x=4\)
b \(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c \(ĐKXĐ:x\ne2;x\ne-2\)
\(\Rightarrow\left(x+2\right)^2-6\left(x-2\right)=x^2\Leftrightarrow x^2+4x+4-6x+12=x^2\Leftrightarrow-2x+16=0\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(TM\right)\)
Đúng 2
Bình luận (1)
a) Ta có: 3x-12=0
\(\Leftrightarrow3x=12\)
hay x=4
Vậy: S={4}
b) Ta có: (x-2)(2x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{-3}{2}\right\}\)
Đúng 2
Bình luận (1)
tìm x
a)x^2-x-12=0
b)x^2+3x-18=0
c)8x^2+30x+7=0
d)x^3-11x^2+30x=0
e)x^3-7x^2+15x-25=0
giúp mk vs ah!!!!!
a) Ta có: \(x^2-x-12=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Đúng 0
Bình luận (0)
b) Ta có: \(x^2+3x-18=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời