tìm x thuộc z, biết:
a. ( x- 3) .( x+4 ) =0
b. ( x -2 ). (5-x ) =0
c. x. (x+1 ) =0
d.|2x - 5| =13
e.-12 =| x - 9 | = 3
f. 11-(15 - | x | ) = 1
làm ơn ,hãy giúp tôi!!!!!
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: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
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
Bài 2:
a: =>x=0 hoặc x+3=0
=>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
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
Bài 2: Tìm x, biết :
a) 3x^2-15x=0
b) 2x(x-3)+3-x=0
c) x^3-3x^2-6x-6+8=0
d) (x-2)^2-16(5-2x)^2
cảm ơn trước ạ
\(a,\Rightarrow3x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,Đề.sai\\ d,Sửa:\left(x-2\right)^2-16\left(5-2x\right)^2=0\\ \Rightarrow\left[x-2-4\left(5-2x\right)\right]\left[x-2+4\left(5-2x\right)\right]=0\\ \Rightarrow\left(x-2-20+8x\right)\left(x-2+20-8x\right)=0\\ \Rightarrow\left(9x-22\right)\left(18-7x\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{9}\\x=\dfrac{18}{7}\end{matrix}\right.\)
Tìm x∈Z, biết:
a)x.(x-6)=0
b)(-7-x).(-x+5)=0
c)(x+3).(x-7)=0
d)(x-3).(x2+12)=0
e)(x+1).(2-x) ≥0
f)(x-3).(x-5) ≤0
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
1.
a) (x+3)(2x-8)≥0
b) (x-2)(5-x)>0
c) (x+1)(x+3)(x-4)≥0
d) (2x-4)(x+5)(1-x)<0
Giúp với mn ơi ;-;a) 9(2x+1)^2-4(x+1)^2=0b) (x+1)^2+2(x+1)+1=0c) (x-1)(x^2-9)+x+3=0d) (7-x)^2__________ -(x+5)^2=0 4e) 4x^2+(x-1)^2-(2x+1)^2=0f) x^3+1=(x+1)(2-x)
Rối quá bn ơi
Bn ko dùng dấu enter để xuống dòng à
Mk sẽ giúp bn 3 câu đầu thôi còn 3 câu sau thì..... chịu vì ko biết làm sao
a) \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)
(=)\(\left[9\left(2x+1\right)-4\left(x+1\right)\right]\left[9\left(2x+1\right)+4\left(x+1\right)\right]=0\)
(=)\(\left(18x+9-4x-4\right)\left(18x+9+4x+4\right)=0\)
(=)\(\left(14x+5\right)\left(22x+12\right)=0\)
(=)\(\left(14x+5\right)=0\) hoặc \(\left(22x+12\right)=0\)
1) \(14x+5=0\Rightarrow x=-\dfrac{5}{14}\)
2) \(22x+12=0\Rightarrow x=-\dfrac{6}{11}\)
Vậy........
b)\(\left(x+1\right)^2+2\left(x+1\right)+1=0\)
(=)\(\left(x+1\right)^2+2\left(x+1\right).1+1^2=0\)
(=) \(\left(x+2\right)^2=0\)
(=) \(x+2=0\Rightarrow x=-2\)
Vậy........
c) \(\left(x-1\right)\left(x^2-9\right)+x+3=0\)
(=) \(\left(x-1\right)\left(x-3\right)\left(x+3\right)+x+3=0\)
(=) \(\left(x+3\right)\left(x-3+1\right)\left(x-1\right)=0\)
(=) \(\left(x+3\right)\left(x-2\right)\left(x-1\right)=0\)
(=) \(x+3=0\) hoặc \(x-2=0\) hoặc \(x-1=0\)
1)\(x+3=0\Rightarrow x=-3\)
2)\(x-2=0\Rightarrow x=2\)
3) \(x-1=0\Rightarrow x=1\)
Vậy.........
Tìm x:
a)(x+2)^2-2(x+2)(x-5)=0
b)2x^2+3x-5=0
c)x+2√2x^2+2x^3=0
d)(3x-1)^2-4(x+5)^2=0
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
bài 1 tìm x
a)6x^2-72x=0
b)-2x^4+16x=0
c)x(x-5)-(x-3)^2=0
d)(x-2)^3-(x-2)(x^2+2x+4)=0
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)