Tìm điều kiện của x để:
a, 3 ( x-1 ) > hoặc = 0
b,-52.( 6 - x )< 0
c,x2 (8 - 2x) > hoặc = 0
d,( x - 5 ) ( 7 - x ) < 0
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13 tháng 7 2021
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
13 tháng 7 2021
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
19 tháng 4 2023
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
=>\(x\in\left\{0;1;-1\right\}\)
8 tháng 9 2021
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
MH
8 tháng 9 2021
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
7 tháng 10 2021
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: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
TG
14 tháng 1 2021
a) (x - 7)(2x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy: S = {7; -4}
b) Tương tự câu a
c) (x - 1)(2x + 7)(x2 + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)
Mà: x2 + 2 > 0 với mọi x
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)
d) (2x - 1)(x + 8)(x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)
TT
14 tháng 1 2021
a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{7;-4\right\}\)
b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
DL
2
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
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) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
10 tháng 8 2023
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
19 tháng 10 2021
\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
