Tìm x :
a/ ( x + 6 )4 - ( x + 6 )6 =0
b/ | x + 3 | . ( x - 7 ) < 0
K
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Những câu hỏi liên quan
30 tháng 10 2021
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
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
M
5
21 tháng 10 2021
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
21 tháng 10 2021
\(2\left(x+3\right)+x\left(3+x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
BT
14 tháng 10 2023
a) (3x-15)7 = 0
3x-15 = 0
3x = 0+15
3x = 15
x = 15:3
x = 5
b) 42x-6 = 1
2x-6 = 0
2x = 0+6
2x = 6
x = 6:2
x = 3
c) Tớ ko bít
d) (x - 6)3 = (x - 6)2
Th1:
x - 6 = 1
x = 1 + 6
x = 7
Th2:
x - 6 = 0
x = 6
Vậy x = 7
x = 6
--thodagbun--
BN
14 tháng 10 2023
a, (3x-15)^7=0 <=> 3x-15=0 <=> x=5
b, 42x+6=1 <=> 16x=-5 <=>x=-5/16
c, \(\dfrac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=1\)
TH1: 10x-20 = 0 <=> x=2
TH2: 3-x=1 <=> x=2
Vậy x=2
d, (x-6)^3 = (x-6)^2
<=> (x-6)^2.[(x-6)-1]=0
<=> (x-6)^2=0 hoặc (x-6)-1=0
<=> x=6 hoặc x=7
L
19 tháng 5 2022
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
TY
2
TD
14 tháng 2 2022
a)\(\dfrac{4}{x}=\dfrac{x}{16}\)
<=>\(x^2=4.16=64\)
<=>\(x=\pm8\)
<=>x=-8(vì x<0)
b)\(\dfrac{x}{-24}=\dfrac{-6}{x}\)
<=>\(x^2=\left(-24\right)\left(-6\right)=144\)
<=>\(x=\pm12\)
<=>x=12(Vì x>0)
HD
3
11 tháng 12 2021
\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)
\(b,\left(10-x\right)\left(3-x\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)
11 tháng 12 2021
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
6T
1
8 tháng 12 2021
\(a,x\in\left\{-5;-4;-3;-2;-1\right\}\\ b,x\in\left\{-3;-2;-1;...;5;6\right\}\\ c,x\in\left\{-4;-3;...;3;4\right\}\\ d,x\in\left\{-3;-2;-1;0;1;2\right\}\)
KV
1 tháng 1
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
a,(x+6)^4-(x+6)^6=0
<=>(x+6)^4-(x+6)^4.(x+6)^2=0
<=>(x+6)^4.[1-(x+6)^2]=0
TH1:(x+6)^4=0=>x+6=0=>x=-6
TH2:1-(x+6)^2=0=>(x+6)^2=1=>x+6=1 hoặc x+6=-1
=>x=-5 hoặc x=-7
Vậy x E {-7;-6;-5}
Tick nhé