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\(a,x\in\left\{-5;-4;-3;-2;-1\right\}\\ b,x\in\left\{-3;-2;-1;...;5;6\right\}\\ c,x\in\left\{-4;-3;...;3;4\right\}\\ d,x\in\left\{-3;-2;-1;0;1;2\right\}\)
Tìm x εIN biết
a) 390 - (x-8) = 168:13
b) (x-140) : 7 = 27 - 24
c) x- 6 :2 - ( 48 - 24 ) :2 :6 - 3 = 0
d) x+5.2-(32+16.3:6-15)=0
b) \(\left(x-140\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(x-140=21\)
\(x=161\)
vay \(x=161\)
c) \(x-6:2-\left(48-24\right):2:6-3=0\)
\(x-3-24:2:6-3=0\)
\(x-3-2-3=0\)
\(x-8=0\)
\(x=8\)
vay \(x=8\)
d) \(x+5.2-\left(32+16.3:6-15\right)=0\)
\(x+10-\left(32+8-15\right)=0\)
\(x+10-25=0\)
\(x-15=0\)
\(x=15\)
vay \(x=15\)
a) \(390-\left(x-8\right)=168:13\)
\(390-x+8=\frac{168}{13}\)
\(x+8=390-\frac{168}{13}\)
\(x+8=\frac{5070}{13}-\frac{168}{13}\)
\(x+8=\frac{4902}{13}\)
\(x=\frac{4902}{13}-8\)
\(x=\frac{4798}{13}\)
vay \(x=\frac{4798}{13}\)
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
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
\(a)2\left(4x-8\right)-7\left(3+x\right)=|-4|\left(3-2\right)\)
\(\Leftrightarrow8x-16-21-7x=4\)
\(\Leftrightarrow\left(8x-7x\right)=16+21+4\)
\(\Leftrightarrow x=41\)
\(b)3|x-3|=15-|-9|+7|-3|-\left(-21\right)\)
\(\Leftrightarrow3|x-3|=15-9+7.3+21\)
\(\Leftrightarrow3|x-3|=15-9+21+21\)
\(\Leftrightarrow3|x-3|=48\)
\(\Leftrightarrow|x-3|=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=-16\\x-3=16\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-13\\x=19\end{cases}}\)
\(c)4\left(x-5\right)-7\left(5-x\right)+10\left(5-x\right)=-3\)
\(\Leftrightarrow4x-20-35+7x+50-10x=-3\)
\(\Leftrightarrow4x+7x-10x=20+35-50-3\)
\(\Leftrightarrow x=2\)
\(d)6|x-3|-9|x-3|=-21\)
\(\Leftrightarrow|x-3|\left(6-9\right)=-21\)
\(\Leftrightarrow-3|x-3|=-21\)
\(\Leftrightarrow|x-3|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=-7\\x-3=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=10\end{cases}}\)
ủa cho em hỏi ý a,b,c đâu ạ . chứ chị giải kiểu vậy em hong có hỉu , mong chị trả lời em ạ
a) \(x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy x=0 hoặc x=-2
b) \(x^2\left(x-5\right)+2\left(x-5\right)=0\)
\(\Rightarrow x-5=0\Rightarrow x=5\)
Vậy x=5
a) x(x-2)=0
=> x=0 hoặc x-2=0
=> x=0 hoặc x=0+2
=> x=0 hoặc x = 2
a, \(x\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}}\)
b, \(\left(2x-6\right)\left(-4x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\-4x-8=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
a, => x = 0 hoặc x + 5 = 0
=> x = 0 hoặc x = -5
b, => 2x - 6 = 0 hoặc -4x - 8 = 0
=> x = 3 hoặc x = -2
Tk mk nha