tìm x
a) (5-x).(x-1)<0
b) (x-4).\(\left(x+\dfrac{1}{2}\right)>hoặc=0\)
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\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(a,Sửa:2021x-1+2022x\left(1-2021x\right)=0\\ \Leftrightarrow\left(2021x-1\right)\left(1-2022x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2021}\\x=\dfrac{1}{2022}\end{matrix}\right.\)
a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)
\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-5x-15=0\)
\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
c) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
a) \(\Leftrightarrow\left[{}\begin{matrix}x-3,5=7,5\\x-3,5=-7,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-4\end{matrix}\right.\)
b) \(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{2}\\x+\dfrac{4}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{10}\\x=-\dfrac{13}{10}\end{matrix}\right.\)
c) \(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
d) \(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\4,5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=4,5\end{matrix}\right.\)(vô lý)
Vậy \(S=\varnothing\)
Nói chung cả 3 câu :
Vì GTTĐ luôn lớn hơn hoặc bằng 0
=> tất cả các số hạng đều bằng 0
sau đó tính ra là xong
a) | x - 1| + | y - 3| = 0
=> |x -1| = 0 => x = 1
|y-3| = 0 => y = 3
KL:...
b) | x - 1 | + |x-3| + |x-5| = 0
Ta thấy: \(\left|x-1\right|;\left|x-3\right|;\left|x-5\right|\ge0.\)
=> | x - 1 | = 0 => x = 1 mà | 1-3| không bằng 0 (Loại)
...
ko tìm được x
c) \(\left|x-2018y\right|+\left|x-2018\right|\le0\)
mà \(\left|x-2018y\right|;\left|x-2018\right|\ge0\)
=> | x - 2018y| + |x-2018| = 0
=> | x - 2018| = 0 => x = 2018
=> |x-2018y| = 0 => |2018-2018y| = 0 => y = 1
KL:...
Bài 2:
a, |x-1| -x +1=0
|x-1| = 0-1+x
|x-1| = -1 + x
\(\orbr{\begin{cases}x-1=-1+x\\x-1=1-x\end{cases}}\)
\(\orbr{\begin{cases}x=-1+x+1\\x=1-x+1\end{cases}}\)
\(\orbr{\begin{cases}x=x\\x=2-x\end{cases}}\)
x = 2-x
2x = 2
x = 2:2
x=1
b, |2-x| -2 = x
|2-x| = x+2
\(\orbr{\begin{cases}2-x=x+2\\2-x=2-x\end{cases}}\)
2-x = x+2
x+x = 2-2
2x = 0
x = 0
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a. x= 1;2
b. x= 1;2;3;4;5;6
c. x= 6;7;8;9;...
d. x= 6;7;8;9;...
e. x= 1;2;3
a) x thuộc{1;-1;2;-2}
b)x thuộc {1;-1;2;-2;3;-3;4;-4;5;-5;6;-6}
c) x thuộc {6;-6;7;-7;...}
d) x thuộc {6:-6:7:-7;...}
f) x thuộc { 2;3;4;5;...}
e) x thuộc {0;1;2;3}
g) x thuộc {0;1}
`a,(5-x)(x-1) < 0`
`<=>5-x<0` hoặc `x-1<0`
`<=>5 <x` hoặc `x<1`
Vậy `S={x|5<x;x<1}`
`b,(x-4)(x+1/2) >= 0`
`<=>TH1 : {(x-4>=0),(x+1/2 >=0):}<=>{(x>=4(TM)),(x>= -1/2(L)):}`
`<=>TH2 :{(x-4<=0),(x+1/2 <= 0):} <=>{(x<=4(L)),(x<=-1/2(TM)):}`
`=>x<= -1/2` hoặc `x>=4`
Vậy `S={x|x<= -1/2 ; x>=4}`