việt 5 giá trị thích hợp của x sao cho:
a) 1<x<2
b)10,1<x<10,2
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a: |2x-3|=|1-x|
=>\(\left[{}\begin{matrix}2x-3=1-x\\2x-3=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=3+1\\2x-x=-1+3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=4\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
b: \(x^2-4x< =5\)
=>\(x^2-4x-5< =0\)
=>\(x^2-5x+x-5< =0\)
=>\(x\left(x-5\right)+\left(x-5\right)< =0\)
=>\(\left(x-5\right)\left(x+1\right)< =0\)
TH1: \(\left\{{}\begin{matrix}x-5>=0\\x+1< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\x< =-1\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-5< =0\\x+1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =5\\x>=-1\end{matrix}\right.\)
=>-1<=x<=5
c: 2x(2x-1)<=2x-1
=>\(\left(2x-1\right)\cdot2x-\left(2x-1\right)< =0\)
=>\(\left(2x-1\right)^2< =0\)
mà \(\left(2x-1\right)^2>=0\forall x\)
nên \(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>\(x=\dfrac{1}{2}\)
1)
( y - 1/5 ) : 2/3 = 3/4
y - 1/5 = 3/4 x 2/3
y - 1/5 = 1/2
y = 1/2 + 1/5
y = 7/10.
2)
7/4 - y x 5/6 = 1/2 + 1/3
7/4 - y x 5/6 = 5/6
7/4 - y = 5/6 : 5/6
7/4 - y = 1
y = 7/4 - 1
y = 3/4
C1 :(y-1/5):2/3=3/4
y-1/5=3/4.2/3
y-15=1/2
y=1/2+15
y=31/2
C2:
7/4-y.5/6=1/2+1/3
7/4-y.5/6=5/6
y.5/6=7/4-5/6
y.5/6=11/12
y=11/12:5/6
y=11/10
x=4,1
x=4,2
x=4,3
x=4,4
x=4,5
x=4,6
x=4,7
x=4,8
x=4,9
chuc ban hoc gioi
a, Gọi I là trọng tâm của ΔABC
⇒ \(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\)
MA2 + MB2 + MC2 = k2
⇔ 3MI2 + 2\(\overrightarrow{MI}\left(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\right)+AB^2+AC^2+BC^2\) = k2
⇔ 3MI2 = k2 - 1014
⇔ MI = \(\sqrt{\dfrac{k-1014}{3}}\) = const
Vậy M thuộc \(\left(I;\sqrt{\dfrac{k-1014}{3}}\right)\)
\(a,\Leftrightarrow-4+k=-3\Leftrightarrow k=1\\ b,\Leftrightarrow-3\left(2k-18\right)=40\\ \Leftrightarrow2k-18=-\dfrac{40}{3}\Leftrightarrow k=\dfrac{7}{3}\\ c,\Leftrightarrow10+18=9\left(2+k\right)\\ \Leftrightarrow k+2=\dfrac{28}{9}\Leftrightarrow k=\dfrac{10}{9}\)
\(x^{2022}\ge0\Leftrightarrow A=2021-x^{2022}\le2021\\ A_{max}=2021\Leftrightarrow x=0\)
a) 1, 1 ; 1,2 ; 1,3 ; 1,4 ; 1,5
b) 10,11 ; 10,12 ; 10,13 ; 10,14 ; 10,15