(x-1)2=1
Giải giúp mik vs, please
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`[2-x]/x >= 1`
`<=>[2-x-x]/x >= 0`
`<=>[2-2x]/x >= 0`
`<=>0 < x <= 1`
`->\bb B`
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
Hệ số góc bằng 2 `=> 3a+4 = 2 <=> a=-2/3`
Tung độ gốc bằng 1 `=> -b+3=1 <=> b=2`
`=> y=-2/3 x +2`
\(\left|x^2-1\right|=2x+1\left(dk:2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\le-\dfrac{1}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+1\\x^2-1=-2x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1-2x-1=0\\x^2-1+2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-2=0\\x^2+2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-2+3=3\\x.\left(x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=3\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1^2\right)-\left(\sqrt{3}\right)^2=0\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1-\sqrt{3}\right).\left(x-1+\sqrt{3}\right)=0\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1-\sqrt{3}\left(loai\right)\\x=1+\sqrt{3\left(loai\right)}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\left(loai\right)\\x=-2\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy x = -2
uk 276 nha mik tính là vậy còn ko biết đúng ko nữa cho mik 1 k nha hihi / HT/
a)(x-3)(x+3)-(x+5)2+(x+1)(x+2)
=x2-9-x-10x-25+x2+2x+x+2
=2x2-8x-32
b)2 . 25 - 8 . 5 - 32=78
ta có: M=10^2020 +1 / 10^2019 +1
=> M/10= 10^2020 +1 / 10( 10^2019 +1 )
= 10^2020+1/ 10^2020 +10
=> 10/A= 10^2020 +10/10^2020 +1
=(10^2020 +1) +9/ 10^2020+1
=10^2020+1 /10^2020+1 + 9/10^2020+1
=1+ 9/10^2020+1
ta lại có: N=10^2021 +1/10^2020 +1
=> N/10= 10^2021+1/ 10(10^2020+1)
= 10^2021+1 / 10^2021+10
=> 10/N=10^2021+10 / 10^2021+1
=(10^2021+1) +9/10^2021+1
=10^2021+1/10^2021+1 +9/10^2021+1
=1+ 9/10^2021+1
ta thấy: 10/M>10N
=>M<N
\(M=\dfrac{10^{2020}+1}{10^{2019}+1}=1-\dfrac{9}{10^{2019}+1}\)
\(N=\dfrac{10^{2021}+1}{10^{2020}+1}=1-\dfrac{9}{10^{2020}+1}\)
Ta có: \(10^{2019}+1< 10^{2020}+1\)
\(\Leftrightarrow\dfrac{9}{10^{2019}+1}>\dfrac{9}{10^{2020}+1}\)
\(\Leftrightarrow-\dfrac{9}{10^{2019}+1}< -\dfrac{9}{10^{2020}+1}\)
\(\Leftrightarrow M< N\)
a,=(x\(^2\)-6x+9)+10-9
=(x-3)\(^2\)+1
Mà(x-3)\(^2\)\(\ge\)0
nên (x-3)\(^2\)+1>0
b,= -(-4x+x\(^2\))-5
= -(4-4x+x\(^2\))-5+4
= -(2-x)\(^2\)-1
Mà -(2-x)\(^2\)\(\le\)0
nên -(2-x)\(^2\)-1< 0
Võ Hoàng Tiên: Cảm ơn pạn nhiều lắm =)) nek :3 Hí Hí :) Thankssssss
x = 2 nha bn
Cảm ơn bn. mik lm đc r nka