Mọi người giúp mình với ạ. Cảm ơn rất nhiều ạ.
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12 tháng 3 2022
Bài 1 :
Thay x = 2 ; y = -1/2 ta được
\(B=-8+2.4\left(-\dfrac{1}{2}\right)-4.2.\left(\dfrac{1}{4}\right)+2\left(-\dfrac{1}{2}\right)-3\)
\(=-8-4-2-1-3=-18\)
6 tháng 2 2021
a) Ta có: \(\left(2x-3\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};-\dfrac{4}{3}\right\}\)
b) Ta có: \(x^3-3x^2+3x-1=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-2x+1-x-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-3x\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
Vậy: S={0;1;3}
c) Ta có: \(x^2+x=2x+2\)
\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: S={-1;2}
d) Ta có: \(\left(x-1\right)^2=2\left(x^2-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1-2x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)Vậy: S={1;-3}
e) Ta có: \(2\left(x+2\right)^2-x^3-8=0\)
\(\Leftrightarrow2\left(x+2\right)^2-\left(x^3+8\right)=0\)
\(\Leftrightarrow2\left(x+2\right)\cdot\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x+4-x^2+2x-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\cdot\left(-x^2+4x\right)=0\)
\(\Leftrightarrow-x\left(x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\end{matrix}\right.\)
Vậy: S={0;-2;4}
31 tháng 10 2021
a: \(P=-\left|5-x\right|+2019\le2019\forall x\)
Dấu '=' xảy ra khi x=5
11 tháng 7 2023
a: P(x)=2x^3-2x^3+x^2+3x^2-4x^2-3x+5x+1=-3x+6
b: P(0)=-3*0+6=6
P(-1)=6+3=9
P(1/3)=-1+6=5
c: P(x)=0
=>-3x+6=0
=>-3x=-6
=>x=2
P(x)=1
=>-3x+6=1
=>-3x=-5
=>x=5/3
TN
Mọi người ơi giúp mình với,mình cảm ơn rất nhiều ạ
2
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22 tháng 1 2022
used to deliver
used to be
used to go
used to drive
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used to believe
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LP
22 tháng 1 2022
1. used to deliver
2. used to be
3. used to go
4. used to drive
5. used to spend
6. used to believe
7.used to work
8. used to serve
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