tìm x E N [2x-1]2023=[2x-1]2024
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\(C=16x^2-8x+2024\)
\(\Rightarrow C=16x^2-8x+1+2023\)
\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)
\(\Rightarrow Min\left(C\right)=2023\)
\(D=-25x^2+50x-2023\)
\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)
\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=1998\)
\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)
\(\Rightarrow Max\left(B\right)=200\)
\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)
\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)
\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)
\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)
\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)
\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)
\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)
\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)
\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)
\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(F\right)=48\)
a) \(\left(x-2024\right)^{2023}=1\)
\(\Rightarrow\left(x-2024\right)^{2023}=1^{2023}\)
\(\Rightarrow x-2024=1\)
\(\Rightarrow x=2025\)
b) \(\left(2x-1\right)^5=32\)
\(\Rightarrow\left(2x-1\right)^5=2^5\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
c) \(5< 2^x< 100\)
\(\Rightarrow4=2^2< 5< 2^x< 100< 128=2^7\)
\(\Rightarrow2< x< 7\)
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0
P = (x^2 + 2x) - 2024
= (x^2 + 2x + 1) - 1 - 2024
= (x + 1)^2 - 2025
Với mọi giá trị của x, (x + 1)^2 luôn lớn hơn hoặc bằng 0. Do đó, giá trị nhỏ nhất của P là khi (x + 1)^2 đạt giá trị nhỏ nhất, tức là bằng 0.
Khi (x + 1)^2 = 0, ta có x + 1 = 0, từ đó suy ra x = -1.
Vậy, giá trị nhỏ nhất của biểu thức P là P = (-1 + 1)^2 - 2025 = -2025.
Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>\(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)
\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)
\(=2^{2023}-1\)
\(\left(2x+4\right)^{2024}+\left(\left|3y-9\right|\right)^{2023}=0\) (*)
Ta có: \(\left(2x+4\right)^{2024}\ge0\forall x\) (vì có số mũ chẵn) (1)
\(\left(\left|3y-9\right|\right)^{2023}\ge0\forall y\) (vì giá trị tuyệt đối luôn ≥0) (2)
Từ (1) và (2) ta có:
\(\Rightarrow\left\{{}\begin{matrix}2x+4=0\\3y-9=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)
Vậy: ...
a, 7\(x\).(2\(x\) + 10) = 0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\){-5; 0}
b, - 9\(x\) : (2\(x\) - 10) = 0
- 9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
d, (\(x\) + 2023).(\(x\) - 2024) = 0
\(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-2023; 2024}
\(\left(2x-1\right)^{2023}=\left(2x-1\right)^{2024}\) \((x\in\mathbb{N})\)
\(\Rightarrow\left(2x-1\right)^{2023}-\left(2x-1\right)^{2024}=0\)
\(\Rightarrow\left(2x-1\right)^{2023}\left[1-\left(2x-1\right)\right]=0\)
\(\Rightarrow\left(2x-1\right)^{2023}\left(1-2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)^{2023}\left(2-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\2-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy \(x=1\) là giá trị cần tìm.
\(\left(2x-1\right)^{2023}=\left(2x-1\right)^{2024}\)
=>\(\left(2x-1\right)^{2024}-\left(2x-1\right)^{2023}=0\)
=>\(\left(2x-1\right)^{2023}\left[\left(2x-1\right)-1\right]=0\)
=>\(\left(2x-1\right)^{2023}\cdot\left(2x-2\right)=0\)
=>\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)