(2x-5)^(2025)=(2x-5)^(2023)
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11 tháng 4 2023
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.
AD
1
22 tháng 8 2023
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) \(5< 2^x< 100\)
\(\Rightarrow4=2^2< 5< 2^x< 100< 128=2^7\)
\(\Rightarrow2< x< 7\)
8 tháng 10 2023
\(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>x=1 và y=-1
\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)
TH
Thầy Hùng Olm
Manager
VIP
23 tháng 6 2023
2 x A = 1 - \(\dfrac{1}{2027}\)
\(A=\dfrac{1013}{2027}\)
LN
0
T
3 tháng 1 2024
a) Ta có: \(\left|x+5\right|\ge0\forall x\)
\(\Rightarrow\left|x+5\right|+2023\ge2023\forall x\)
\(\Rightarrow A\ge2023\forall x\)
Dấu \("="\) xảy ra khi: \(x+5=0\Leftrightarrow x=-5\)
Vậy \(Min_A=2023\) khi \(x=-5\).
b) Ta có: \(\left\{{}\begin{matrix}\left|2x+6\right|\ge0\forall x\\\left|y+3x\right|\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|\ge0\forall x,y\)
\(\Rightarrow\left|2x+6\right|+\left|y+3x\right|+25\ge25\forall x,y\)
\(\Rightarrow B\ge25\forall x,y\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}2x+6=0\\y+3x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-6\\y=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-6:2=-3\\y=-3\cdot\left(-3\right)=9\end{matrix}\right.\)
Vậy \(Min_B=25\) khi \(x=-3;y=9\).
c) Ta có: \(\left\{{}\begin{matrix}\left|12-3x\right|\ge0\forall x\\\left|-y-4x\right|\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|\ge0\forall x,y\)
\(\Rightarrow\left|12-3x\right|+\left|-y-4x\right|-12\ge-12\forall x,y\)
\(\Rightarrow C\ge-12\forall x,y\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}12-3x=0\\-y-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=12\\y=-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=12:3=4\\y=-4\cdot4=-16\end{matrix}\right.\)
Vậy \(Min_C=-12\) khi \(x=4;y=-16\).
\(\mathit{Toru}\)
16 tháng 7 2023
\(C=16x^2-8x+2024\)
\(\Rightarrow C=16x^2-8x+1+2023\)
\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)
\(\Rightarrow Min\left(C\right)=2023\)
\(D=-25x^2+50x-2023\)
\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)
\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=1998\)
\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)
\(\Rightarrow Max\left(B\right)=200\)
\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)
\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)
\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)
\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)
\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)
\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)
\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)
\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)
\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)
\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(F\right)=48\)
❓❤❓♂
(2\(x\) - 5)\(^{2025}\) = (2\(x-5\))\(^{2023}\)
(2\(x\) - 5)\(^{2025}\) - (2\(x-5\)) = 0
(2\(x-5\))\(^{2023}\) .[(2\(x-5\))\(^2\) - 1] = 0
\(\left[\begin{array}{l}2x-5=0\\ \left(2x-5\right)^2=1\end{array}\right.\)
\(\left[\begin{array}{l}2x=5\\ 2x-5=-1\\ 2x-5=1\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac52\\ 2x=6\\ 2x=4\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac52\\ x=3\\ x=2\end{array}\right.\)
Vậy \(\in\left\lbrace\frac52;2;3\right\rbrace\)