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:

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.

Em xem lại số mũ của 2x - 5y nhé

2023 hay 2024?

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) \(5< 2^x< 100\)

\(\Rightarrow4=2^2< 5< 2^x< 100< 128=2^7\)

\(\Rightarrow2< x< 7\)

b , x = 3/2 a và b mình ko biết

\(5x^2+5y^2+8xy-2x+2y+2=0\)

=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

=>x=1 và y=-1

\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)

E kh hiểu lắm ạ="))

2 x A = 1 - \(\dfrac{1}{2027}\)

 \(A=\dfrac{1013}{2027}\)

2 x A = 1- 1/2027

A=1013/2027

Ta có: P=|x+2023|-|2025-x|

=>P=|x+2023|-|x-2025|

TH1: x<-2023

=>x+2023<0; x-2025<0

=>P=-x-2023-(-x+2025)

=-x-2023+x-2025

=-4048(1)

TH2: -2023<=x<2025

=>x+2023>=0; x-2025<0

=>P=x+2023-(-x+2025)

=2x-2

Vì P=2x-2 là hàm số đồng biến trên R nên P lớn nhất khi x lớn nhất

-2023<=x<2025 nên x không có giá trị lớn nhất

=>P không có giá trị lớn nhất

TH3: x>=2025

=>x+2023>0; x-2025>=0

=>P=x+2023-(x-2025)=4048(2)

Từ (1),(2) suy ra \(P_{\max}=4048\) khi x>=2025

Sửa đề: Tìm x

a: \(x\left(6-x\right)^{2023}=\left(6-x\right)^{2023}\)

=>\(x\left(6-x\right)^{2023}-\left(6-x\right)^{2023}=0\)

=>\(\left(6-x\right)^{2023}\left(x-1\right)=0\)

=>\(\left[\begin{array}{l}6-x=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=1\end{array}\right.\)

b: \(5^{x}+5^{x+2}=650\)

=>\(5^{x}+5^{x}\cdot25=650\)

=>\(5^{x}\left(1+25\right)=650\)

=>\(5^{x}=\frac{650}{26}=25=5^2\)

=>x=2

c: \(2^{x+2}-2^{x}=96\)

=>\(2^{x}\cdot2^2-2^{x}=96\)

=>\(2^{x}\left(2^2-1\right)=96\)

=>\(2^{x}\cdot3=96\)

=>\(2^{x}=\frac{96}{3}=32=2^5\)

=>x=5

d: \(10^{x}:5^{y}=20^{y}\)

=>\(10^{x}=20^{y}\cdot5^{y}=100^{y}=\left(10\right)^{2y}\)

=>x=2y

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 \(x=4;y=-16\).

\(\mathit{Toru}\)

a: \(x^2-2x+3\)

\(=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)

=>\(A=\frac{37}{x^2-2x+3}\le\frac{37}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(x^2-5x+10\)

\(=x^2-5x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac52\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)

=>\(\frac{26}{x^2-5x+10}\le26:\frac{15}{4}=26\cdot\frac{4}{15}=\frac{104}{15}\forall x\)

=>\(B=-\frac{26}{x^2-5x+10}\ge-\frac{104}{15}\forall x\)

Dấu '=' xảy ra khi \(x-\frac52=0\)

=>\(x=\frac52\)

c: \(x^2-x+6\)

\(=x^2-x+\frac14+\frac{23}{4}\)

\(=\left(x-\frac12\right)^2+\frac{23}{4}\ge\frac{23}{4}\forall x\)

=>\(\frac{2023}{x^2-x+6}\le2023:\frac{23}{4}=2023\cdot\frac{4}{23}=\frac{8092}{23}\forall x\)

=>\(C=-\frac{2023}{x^2-x+6}\ge-\frac{8092}{23}\forall x\)

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

d: \(x^2+x+5\)

\(=x^2+x+\frac14+\frac{19}{4}\)

\(=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

=>\(D=\frac{0.75}{x^2+x+5}\le\frac34:\frac{19}{4}=\frac{3}{19}\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

e: \(2x^2-x+37=2\left(x^2-\frac12x+\frac{37}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{295}{16}\right)=2\left(x-\frac14\right)^2+\frac{295}{8}\ge\frac{295}{8}\forall x\)

=>\(\frac{13}{2x^2-x+37}\le13:\frac{295}{8}=\frac{104}{295}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

f: \(3x^2-x+19\)

\(=3\left(x^2-\frac13x+\frac{19}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}+\frac{227}{36}\right)=3\left(x-\frac16\right)^2+\frac{227}{12}\ge\frac{227}{12}\forall x\)

=>\(\frac{61}{3x^2-x+19}\le61:\frac{227}{12}=61\cdot\frac{12}{227}=\frac{732}{227}\forall x\)

=>\(-\frac{61}{3x^2-x+19}\ge-\frac{732}{227}\forall x\)

Dấu '=' xảy ra khi x-1/6=0

=>x=1/6

\(C=16x^2-8x+2024\)

\(\Rightarrow C=16x^2-8x+1+2023\)

\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)

\(\Rightarrow Min\left(C\right)=2023\)

\(D=-25x^2+50x-2023\)

\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)

\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(D\right)=1998\)

\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)

\(\Rightarrow Max\left(B\right)=200\)

\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)

\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)

\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)

\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)

\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)

\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)

\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)

\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)

\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)

\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(F\right)=48\)