Từ 4 chữ số: 1; 2; 3 và 4 có thể lập được 4*3*2*1=24 số có 4 chữ số khác nhau. Nên nếu viết 25 số tự nhiên có 4 chữ số sẽ có ít nhất 2 số bằng nhau

P/s: Đề bài chắc muốn hổi 25 số tự nhiên có 4 chữ số?  

đề bài chưa rõ, 2 số bằng nhau hay 2 chữ số

\(\dfrac{2\left(2n+1\right)}{n}=\dfrac{2.2n+2}{n}=\dfrac{4n+2}{n}\)

Mai mik đi hk r mn ng ơi giúp mik đi

Mn ơi mik gấp lắm r giúp mik đi

Cái cuối là bằng 2 nha mn mik cần gấp mong mn giúp 

a)xét tam giác ADB và tam giác AEC

chung A

E=D=90

=>ADB đồng dạng vs aAEC(gg)

=>AE.AB=AD.AC

b)xét 2 tam giác có

AE/AC=AD/AB

chung A

=>ABE đồng dạng với ACB

kb ko bn

`(x -5).(x+7) + (x+4).(2-x)`

` = x^2 +2x-35 -x^2 -2x + 8`

`= (x^2 - x^2) + (2x - 2x) + (-35 +8)`

`=0 + 0 - 27`

`=-27`

`=>` Biểu thức trên không phụ thuộc vào biến `x`

\(\left(x-5\right)\left(x+7\right)+\left(x+4\right)\left(2-x\right)=x^2+2x-35+2x-x^2+8-4x=-27\)

Vậy bth ko phụ thuộc biến x

Gọi x là tuổi con hiện nay, vậy tuổi bố hiện nay là: 4x

5 năm sau tuổi con là: x + 5 và tuổi bố khi đó là: 4x + 5

Ta có PT: 4x + 5 = 3.(x + 5) => x = 10

Vậy hiện nay: Con 10 tuổi và bố 40 tuổi

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^3=0\)

\(\Rightarrow\left(a+b\right)^3+3\left(a+b\right)c\left(a+b+c\right)+c^3=0\)

\(\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)+3ab\left(a+b\right)=0\)

\(\Rightarrow3\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Rightarrow3\left(a+b\right)\left[c\left(b+c\right)+a\left(b+c\right)\right]=0\)

\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}c=0\\a=0\\b=0\end{matrix}\right.\)
*\(a=-b;c=0\Rightarrow P=a^{2021}-a^{2021}+0^{2021}=0\)

*\(b=-c;a=0\Rightarrow P=0^{2021}+b^{2021}-b^{2021}=0\)

*\(c=-a;b=0\Rightarrow P=a^{2021}+0^{2021}-a^{2021}=0\)

Vậy \(P=0\)

Đặt \(x^3=a^3;27y^3=b^3;8z^3=c^3\)

\(\Rightarrow a^3-b^3-c^3=3abc\)

\(\Rightarrow a^3-b^3-c^3-3abc=0\)

\(\Rightarrow a^3-\left(b+c\right)^3+3bc\left(b+c\right)-3abc=0\)

\(\Rightarrow\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2\right]-3bc\left(a-b-c\right)=0\)

\(\Rightarrow\left(a-b-c\right)\left(a^2+ab+ac+b^2+2bc+c^2-3bc\right)=0\)

\(\Rightarrow\left(a-b-c\right)\left(a^2+b^2+c^2+ab-bc+ca\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a^2+b^2+c^2+ab-bc+ca=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c+a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a=-b=-c\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3y-2z=0\\x=-3y=-2z\end{matrix}\right.\)

*\(x-3y-2z=0\) :

\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}=\dfrac{2z.x.3y}{6xyz}=1\)

*\(x=-3y=-2z\) :

\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}\dfrac{\left(x+x\right)\left(3y+3y\right)\left(-2z-2z\right)}{6xyz}=\dfrac{2x.6y.\left(-4\right)z}{6xyz}=-8\)

Mk sửa lại biểu thức P :\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}\)

Ta có : x³ - 27y³ - 8z³ = 18xyz 

<=> (x - 3y)³ + 9xy(x - 3y) - 8z³ = 18xyz

<=> [(x - 3y)³ - (2z)³] + 9xy(x - 3y - 2z) = 0

<=> (x - 3y - 2z)[(x - 3y)² + (x - 3y).2z + 4z²] + 9xy(x - 3y - 2z) = 0

<=> (x - 3y - 2z)[(x - 3y)² + (x - 3y).2z + 4z² + 9zy] = 0

<=> \(\left(x-3y-2z\right)\left\{\left[\dfrac{1}{4}\left(x-3y\right)^2+\left(x-3y\right).2z+4z^2\right]+\dfrac{3}{4}\left(x-3y\right)^2+9xy\right\}=0\)

<=> \(\left(x-3y-2z\right)\left\{\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2\right\}=0\)

<=> \(\left[{}\begin{matrix}x-3y-2z=0\\\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2=0\end{matrix}\right.\)

THI1 x - 3y - 2z = 0

<=> x = 3y + 2z

Khi đó \(P=\dfrac{2z.x.3y}{6xyz}=1\)

TH2 \(\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}\left(x-3y\right)+2z=0\\x+3y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2z=3y\\x=-3y\end{matrix}\right.\Leftrightarrow x=-3y=-2z\)

Khi đó P = \(\dfrac{\left(-6y\right).\left(-2x\right).\left(-4z\right)}{xyz}=-48\)

\(a^3+8b^3+1=6ab\)

\(\Rightarrow\left(a+2b\right)^3-6a^2b-12ab^2+1-6ab=0\)

\(\Rightarrow\left(a+2b\right)^3+1-6ab\left(a+2b+1\right)=0\)

\(\Rightarrow\left(a+2b+1\right)\left[\left(a+2b\right)^2-\left(a+2b\right)+1\right]-6ab\left(a+2b+1\right)=0\)

\(\Rightarrow\left(a+2b+1\right)\left(a^2+4ab+4b^2-a-2b+1-6ab\right)=0\)

\(\Rightarrow\left(a+2b+1\right)\left(a^2-2ab+4b^2-a-2b+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\a^2-2ab+4b^2-a-2b+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a^2-2a\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b\right)+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a^2-2a+1-1\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b+\dfrac{1}{4}-\dfrac{1}{4}\right)+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a-1\right)^2-\dfrac{1}{2}+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2-\dfrac{1}{2}+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a-1\right)^2+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\a=1;b=\dfrac{1}{2}\end{matrix}\right.\)

*\(a+2b+1=0\Rightarrow a+2b=-1\)

*\(a=1;b=\dfrac{1}{2}\Rightarrow a+2b=1+2.\dfrac{1}{2}=2\)

.

\(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow3+2\left(ab+bc+ca\right)=9\Rightarrow ab+bc+ca=3\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\)

\(\Rightarrow a=b=c\) mà \(a+b+c=3\Rightarrow a=b=c=1\)

\(P=\left(2a-3\right)^{2021}+\left(2b-3\right)^{2021}+\left(2c-3\right)^{2021}=\left(2.1-3\right)^{2021}+\left(2.1-3\right)^{2021}+\left(2.1-3\right)^{2021}=\left(-1\right)^{2021}+\left(-1\right)^{2021}+\left(-1\right)^{2021}=-1-1-1=-3\)

ai giúp em với ạ