Toán

\(9-\left(x-y\right)^2\)

\(=3^2-\left(x-y\right)^2\)

\(=\left(3-x+y\right)\left(3+x-y\right)\)

____

\(\left(x-y\right)^2-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

____

\(\left(x+2\right)^2-y^2\)

\(=\left[\left(x+2\right)-y\right]\left[\left(x+2\right)+y\right]\)

\(=\left(x-y+2\right)\left(x+y+2\right)\)

____

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

\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)

\(=2x\left(4x+2\right)\)

\(=4x\left(2x+1\right)\)

____

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

____
\(\left(2xy+1\right)^2-\left(2x+y\right)^2\)

\(=\left(2xy+1-2x-y\right)\left(2xy+1+2x+y\right)\)

\(=\left[2x\left(y-1\right)-\left(y-1\right)\right]\left[2x\left(y+1\right)+\left(y+1\right)\right]\)

\(=\left(y-1\right)\left(2x-1\right)\left(2x+1\right)\left(y+1\right)\)

Các tập hợp tạo thành được:

\(\left\{a;b;c\right\};\left\{a;b;d\right\};\left\{a;b;đ\right\};\left\{a;b;e\right\};\left\{a;b;\text{ê}\right\};\\ \left\{a;c;d\right\};\left\{a;c;đ\right\};\left\{a;c;e\right\};\left\{a;c;\text{ê}\right\};\left\{a;d;đ\right\};\\ \left\{a;d;e\right\};\left\{a;d;\text{ê}\right\};\left\{a;đ;e\right\};\left\{a;\text{đ};\text{ê}\right\};\left\{a;e;\text{ê}\right\}\)

Có thể tạo thành 15 tập hợp  

\(10x\left(x-y\right)-6y\left(y-x\right)\)

\(=10x\left(x-y\right)+6y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+6y\right)\)

\(=2\left(x-y\right)\left(5x+3y\right)\)

___

\(3x\left(x-2y\right)+6y\left(2y-x\right)\)

\(=3x\left(x-2y\right)-6y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(3x-6y\right)\)

\(=2\left(x-2y\right)^2\)

___

\(20x\left(x+y\right)-8y\left(y+x\right)\)

\(=\left(x+y\right)\left(20x-8y\right)\)

\(=4\left(x+y\right)\left(5x-2y\right)\)

___

\(xy^2\left(x-3\right)+4x\left(3-x\right)\)

\(=xy^2\left(x-3\right)-4x\left(x-3\right)\)

\(=\left(x-3\right)\left(xy^2-4x\right)\)

\(=x\left(x-3\right)\left(y^2-4\right)\)

\(=x\left(x-3\right)\left(y+2\right)\left(y-2\right)\)

____

\(2x\left(x+y\right)-6x^2\left(x+y\right)\)

\(=\left(x+y\right)\left(2x-6x^2\right)\)

\(=2x\left(x+y\right)\left(1-3x\right)\)

____

\(9x^2\left(y+z\right)+3x\left(y+z\right)\)

\(=\left(y+z\right)\left(9x^2+3x\right)\)

\(=3x\left(y+z\right)\left(3x+1\right)\)

Mik đang cần gấp giúp mik với

Số lượng số chẵn liên tiếp từ 312 đến 324: 312, 314, 316, 318, 320, 322, 324 

⇒ Có 8 số chẵn  

\(A=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=1+\dfrac{3}{x-2}\)

A là số nguyên khi: \(\dfrac{3}{x-2}\) nguyên 

3 ⋮ x - 2

\(\Rightarrow x-2\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{3;1;5;-1\right\}\)

Ta có: \(A=\left\{a;b;c;f;w\right\}\)

Nên a,b thuộc tập hợp A r không thuộc A

\(B=\left\{b;d;y;t;u;v\right\}\)

Nên b thuộc tập hợp B còn a,r không thuộc B

_______

Gọi tập hợp đó là C:
\(C=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

Gọi tập hợp đó là S:
\(S=\left\{0;2;4;6;8;10;12;14;16;18;20;22;24;26;28;30;32;34;36;38\right\}\)

Viết các số từ bé đến lớn:

\(1,2,3,4,5\)

Viết các số từ lớn đến bé:

\(5;4;3;2;1\)

thanks

từ bé đến lớn : 1,2,3,4,5

từ lớn đến bé : 5,4,3,2,1

a) \(\sqrt[]{x^2-2x+4}=2x-2\)

\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)

Giải pt \(3x^2-6x=0\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=2\)

c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(\sqrt{x^2-6x+5}=2\)

ĐK: \(\left(x-5\right)\left(x-1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\)

\(\Leftrightarrow x^2-6x+5=2^2\)

\(\Leftrightarrow x^2-6x+1=0\)

\(\Rightarrow\Delta=\left(-6\right)^2-4\cdot1\cdot1=32>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{6+\sqrt{32}}{2}=3+2\sqrt{2}\left(tm\right)\\x_2=\dfrac{6-\sqrt{32}}{2}=3-2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=3+2\sqrt{2}\)

\(\sqrt[]{x^2-6x+5}=2\left(2>0\right)\)

\(\Leftrightarrow x^2-6x+5=4\)

\(\Leftrightarrow x^2-6x+1=0\left(1\right)\)

\(\Delta'=9-1=8\)

\(\Rightarrow\sqrt[]{\Delta'}=2\sqrt[]{2}\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt[]{2}\\x=3-2\sqrt[]{2}\end{matrix}\right.\)

\(\sqrt[]{A}=B\Leftrightarrow\left\{{}\begin{matrix}B\ge0\\A=B^2\end{matrix}\right.\)

Anh Pop xem lại dùm em bài này nhé! Bị nhầm công thức rồi anh. Cảm ơn anh!

Xét hiệu: \(C-D=\dfrac{2^{2008}-3}{2^{2007}-1}-\dfrac{2^{2007}-3}{2^{2006}-1}\)

\(=\dfrac{\left(2^{2008}-3\right)\left(2^{2006}-1\right)-\left(2^{2007}-1\right)\left(2^{2007}-3\right)}{\left(2^{2007}-1\right)\left(2^{2006}-1\right)}\)

\(=\dfrac{2^{4014}-2^{2008}-3\cdot2^{2006}+3-2^{4014}+3\cdot2^{2007}+2^{2007}-3}{\left(2^{2007}-1\right)\left(2^{2006}-1\right)}\)

\(=\dfrac{3\left(2^{2007}-2^{2006}\right)-\left(2^{2008}-2^{2007}\right)}{x}\) với \(x=\left(2^{2007}-1\right)\left(2^{2006}-1\right)\)

\(=\dfrac{3\cdot2^{2006}-2^{2007}}{x}\)

\(=\dfrac{3\cdot2^{2007}-2\cdot2^{2006}}{x}=\dfrac{2^{2006}}{x}\)

Dễ thấy: \(\left\{{}\begin{matrix}2^{2006}>0\\\left\{{}\begin{matrix}2^{2007}-1>0\\2^{2006}-1>0\end{matrix}\right.\Rightarrow x=\left(2^{2007}-1\right)\left(2^{2006}-1\right)>0\end{matrix}\right.\)

Nên: \(C-D=\dfrac{2^{2006}}{x}>0\Leftrightarrow C>D\).

Vậy: \(C>D.\)