Cái này ghi dấu "=" xảy ra chắc kì lắm nhỉ @@

BT1:

a) \(\)Ta có: \(x-y=-\left(-x+y\right)\)

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

Hai biểu thức chỉ bằng nhau khi x - y = y - x = 0 hay x = y

b) Ta có: \(\left(x+1\right)^2=x^2+2x+1\)

Hai biểu thức chỉ bằng nhau khi 2x = 0 hay x = 0

c) Nhìn vô là bt ko bằng nhau r`

Hai biểu thức chỉ bằng nhau khi x - y = y - x hay x = y

https://hoc24.vn/hoi-dap/question/418773.html

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

\(\Leftrightarrow x^2-9=x^2-4x+4-1\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=3\)

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

\(\Leftrightarrow4x^2-4x+1=4x^2-25+2\)

\(\Leftrightarrow4x=24\)

\(\Leftrightarrow x=6\)

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

\(\Leftrightarrow x^3+3x^2+3x+1=x^3-2x^2+5x^2-5\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-2\)

d) \(\left(x^2-3x+9\right)\left(x+3\right)=\left(x-2\right)^3+\left(3x-1\right)\left(2x+3\right)\)

\(\Leftrightarrow x^3+3^3=x^3-6x^2+12x-8+6x^2+7x-3\)

\(\Leftrightarrow x^3+27=x^3+19x-11\)

\(\Leftrightarrow19x=38\)

\(\Leftrightarrow x=2\)

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

\(\Rightarrow\left(x-2\right)^4-\left(x-2\right)^2=0\)

\(\Rightarrow\left(x-2\right)^2\left[\left(x-2\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-2\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

Vậy x = 2; x = 3 hoặc x = 1.

a) Ta có: 3³⁹ < 3⁴⁰ ; 11²¹ > 11²⁰

mà 3⁴⁰ = (3²)²⁰ = 9²⁰

Vì 9 < 11 nên 9²⁰ < 11²⁰

Vậy 3³⁹ < 11²¹.

b) Ta có: 3⁹⁹ > 3⁹² ; 11²¹ < 11²³

mà 3⁹² = (3⁴)²³ = 81²³

Vì 81 > 11 nên 81²³ > 11²³

Vậy 3⁹⁹ > 11²³.

a) \(x^2+\left(a+b\right)xy+aby^2\)

\(=x^2+axy+bxy+aby^2\)

\(=x\left(x+ay\right)+by\left(x+ay\right)\)

\(=\left(x+ay\right)\left(x+by\right)\)

b) \(a^2-\left(c+d\right)ab+cdb^2\)

\(=a^2-abc-abd+cdb^2\)

\(=a\left(a-bc\right)-bd\left(a-bc\right)\)

\(=\left(a-bc\right)\left(a-bd\right)\)

c) Sửa đề: \(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2+a^2xy+b^2xy\)

\(=abx^2+b^2xy+a^2xy+aby^2\)

\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)

\(=\left(ax+by\right)\left(bx+ay\right)\)

d) Sửa đề: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)

\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2abxy+b^2x^2\)

\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)

\(=y^2\left(x^2+a^2\right)+b^2\left(x^2+a^2\right)\)

\(=\left(x^2+a^2\right)\left(y^2+b^2\right)\)

x . 2011 - x = 2011 . 2009 + 2011

\(\Rightarrow\) x (2011 - 1) = 2011 (2009 + 1)

\(\Rightarrow\) x . 2010 = 2011 . 2010

\(\Rightarrow\) x = 2011

Điều kiện: x - 3 \(\ne\) 0

*Để P = 0 thì: x + 2 = 0

\(\Rightarrow\) x = -2

*Để P > 0 thì: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)

*Để P < 0 thì: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\x>3\end{matrix}\right.\left(\text{loại}\right)\end{matrix}\right.\)

tổng của 3 số là: 18.3=54

số thứ 2 là: 28:2=14

số thứ 3 là: 54-14-28=12

lưu ý . là nhân đó tick mình nhé