45:8=5 (dư 5)

45 là đúng

45 nha

\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+4x^2+3}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+1\right)\left(x^2+3\right)}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(=\frac{x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(=\frac{x^2}{x^4-x^2+1}\)

\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+1\right)\left(x^2+3\right)}\)

\(=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)

\(=\frac{x^4+2+\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)

\(\frac{x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}=\frac{x^2}{x^4-x^2+1}\)

Vậy \(M=\frac{x^2}{x^4-x^2+1}\forall x\)

a.\(x^2+7x+6\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

Sửa đề:.\(x^4+2008x^2+2007x+2008\)

\(=x^4+x^2+1+2007x^2+2007x+2007\)

\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

Trả lời:

a, x² + 7x + 6

= x² + x + 6x + 6

= ( x² + x ) + ( 6x + 6 )

= x ( x + 1 ) + 6 ( x + 1 )

= ( x + 6 ) ( x + 1 )

\(x^3+x^2+a-x⋮\left(x+1\right)^2\)

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

Để \(x^3+x^2+a-x⋮\left(x+1\right)^2\)thì   \(a+1=0\) \(\forall a\)

\(\Rightarrow a=-1\)

Để \(x^3+x^2+a-x⋮\left(x+1\right)^2\Rightarrow a-1=0\Leftrightarrow a=1\)

Trả lời:

1, A = x² - 20x + 101

= ( x² - 20x + 100 ) + 1

= ( x - 10 )² + 1

Ta có: \(\left(x-10\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x-10\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra khi x - 10 = 0 <=> x = 10

Vậy GTNN của A bằng 1 khi x = 10.

2, B = 4x - x² + 3

= ( - x² + 4x - 4 ) + 7

= - ( x² - 4x + 4 ) + 7

= - ( x - 2 )² + 7

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Leftrightarrow-\left(x-2\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x-2\right)^2+7\le7\forall x\)

Dấu "=" xảy ra khi x - 2 = 0 <=> x = 2

Vậy GTLN của B bằng 7 khi x = 2

A=x^2-20x+101=x^2-20x+100+1=(x-10)^2+1>=1

dấu "=" xẩy ra <=> x-10=0<=>x=10;

Vậy...

B=4x-x^2+3=-x^2+4x-4+7=-(x-2)^2+7<=7

dấu "=" xẩy ra <=> x-2=0<=>x=2;

Vậy...

Số cần tìm là : 211

cm định lí 4 điểm giúp mk nha

Xét tứ giác ABCD có cạnh đối diện AD và BC cắt nhau tại O

Gọi D₁ và C₁ lần lượt là các điểm đối xứng của C và D qua O

Khi đó:\(\hept{\begin{cases}AC_1=AC\\BD_1=BD\\C_1D_1=CD\end{cases}}\)

Áp dụng định lí ta có:

Tứ giác \(ABC_1D_1:AD_1\perp BC_1\)

\(\Leftrightarrow AB^2+C_1D_1^2=AC_1^2+BD_1^2\)

\(\Rightarrow AD\perp BC\)

\(\Leftrightarrow AB^2+CD^2=AC^2+BD^2\)

Cre:h