\(a,\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x^2-6x}\)

\(=\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x\left(x-6\right)}\)

\(=\frac{7\left(x-6\right)\left(x+6\right)-x\left(x-6\right)+36\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)}\)

\(=\frac{7\left(x^2-6\right)-x^2+6x+36x+216}{x\left(x^2-6\right)}\)

\(=\frac{7x^2-42-x^2+6x+36x+216}{x\left(x^2-6\right)}\)

\(=\frac{6x^2+42x+216}{x\left(x^2-6\right)}\)

\(=\frac{6\left(x^2+7x+36\right)}{x\left(x^2-6\right)}\)

Đề sai nhé, phải là như này nè :

\(b,\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)

\(=\frac{x^2+x+1-\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)

\(=\frac{x^2+x+1-x^2+x-1}{x^4+x^2+1}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)

\(=\frac{2x}{x^4+x^2+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)

\(=\frac{2x\left(x^4-x^2+1\right)-2x\left(x^4+x^2+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)}+\frac{4x^3}{x^8-x^4+1}\)

\(=\frac{2x^5-2x^3+2x-2x^5-2x^3-2x}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}\)

\(=-\frac{4x^3}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}=0\)

Cho \(\Delta ABC\)Cân tại \(A\)sao lại có \(AB=10cm\)và \(AC=15cm\)thế em ? 

Bạn Cao Kỳ Duyên : Đề sai kìa bạn !

Tam giác ABC cân tại A mà sao AB=10cm, AC=15cm. Như vậy AB đâu bằng AC đâu.

Khái niệm của tam giác cân là : Tam giác cân là tam giác có hai cạnh bằng nhau. Trong một tam giác cân hai góc ở đáy bằng nhau. Nếu một tam giác có hai góc bằng nhau thì là tam giác cân. Tam giác vuông cân là tam giác vuông có hai cạnh góc vuông bằng nhau .

a) \(\frac{3x^2+5x+1}{x^3-1}-\frac{1-x}{x^2+x+x}-\frac{3}{x-1}\)

\(=\frac{3x^2+5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{3x^2+5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{1-x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2+3x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{3x^2+5x+1-1+x^2-3x^2-3x-3}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

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

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

\(a,\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

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

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

\(=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

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

\(b,\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)

\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}-\frac{3}{\left(x-3\right)\left(x-3\right)}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)

\(=\frac{18-3x-9-x^2+3x}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-x^2+9}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)

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

Ta có : n^3 - n^2 + n - 1 = n^2(n - 1) + (n - 1) = (n^2 + 1)(n - 1).
Để n^3 - n^2 + n - 1 là số nguyên tố thì ta có 2 TH :
TH1 : n^2 + 1 = 1 ; n - 1 nguyên tố => không có n thỏa mãn.
TH2 : n^2 + 1 nguyên tố, n - 1 = 1 => n = 2 (chọn)
Vậy n = 2 để n^3 - n^2 + n - 1 nguyên tố

Đặt A=x^4-x^3+3x^2-2x+2

=(x^4+3x^2+2)-(x^3+2x)

=(x^4+x^2+2x^2+2)-x(x^2+2)

=(x^2+1)(x^2+2)-x(x^2+2)

=(x^2+2)(x^2-x+1)

Ta có x^2+2>=2>0;

x^2-x+1=(x^2-x+1/4)+3/4 =(x-1/2)^2+3/4>=3/4>0 

=> A>0  

1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0

Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)

b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)

B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B =  \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B = \(\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B = \(\frac{x+3y}{x\left(x-3y\right)}\)

\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)

\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)

x^2-x+1 2x^4-3x^3+5x^2-4x+3 2x^2-x+2 2x^4-2x^3+2x^2 -x^3+3x^2-4x+3 -x^3+x^2-x 2x^2-3x+3 2x^2-2x+2 -x+1

Vậy (2x⁴-3x³+5x²-4x+3):(x²-x+1) = 2x² -x + 2 dư -x+1

Giả sử:

\(a>b>c\Rightarrow a-b>0,b-c>0,a-c>0\)

Ta có:

\(\hept{\begin{cases}a^2+b^2+c^2\ge a^2+c^2\\\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}\ge\frac{\left(\frac{1}{a-b}+\frac{1}{b-c}\right)^2}{2}\ge\frac{8}{\left(a-c\right)^2}\end{cases}}\)

Từ đây ta có:

\(VT\ge\left(a^2+c^2\right).\frac{9}{\left(c-a\right)^2}\)

Ta chứng minh

\(\left(a^2+c^2\right).\frac{9}{\left(c-a\right)^2}\ge\frac{9}{2}\)

\(\Leftrightarrow\left(a+c\right)^2\ge0\)(Đúng)

Vậy ta có điều phải chứng minh là đúng. Dấu = xảy ra khi a = - c; b = 0 và các hoán vị của nó