Bài 2 :

a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

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

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

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

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

\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)

\(A=\frac{x-2}{x+2}\)

c) Thay x = 4 ta có :

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

Vậy.........

\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)

\(\left(5x-2\right)\left(25x^2+10x+4\right)\)

\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)

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

\(=125x^3-8\)

\(A=x^2-4x+1\)

\(A=x^2-4x+4-3=\left(x-4\right)^2-3\ge-3\)

Dấu = xảy ra khi \(x-4=0\)

\(\Rightarrow x=4\)

Vậy ...

b) \(4x^2+4x+11\)

\(\left(2x\right)^2+4x+1^2+10=\left(2x+1\right)^2+10\ge10\)

dấu = xảy ra khi \(2x+1=0\)

\(x=-\frac{1}{2}\)

Vậy....

A=\(x^2-4x+1\)

=\(x^2-2.x.2+4-4+1\)

=\(\left(x-2\right)^2-3\ge-3\)              ( vì (x-2)^2 \(\ge0\)) 

Dấu "=" xảy ra khi và chỉ khi x-2=0

                                          <=> x=2

vậy GTNN của A là -3 khi x=2

B= \(4x^2+4x+11\)

 =\(\left(2x\right)^2+2.2x.1+1-1+11\)

 = \(\left(2x+1\right)^2-10\ge-10\)         ( vì \(\left(2x+1\right)^2\ge0\))

Dâu '=' xayr ra khii và chỉ khi 2x+1=0

                                      <=> 2x=-1

                                      <=> x = \(\frac{-1}{2}\)

Vậy GTNN của B là -10 khi x=\(\frac{-1}{2}\)

Chúc bạn buổi tối vui vẻ

\(\frac{x}{a}+\frac{y}{b}=1\)

\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}\right)^3=1\)

\(\Leftrightarrow\frac{x^3}{a^3}+\frac{y^3}{b^3}+3\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1\)

\(\Leftrightarrow\frac{x^3}{a^3}+\frac{y^3}{b^3}-6=1\)

\(\Leftrightarrow\frac{x^3}{a^3}+\frac{y^3}{b^3}=7\)

                     đpcm

để A xác định

\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)

\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)

\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)

\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)

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

C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)

d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)

\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)

\(x^2-4\inℕ\)

\(\Rightarrow x^2\in\left\{0;4;12\right\}\)

Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn

mk học lớp 6 mong mn thông cảm nếu có sai sót

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

\(a,\frac{x+2}{x-1}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)

\(=\frac{-x-2}{1-x}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)

\(=\frac{-x-2}{1-x}+\frac{-\left(x-9\right)}{1-x}+\frac{-\left(x-9\right)}{1-x}\)

\(=\frac{-x-2-x+9-x+9}{1-x}=\frac{-3x+16}{1-x}\)

Câu b,c mk chưa học, bn thông cảm

Còn câu a, nếu sai thì xin lượng thứ :))

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

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

a, Ta có: \(\frac{2x^2-1}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}\)       ĐKXĐ: \(x\ne1\)

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

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

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

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

=\(x\)

b, Ta có: \(\frac{x+3}{x^2-1}-\frac{x+1}{x^2-x}\)     ĐKXĐ \(x\ne\pm1\)

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

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

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

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

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

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

=\(\frac{1}{x+1}\)

Chúc bạn học tốt

Buổi tối vui vẻ

a)\(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)

\(\frac{-5}{2\left(2+y\right)}+\frac{y-2}{y\left(2+y\right)}\)

\(\frac{-5y}{2y\left(2+y\right)}+\frac{2y-4}{2y\left(2+y\right)}\)

\(\frac{-5y+2y-4}{2y\left(2+y\right)}\)

\(\frac{-3y-4}{2y\left(2+y\right)}\)

b)\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)

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

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

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

\(\frac{x-4}{x\left(x-2\right)}\)

\(a,\frac{6}{x^2+4x}+\frac{2}{2x+8}=\frac{6}{x\left(x+4\right)}+\frac{2}{2\left(x+4\right)}\)

                                 \(=\frac{6.2}{2x\left(x+4\right)}+\frac{2.x}{2x\left(x+4\right)}=\frac{12}{2x\left(x+4\right)}+\frac{2x}{2x\left(x+4\right)}\)

                                  \(=\frac{12+2x}{2x\left(x+4\right)}=\frac{2\left(6+x\right)}{2x\left(x+4\right)}=\frac{x+6}{x\left(x+4\right)}\)

\(b,\frac{3-2x}{x^2-9}+\frac{1}{2x-6}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}+\frac{1}{2\left(x-3\right)}\)

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

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

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

P/s : Cs sai sót mong thông cảm.

\(b,\frac{3-2x}{x^2-3^2}+\frac{1}{2x-6}=\frac{3-2x}{\left(x-3\right).\left(x+3\right)}+\frac{1}{2.\left(x-3\right)}\)

\(=\frac{6-4x}{2.\left(x-3\right).\left(x+3\right)}+\frac{x+3}{2.\left(x-3\right).\left(x+3\right)}=\frac{9-3x}{2.\left(x-3\right).\left(x+3\right)}=\frac{-3.\left(x-3\right)}{2.\left(x-3\right).\left(x+3\right)}=-\frac{3}{2.\left(x-3\right)}\)