\(\frac{2x-3}{2}>\frac{8x-11}{6}\)

\(\Leftrightarrow\frac{3\left(2x-3\right)}{6}>\frac{8x-11}{6}\)

\(\Leftrightarrow6x-9>8x-11\)

\(\Leftrightarrow6x-8x>-11+9\)

\(\Leftrightarrow-2x>-2\)

\(\Leftrightarrow x< 1\)

Vậy nghiệm của bất phương trình là x < 1

a.2x#+_2 . quy đồng khử mẫu tchung : (x+2)(x+1)+(x-1)(x-2)--->2x^2 + 4=2(x^2+2). --> s={x thuộc R/ X#+_2}

 a) ĐKXĐ \(\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)

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

 \(\Leftrightarrow x^2+3x+2+x^2-3x+2-2x^2-4=0\)

 \(\Leftrightarrow0x=0\)(vô số nghiệm)

nghiệm x thỏa mãn phương trình S \(\in\)R  với   \(\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)

 b) ĐKXĐ  \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

\(\Rightarrow\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}=\frac{1}{2x\left(x-2\right)}-\frac{7}{8x}\) 

 \(\Rightarrow2\left(5-x\right)-x-4\left(x-1\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow10-2x-x-4x+4+7x-14=0\) 

 \(\Leftrightarrow0x=0\)(phương trìh vô số nghiệm)

nghiệm x thỏa mãn phương trình S \(\in\)R  với   \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

Ta có \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{x\left(32-8x+2x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)

Chuyển hết sang vế phải quy đồng ta được:

\(\frac{16x^2+4x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)}{6\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow\frac{16x^2+8x^2+4x-48x^2+6x+1}{6\left(2x+1\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow24x^2-10x-1=0\Leftrightarrow\left(12x+1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{12}\\x=\frac{1}{2}\end{cases}}\)

a) ĐKXĐ: x khác +2

\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)

<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)

<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22

<=> x^2 - 7x - 2 = 2x - 22

<=> x^2 - 7x - 2 - 2x + 22 = 0

<=> x^2 - 9x + 20 = 0

<=> (x - 4)(x - 5) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 hoặc x = 5

làm nốt đi 

\(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

\(\Leftrightarrow\)\(\frac{5\left(5x+2\right)}{30}-\frac{10\left(8x-1\right)}{30}=\frac{6\left(4x+2\right)}{30}-\frac{150}{30}\)

\(\Leftrightarrow\)\(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow\)\(25x+10-80x+80=24x+12-150\)

\(\Leftrightarrow\)\(79x=240\)

\(\Leftrightarrow\)\(x=\frac{240}{79}\)

ủa bạn ơi, ở hàng thứ 4 là -10(8x - 1) = -80x + 10 chứ???

bài kia của mk nhầm chút nhé, mk làm lại, m.n cho mk xin lỗi ạ

     \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

\(\Leftrightarrow\)\(\frac{5\left(5x+2\right)}{30}-\frac{10\left(8x-1\right)}{30}=\frac{6\left(4x+2\right)}{30}-\frac{150}{30}\)

\(\Leftrightarrow\)\(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow\)\(25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow\)\(79x=158\)

\(\Leftrightarrow\)\(x=2\)

A = 2.(x^2-8x+22)-1/x^2-8x+22 = 2 - 1/x^2-8x+22

Có : x^2-8x+22 = (x^2-8x+16)+6 = (x-4)^2+6 >= 6 => 1/x^2-8x+22 < = 1/6

=> A = 2 - 1/x^2-8x+22 >= 2-1/6 = 11/6

Dấu "=" xảy ra <=> x-4 = 0 <=> x=4

Vậy GTNN của A = 11/6 <=> x=4

k mk nha

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

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

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

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

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

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