a) Theo mình thì chỉ min thôi nhé!

\(A=\frac{8x^2-1}{4x^2+1}+1+11=\frac{12x^2}{4x^2+1}+11\ge11\)

b)Bạn rút gọn lại giùm mìn, lười quy đồng lắm:(

\(a.=\frac{4x\left(x^2-2x+1\right)}{x^2-1x-5x+5}\)

\(=\frac{4x\left(x-1\right)^2}{x\left(x-1\right)-5\left(x-1\right)}\)

\(=\frac{4x\left(x-1\right)^2}{\left(x-5\right)\left(x-1\right)}\)

\(=\frac{4x\left(x-1\right)}{x-5}\)

b) \(\frac{4x^3-64x}{x^2-7x+12}\)

\(=\frac{4x\left(x^2-16\right)}{x^2-3x-4x+12}\)

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

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

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

c) \(\frac{x^2-6x+8}{x^3-8}\)

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

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

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

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

Giải:

a) $\frac{}{}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}$⇔⇔ 9x² + 12x + 4 - 18x + 12 = 9x² ⇔ 9x² + 12x + 4 - 18x + 12 - 9x² = 0

⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)

Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .

b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)

⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4

⇔ 8 = 4 ( vô lí)

Vậy phương trình trên vô nghiệm.

Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!

khó qua mik mới hc lớp 7 thôi

a) ĐKXĐ: \(x\notin\left\{\frac{1}{2};\frac{-1}{2}\right\}\)

Ta có: \(\frac{1+8x}{8x+4}=\frac{2x}{6x-3}-\frac{8x^2}{3-12x^2}\)

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

\(\Leftrightarrow\frac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}=\frac{2x\cdot4\cdot\left(2x+1\right)}{12\left(2x+1\right)\left(2x-1\right)}+\frac{32x^2}{12\left(2x-1\right)\left(2x+1\right)}\)

Suy ra: \(3\left(8x+1\right)\left(2x-1\right)=8x\left(2x+1\right)+32x^2\)

\(\Leftrightarrow3\left(16x^2-8x+2x-1\right)=16x^2+8x+32x^2\)

\(\Leftrightarrow3\left(16x^2-6x-1\right)=48x^2+8x\)

\(\Leftrightarrow48x^2-18x-3-48x^2-8x=0\)

\(\Leftrightarrow-26x-3=0\)

\(\Leftrightarrow-26x=3\)

hay \(x=-\frac{3}{26}\)

Vậy: \(S=\left\{-\frac{3}{26}\right\}\)

b) Ta có: \(\left(x-2\right)\left(x-3\right)< \left(x-4\right)^2-2\left(x+3\right)\)

\(\Leftrightarrow x^2-5x+6< x^2-8x+16-2x-6\)

\(\Leftrightarrow x^2-5x+6< x^2-10x+10\)

\(\Leftrightarrow x^2-5x+6-x^2+10x-10< 0\)

\(\Leftrightarrow5x-4< 0\)

\(\Leftrightarrow5x< 4\)

hay \(x< \frac{4}{5}\)

Vậy: S={x|\(x< \frac{4}{5}\)}

\(a,\frac{x^2-8x+15}{x^2-6x+9}\)

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

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

\(=\frac{x-5}{x-3}\)

b) \(\frac{2x^2+3x-2}{x^2+x-2}\)

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

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

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

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

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

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