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\(a.=\frac{4x\left(x^2-2x+1\right)}{x^2-1x-5x+5}\)
\(=\frac{4x\left(x-1\right)^2}{x\left(x-1\right)-5\left(x-1\right)}\)
\(=\frac{4x\left(x-1\right)^2}{\left(x-5\right)\left(x-1\right)}\)
\(=\frac{4x\left(x-1\right)}{x-5}\)
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) ⇔⇔ 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!
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) \(\frac{2x^2+3x-2}{x^2+x-2}\)
\(=\frac{2x^2-4x+x-2}{x^2+2x-x-2}\)
\(=\frac{2x\left(x-2\right)+\left(x-2\right)}{x\left(x+2\right)-\left(x+2\right)}\)
\(=\frac{\left(2x+2\right)\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}\)
a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)
Câub mô