Bài 2:c: ĐKXĐ: x<>0\(\left(14x^3+12x^2-14x\right):2x=\left(x+2\right)\left(3x-4\right)\) =>\(\frac{14x^3}{2x}+\frac{12x^2}{2x}-\frac{14x}{2x}=3x^2-4x+6x-8\) =>\(7x^2+6x-7-3x^2-2x+8=0\) =>\(4x^2+4x+1=0\) =>\(\left(2x+1\right)^2=0\) =>2x+1=0=>2x=-1=>\(x=-\frac12\) d: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\) =>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\) =>\(24x^2-26x-5-\left(24x^2-23x-12\right)=15\) =>\(24x^2-26x-5-24x^2+23x+12=5\) =>-3x+7=5=>-3x=-2=>\(x=\frac23\) Bài 1:a: \(\left(x-2\right)\left(2x-1\right)-\left(2x-3\right)\left(x-1\right)-2\) \(=2x^2-x-4x+2-\left(2x^2-2x-3x+3\right)-2\) \(=2x^2-5x+2-\left(2x^2-5x+3\right)-2=2x^2-5x-2x^2+5x-3=-3\) b: \(x\left(x+3y+1\right)-2y\left(x-1\right)-x\left(x+y+1\right)\) \(=x^2+3xy+x-2xy+2y-x^2-xy-x\) =2yCâu trả lời: Bài 2:c: ĐKXĐ: x<>0\(\left(14x^3+12x^2-14x\right):2x=\left(x+2\right)\left(3x-4\right)\) =>\(\frac{14x^3}{2x}+\frac{12x^2}{2x}-\frac{14x}{2x}=3x^2-4x+6x-8\) =>\(7x^2+6x-7-3x^2-2x+8=0\) =>\(4x^2+4x+1=0\) =>\(\left(2x+1\right)^2=0\) =>2x+1=0=>2x=-1=>\(x=-\frac12\) d: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\) =>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\) =>\(24x^2-26x-5-\left(24x^2-23x-12\right)=15\) =>\(24x^2-26x-5-24x^2+23x+12=5\) =>-3x+7=5=>-3x=-2=>\(x=\frac23\) Bài 1:a: \(\left(x-2\right)\left(2x-1\right)-\left(2x-3\right)\left(x-1\right)-2\) \(=2x^2-x-4x+2-\left(2x^2-2x-3x+3\right)-2\) \(=2x^2-5x+2-\left(2x^2-5x+3\right)-2=2x^2-5x-2x^2+5x-3=-3\) b: \(x\left(x+3y+1\right)-2y\left(x-1\right)-x\left(x+y+1\right)\) \(=x^2+3xy+x-2xy+2y-x^2-xy-x\) =2y