Giúppohvvbh

B2: a) $\frac{24x^8}{25y^3}.\frac{5044}{x^3}$ Ta có: \[ \frac{24x^8}{25y^3}.\frac{5044}{x^3} = \frac{24 \times 5044 \times x^8}{25y^3 \times x^3} = \frac{24 \times 5044 \times x^{8-3}}{25y^3} = \frac{24 \times 5044 \times x^5}{25y^3} \] b) $\frac{x^2+5x}{2x-0}.\frac{x^2-10x+25}{3x+15}$ Ta có: \[ \frac{x^2+5x}{2x}.\frac{x^2-10x+25}{3(x+5)} = \frac{x(x+5)}{2x}.\frac{(x-5)^2}{3(x+5)} = \frac{x+5}{2}.\frac{(x-5)^2}{3(x+5)} = \frac{(x-5)^2}{6} \] c) $\frac{7x+21}{5x-5}.\frac{18-6x}{x+3}$ Ta có: \[ \frac{7(x+3)}{5(x-1)}.\frac{6(3-x)}{x+3} = \frac{7(x+3)}{5(x-1)}.\frac{-6(x-3)}{x+3} = \frac{-42(x-3)}{5(x-1)} \] B3: a) $A=\frac{x^2-1}{x+5}.\frac{2x+10}{x^2-x}$ với $x=99$ Ta có: \[ A = \frac{(x-1)(x+1)}{x+5}.\frac{2(x+5)}{x(x-1)} = \frac{(x+1)}{x} = \frac{99+1}{99} = \frac{100}{99} \] b) $B=\frac{x^2-16}{2x+5}.\frac{6}{4-x}$ với $x=-1$ Ta có: \[ B = \frac{(x-4)(x+4)}{2x+5}.\frac{6}{4-x} = \frac{(x-4)(x+4)}{2x+5}.\frac{-6}{x-4} = \frac{-6(x+4)}{2x+5} = \frac{-6(-1+4)}{2(-1)+5} = \frac{-6 \times 3}{3} = -6 \] B1 b) $\frac{5x-5}{x^2-4}:\frac{x-3}{x+2}$ Ta có: \[ \frac{5(x-1)}{(x-2)(x+2)}.\frac{x+2}{x-3} = \frac{5(x-1)}{(x-2)(x-3)} \] a) $\frac{9x^3-4}{3x+1}:\frac{3x+2}{6x^2+2x}$ Ta có: \[ \frac{(3x-2)(3x^2+2x+4)}{3x+1}.\frac{2x(3x+1)}{3x+2} = \frac{2x(3x-2)(3x^2+2x+4)}{3x+2} \] B2: a) $\frac{x^3-8}{x^2-4}:(x^2+2x+4)$ Ta có: \[ \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}.\frac{1}{x^2+2x+4} = \frac{1}{x+2} \] b) $\frac{2x+6x^2}{\sqrt{x^2}+x}:\frac{4x^2+4x+1}{x+1}$ Ta có: \[ \frac{2x(1+3x)}{|x|+x}.\frac{x+1}{(2x+1)^2} = \frac{2x(1+3x)}{2x}.\frac{x+1}{(2x+1)^2} = \frac{(1+3x)(x+1)}{(2x+1)^2} \] B4: b) $\frac{x+1}{x+2}-(\frac{x+2}{x+3}:\frac{x+3}{x+1})$ Ta có: \[ \frac{x+1}{x+2}-(\frac{x+2}{x+3}.\frac{x+1}{x+3}) = \frac{x+1}{x+2}-(\frac{(x+2)(x+1)}{(x+3)^2}) = \frac{(x+1)(x+3)^2-(x+2)(x+1)}{(x+2)(x+3)^2} = \frac{(x+1)((x+3)^2-(x+2))}{(x+2)(x+3)^2} = \frac{(x+1)(x^2+6x+9-x-2)}{(x+2)(x+3)^2} = \frac{(x+1)(x^2+5x+7)}{(x+2)(x+3)^2} \] a) $\frac{x+1}{x+2}:\frac{x+2}{x+3}:\frac{x+3}{x+1}$ Ta có: \[ \frac{x+1}{x+2}.\frac{x+3}{x+2}.\frac{x+1}{x+3} = \frac{(x+1)^2}{(x+2)^2} \] B1 a) $A=\frac{12x+5}{x+9}.\frac{4x+3}{360x+150}+\frac{12x+5}{x+9}.\frac{6-3x}{360x+150}$ Ta có: \[ A = \frac{12x+5}{x+9}.\left(\frac{4x+3}{360x+150}+\frac{6-3x}{360x+150}\right) = \frac{12x+5}{x+9}.\frac{4x+3+6-3x}{360x+150} = \frac{12x+5}{x+9}.\frac{x+9}{360x+150} = \frac{12x+5}{360x+150} \] b) $Bt=\frac{x+3y}{3x+y}.\frac{4x-2y}{x-y}-\frac{x+3y}{3x+y}.\frac{x-3y}{x-y}$ Ta có: \[ Bt = \frac{x+3y}{3x+y}.\left(\frac{4x-2y}{x-y}-\frac{x-3y}{x-y}\right) = \frac{x+3y}{3x+y}.\frac{4x-2y-x+3y}{x-y} = \frac{x+3y}{3x+y}.\frac{3x+y}{x-y} = \frac{x+3y}{x-y} \] B2: Cho biểu thức $A=(1-\frac{4}{x+1}):\frac{9-x^2}{x^2+2x+1}$ a) Tìm x để A có nghĩa Điều kiện: $x \neq -1$, $x \neq 3$, $x \neq -3$ b) Rút gọn A Ta có: \[ A = \left(\frac{x+1-4}{x+1}\right):\frac{(3-x)(3+x)}{(x+1)^2} = \frac{x-3}{x+1}.\frac{(x+1)^2}{(3-x)(3+x)} = \frac{x-3}{x+1}.\frac{(x+1)^2}{-(x-3)(x+3)} = \frac{x+1}{-(x+3)} = -\frac{x+1}{x+3} \] c) Tìm các giá trị nguyên của x để A nhận giá trị nguyên Ta có: \[ A = -\frac{x+1}{x+3} \] Để A nhận giá trị nguyên, $\frac{x+1}{x+3}$ phải là số nguyên. Điều này xảy ra khi $x+1$ chia hết cho $x+3$. Do đó, $x+1 = k(x+3)$ với $k$ là số nguyên. \[ x+1 = kx + 3k \implies x(1-k) = 3k-1 \] Các trường hợp: - $k = 0$: $x = -1$ (loại vì $x \neq -1$) - $k = 1$: $0 = 2$ (loại) - $k = -1$: $x = 2$ - $k = 2$: $x = -7$ - $k = -2$: $x = 7$ - $k = 3$: $x = -\frac{8}{2}$ (loại) Vậy các giá trị nguyên của x để A nhận giá trị nguyên là $x = 2$, $x = -7$, $x = 7$.