giup mk vs mn oi sap nop r
B1
a) `{x^3+2x}/{x^3+1}` + `{2x}/{x^2-x+1}`
b) `({3x}/{1-3x} + {2x}/{3x+1})` : `{6x^2+10x}/{1-6x+9x^2}`
B2 A = `{x^2-10x+25}/{x^2-2x}`
a) Tìm đkxđ
b) Tìm x để A = 2
c) Tìm x nguyên để A nguyên
Giúp với
giup mk vs mn oi sap nop r B1 a) `{x^3+2x}/{x^3+1}` + `{2x}/{x^2-x+1}` b) `({3x}/{1-3x} + {2x}/{3x+1})` : `{6x^2+10x}/{1-6x+9x^2}` B2 A = `{x^2-10x+
By Adalyn
Bài 1:
a,
$\frac{x^3+2x}{x^3+1}+\frac{2x}{x^2-x+1}$
$=\frac{x^3+2x+2x(x+1)}{(x+1)(x^2-x+1)}$
$=\frac{3x^3+4x}{(x+1)(x^2-x+1)}$
b,
$(\frac{-3x}{3x-1}+\frac{2x}{3x+1}):\frac{6x^2+10x}{1-6x+9x^2}$
$=\frac{-3x(3x+1)+2x(3x-1)}{(3x+1)(3x-1)} : \frac{2x(3x+5)}{(1-3x)^2}$
$= \frac{-3x^2-5x}{(3x+1)(3x-1)}.\frac{(3x-1)^2}{2x(3x+5)}$
$= \frac{-x(3x+5)}{(3x+1)(3x-1)}.\frac{(3x-1)^2}{2x(3x+5)}$
$=\frac{-(3x-1)}{2(3x+1)}$
Bài 2:
a,
ĐK: $x\neq 0; x\neq 2$
b,
$A=\frac{(x-5)^2}{x(x-2)}$
$A=2\Rightarrow (x-5)^2= 2x(x-2)$
$\Leftrightarrow x^2-10x+25=2x^2-4x$
$\Leftrightarrow x^2+6x-25=0$
$\Leftrightarrow x=-3\pm \sqrt{34}$
Đáp án:
Đề nó làm sao ấy!
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