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\(\dfrac{x^2-4x+4}{x^2+2x+1}.\dfrac{x+1}{x^2-2x}.\dfrac{6x}{2x+4}\)
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19 tháng 12 2021
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
14 tháng 5 2022
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
29 tháng 7 2021
a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)
\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)
c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)
\(=\dfrac{-x+7x-4}{3x-2}\)
\(=\dfrac{6x-4}{3x-2}=2\)
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
15 tháng 11 2021
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
6 tháng 1 2021
a) Ta có: \(\dfrac{x}{x-1}-\dfrac{2}{x-1}\)
\(=\dfrac{x-2}{x-1}\)
b) Ta có: \(\dfrac{4+4x}{3x^2+6x}+\dfrac{x}{3x+6}\)
\(=\dfrac{4+4x}{x\left(3x+6\right)}+\dfrac{x^2}{x\left(3x+6\right)}\)
\(=\dfrac{x^2+4x+4}{3x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{3x\left(x+2\right)}\)
\(=\dfrac{x+2}{3x}\)
c) Ta có: \(\dfrac{x^2-2x}{x-1}\cdot\dfrac{1}{x}:\dfrac{x^2-4}{x^2-2x+1}\)
\(=\dfrac{x\left(x-2\right)}{x-1}\cdot\dfrac{1}{x}\cdot\dfrac{x^2-2x+1}{x^2-4}\)
\(=\dfrac{x-2}{x-1}\cdot\dfrac{\left(x-1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-1}{x+2}\)
19 tháng 8 2021
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2021
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
29 tháng 1 2024
a: \(x^2+2x+1+4x+4\)
\(=\left(x^2+2x+1\right)+\left(4x+4\right)\)
\(=\left(x+1\right)^2+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+1+4\right)\)
\(=\left(x+1\right)\left(x+5\right)\)
b: Sửa đề: \(2x^3+6x^2+x^2+3x\)
\(=2x^2\left(x+3\right)+x\left(x+3\right)\)
\(=\left(x+3\right)\left(2x^2+x\right)\)
\(=x\left(x+3\right)\left(2x+1\right)\)
c: \(\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+1\)
\(=\dfrac{1}{4}x\left(\dfrac{1}{4}x+1\right)+\left(\dfrac{1}{4}x+1\right)\)
\(=\left(\dfrac{1}{4}x+1\right)\left(\dfrac{1}{4}x+1\right)=\left(\dfrac{1}{4}x+1\right)^2\)
\(\dfrac{x^2-4x+4}{x^2+2x+1}.\dfrac{x+1}{x^2-2x}.\dfrac{6x}{2x+4}=\dfrac{\left(x-2\right)^2}{\left(x+1\right)^2}.\dfrac{x+1}{x\left(x-2\right)}.\dfrac{6x}{2\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{\left(x+1\right)\left(x+2\right)}\)