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5 tháng 3 2021
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
19 tháng 2 2021
\(ĐKXĐ:x\ne-2\)
Ta thấy x=0 ko là nghiệm của phương trình. Do đó \(x\ne0\)
\(\Rightarrow\dfrac{1}{\dfrac{x^2+4x+4}{x}}+\dfrac{5}{\dfrac{x^2+4}{x}}=-2\) (chia cả tử và mẫu của 2 phân số vế trái cho x )
\(\Leftrightarrow\dfrac{1}{x+\dfrac{4}{x}+4}+\dfrac{5}{x+\dfrac{4}{x}}=-2\)
Đặt \(x+\dfrac{4}{x}=t\) (\(t\ne0,t\ne-4\))
\(pt\) trở thành: \(\dfrac{1}{t+4}+\dfrac{5}{t}=-2\) \(\Rightarrow t+5\left(t+4\right)=-2\left(t+4\right)t\Leftrightarrow t+5t+20=-2t^2-8t\Leftrightarrow2t^2+14t+20=0\Leftrightarrow t^2+7t+10=0\) \(\Leftrightarrow\left(t+2\right)\left(t+5\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-2\left(1\right)\\t=-5\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow x+\dfrac{4}{x}=-2\Rightarrow x^2+4=-2x\Leftrightarrow x^2+2x+4=0\Leftrightarrow\left(x+1\right)^2+3=0\left(VL\right)\)
Từ (2) \(\Rightarrow x+\dfrac{4}{x}=-5\Rightarrow x^2+4=-5x\Leftrightarrow x^2+5x+4=0\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-4\left(TM\right)\end{matrix}\right.\) Vậy...
KT
4 tháng 5 2018
*\(\dfrac{x-1}{x+2}\)-\(\dfrac{x}{x+2}\)=\(\dfrac{5x-2}{4-x^2}\).ĐKXĐ: x\(\ne\pm2\)
<=>\(\dfrac{\left(x-1\right)\left(2-x\right)}{4-x^2}\)-\(\dfrac{x\left(2-x\right)}{4-x^2}\)=\(\dfrac{5x-2}{4-x^2}\)
=>2x-\(x^2\)-2+x-2x+\(x^2\)=5x-2
<=>x-2=5x-2
<=>x-5x=2-2
<=>-4x=0
<=> x = 0(TM)
Vậy phương trình có tập nghiệm là S={0}
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2021
ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow x=3\)
S
22 tháng 3 2021
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
22 tháng 3 2021
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
MH
6 tháng 1 2022
⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)
⇔ \(x^2+8x+12=32\)
⇔ \(x^2+8x-20=0\)
⇔ \(\left(x-2\right)\left(x+10\right)=0\)
⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
19 tháng 8 2021
1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(5x^2+3x-9=5x^2-5x\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(tm\right)\)
2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(6x=3x-15\)
\(\Leftrightarrow3x=-15\)
hay \(x=-5\left(loại\right)\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2021
2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)
Vậy pt vô nghiệm.
19 tháng 8 2021
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
19 tháng 8 2021
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2-3x+2-x^2-2x-2+5x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)