giải pt bậc nhất một ẩn
\(\dfrac{2x+1}{6}\)-\(\dfrac{x-2}{4}\)=\(\dfrac{3-2x}{3}\)-x
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HH
2
25 tháng 1 2022
\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{6\left(2x+1\right)}{12}=\dfrac{2x}{12}-\dfrac{3x}{12}\)
\(\Leftrightarrow4x-6\left(2x+1\right)=2x-3x\)
\(\Leftrightarrow4x-12x-6=-x\)
\(\Leftrightarrow4x-12x-6+x=0\)
\(\Leftrightarrow-7x-6=0\)
\(\Leftrightarrow x=-\dfrac{6}{7}\)
16 tháng 2 2021
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{3x}+\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)
Suy ra: \(3x=10\)
\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)
b) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)
Suy ra: \(3x-4x=8\)
\(\Leftrightarrow-x=8\)
hay x=-8(thỏa ĐK)
Vậy: S={-8}
c)ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)
\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)
Suy ra: 2x+3x=10
\(\Leftrightarrow5x=10\)
hay x=2(thỏa ĐK)
Vậy: S={2}
16 tháng 2 2021
d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)
\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)
\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)
\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)
\(\Leftrightarrow\) x = a (TM)
Vậy S = {a}
Chúc bn học tốt!
22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
NL
2
DM
18 tháng 1 2018
Đặt \(u=x^2-2x+2\)
=> Pt tương đương :
\(\dfrac{1}{u}+\dfrac{2}{u+1}=\dfrac{6}{u+2}\)
\(\Leftrightarrow\dfrac{\left(u+1\right)\left(u+2\right)+2u\cdot\left(u+2\right)}{u\left(u+1\right)\left(u+2\right)}=\dfrac{6u\left(u+1\right)}{u\left(u+1\right)\left(u+2\right)}\)
\(\Leftrightarrow\left(u+1\right)\left(u+2\right)+2u\left(u+2\right)=6u\left(u+1\right)\)
\(\Leftrightarrow u^2+3u+2+2u^2+4u=6u^2+6u\)
\(\Leftrightarrow-3u^2+u+2=0\)
\(\Rightarrow\left[{}\begin{matrix}u=1\\u=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2=1\\x^2-2x+2=-\dfrac{2}{3}\end{matrix}\right.\Rightarrow x=1\)
Kết luận \(x=1\)
W
18 tháng 1 2018
\(pt\Leftrightarrow\dfrac{1}{\left(x-1\right)^2+1}+\dfrac{2}{\left(x-1\right)^2+2}=\dfrac{6}{\left(x-1\right)^2+3}\)
Đặt: \(\left(x-1\right)^2=t\ge0\)
\(pt\Leftrightarrow\dfrac{1}{t+1}+\dfrac{2}{t+2}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{t+2+2\left(t+1\right)}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{t+2+2t+2}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{3t+4}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\left(3t+4\right)\left(t+3\right)=6\left(t+1\right)\left(t+2\right)\)
Phân tích ra:v
Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8+20x-12=0\)
\(\Leftrightarrow x=\dfrac{4}{21}\)