(3x-2).(x+1)=18
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TN
0
8 tháng 12 2021
i: \(=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)
8 tháng 12 2021
Bài 1:
\(i,\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}-\dfrac{x+2}{5-x}=\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
\(j,\dfrac{3x\left(x-2\right)}{3x-2}+\dfrac{6x^2}{3x-2}-\dfrac{2\left(2-3x\right)}{2-3x}=\dfrac{3x^2-6x}{3x-2}+\dfrac{6x^2}{3x-2}+\dfrac{4-6x}{3x-2}=\dfrac{3x^2-6x+6x^2+4-6x}{3x-2}=\dfrac{9x^2-12x+4}{3x-2}=\dfrac{\left(3x-2\right)^2}{3x-2}=3x-2\)
\(n,\dfrac{2}{x}+\dfrac{3}{x-1}+\dfrac{1-4x}{x^2-x}=\dfrac{2\left(x-1\right)+3x+1-4x}{x\left(x-1\right)}=\dfrac{2x-2+3x+1-4x}{x\left(x-1\right)}=\dfrac{x-1}{x\left(x-1\right)}=\dfrac{1}{x}\)
Bài 2:
\(j,\dfrac{2}{3x}-\dfrac{1}{2x-2}-\dfrac{x-4}{6x-6x^2}=\dfrac{4\left(x-1\right)}{6x\left(x-1\right)}-\dfrac{3x}{6x\left(x-1\right)}-\dfrac{x-4}{6x\left(1-x\right)}=\dfrac{4x-4-3x+x-4}{6x\left(x-1\right)}=\dfrac{2x-8}{6x\left(x-1\right)}=\dfrac{2\left(x-4\right)}{6x\left(x-1\right)}=\dfrac{x-4}{3x\left(x-1\right)}\)
16 tháng 9 2021
1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)
2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)
\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)
3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)
\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)
NT
3
10 tháng 7 2018
a) \(\left|18-2x\right|=18\Rightarrow\orbr{\begin{cases}18-2x=18\\18-2x=-18\end{cases}\Rightarrow\orbr{\begin{cases}2x=0\\2x=36\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=18\end{cases}}}\)
b) \(\left|1-x\right|=1-x\Rightarrow\orbr{\begin{cases}1-x=1-x\\1-x=-\left(1-x\right)=-1+x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left\{\mp1;\mp2;...\right\}\\-x-x=-1-1\Rightarrow x=1\end{cases}}\)
Làm từng nãy đã , mỏi tay
TT
10 tháng 7 2018
a) ta có \(|\)18 - 2x\(|\)=18
=> 18-2x = 18 hoặc 18 -2x=-18
-2x = 0 -2x = -36
x = 0 x = 18
vậy x = 0 hoặc x=18
4 tháng 3 2022
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
4 tháng 3 2022
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
HN
22 tháng 8 2021
\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)
\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=n\)
\(=\left(n+3\right)n-4\)
\(=n^2+3n-4\)
\(=n^2-n+4n-4\)
\(=n\left(n-1\right)+4\left(n-1\right)\)
\(=\left(n-1\right)\left(n+4\right)\)
\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)
\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
HN
22 tháng 8 2021
\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)
\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)
Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):
\(\left(12n+1\right)n=6\)
\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)
NN
0
15 tháng 8 2020
Ta có: \(6x\left(3x+5\right)-2x\left(3x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)=0\)
\(\Leftrightarrow18x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0\)
\(\Leftrightarrow12x^2-34x-17=0\)
\(\Leftrightarrow12\left(x^2-\frac{34}{12}x-\frac{17}{12}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{17}{12}+\frac{289}{144}-\frac{493}{144}=0\)
\(\Leftrightarrow\left(x-\frac{17}{12}\right)^2=\frac{493}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{17}{12}=\frac{\sqrt{493}}{12}\\x-\frac{17}{12}=-\frac{\sqrt{493}}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17+\sqrt{493}}{12}\\x=\frac{17-\sqrt{493}}{12}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{17+\sqrt{493}}{12};\frac{17-\sqrt{493}}{12}\right\}\)