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28 tháng 6 2020
a) Ta có: 3(x+1)-(4x-3)=5
\(\Leftrightarrow3x+3-4x+3-5=0\)
\(\Leftrightarrow-x+1=0\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(2x+1\right)\left(x-1\right)=x^2-2x+1\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1-x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: S={1;-2}
c) Ta có: |-5x|=3x-16(*)
Trường hợp 1: \(-5x\ge0\Leftrightarrow x\le0\)
(*)\(\Leftrightarrow-5x=3x-16\)
\(\Leftrightarrow-5x-3x=-16\)
\(\Leftrightarrow-8x=-16\)
hay x=2(loại)
Trường hợp 2: \(-5x< 0\Leftrightarrow x>0\)
(*)\(\Leftrightarrow5x=3x-16\)
\(\Leftrightarrow5x-3x=-16\)
\(\Leftrightarrow2x=-16\)
hay x=-8(loại)
Vậy: \(S=\varnothing\)
d) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-2}{x-2}-\frac{2}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{5x+2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{x^2-2x-8+5x+2}{\left(x-2\right)\left(x+2\right)}=0\)
Suy ra: \(x^2+3x-6=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{33}{4}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{33}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{2}=\frac{\sqrt{33}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{33}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{33}-3}{2}\left(tm\right)\\x=\frac{-\sqrt{33}-3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{\sqrt{33}-3}{2};\frac{-\sqrt{33}-3}{2}\right\}\)
LH
1
MV
7 tháng 1 2018
\(x^2-1=\left(x-1\right)\left(2x-3\right)\\ \Leftrightarrow x^2-1-\left(x-1\right)\left(2x-3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(2x-3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1-2x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
1 tháng 9 2017
Ta có : (2x - 1)2 - 25 = 0
=> (2x - 1)2 = 25
=> \(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
KT
26 tháng 4 2018
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
10 tháng 2 2018
a) \(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
\(3x^2+10x-8=5x^2-2x+10\)
\(3x^2-5x^2+10x+2x-8-10=0\)
\(-2x^2+12x-18=0\)
\(x^2-6x+9=0\)
\(\left(x-3\right)^2=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
b) \(\frac{x^2-x-6}{x-3}=0\)
\(\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-6=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{2}-\frac{5}{2}\right)\left(x-\frac{1}{2}+\frac{5}{2}\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
28 tháng 5 2020
a) \(3\left(x-4\right)+5=2\left(x+1\right)-8\)
\(\Leftrightarrow3x-12+5=2x+2-8\)
\(\Leftrightarrow x=1\)
Vậy : \(S=\left\{1\right\}\)
b) \(5\left(x+1\right)^2+2x=5x^2-3\)
\(\Leftrightarrow5x^2+10x+5+2x=5x^2-3\)
\(\Leftrightarrow12x=-8\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy : \(S=\left\{-\frac{2}{3}\right\}\)
c) \(\frac{4\left(x+2\right)}{15}=\frac{13x-9}{40}\)
\(\Leftrightarrow32\left(x+2\right)=3\left(13x-9\right)\)
\(\Leftrightarrow32x-39x=-27-64\)
\(\Leftrightarrow-7x=-91\)
\(\Leftrightarrow x=13\)
Vậy : \(S=\left\{13\right\}\)
