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\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)
=>18x-6+18x-6=4-12x
=>36x-12=4-12x
=>48x=16
hay x=1/3
2: \(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)
=>(2x-1)(3x-4)=0
=>x=1/2 hoặc x=4/3
a)\(2x^3=x^2+2x-1\)
\(\Rightarrow2x^3-x^2-2x+1=0\)
\(\Rightarrow x^2\left(2x-1\right)-\left(2x-1\right)=0\)
\(\Rightarrow\left(x^2-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\pm1\\x=\frac{1}{2}\end{cases}}\)
b)\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-x\right)\)
\(\Rightarrow\left(3x-1\right)\left(x^2+2\right)-6x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(x^2+2-6x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\\Delta_{x^2-6x+2=0}=\left(-6\right)^2-4\cdot1\cdot2=28\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x_{2,3}=\frac{6\pm\sqrt{28}}{2}\end{cases}}\)
1, <=> 3x-1-2x=6 <=> x =7
2, (2x-1)(7-x)=x^2-7x
<=> 14x -2x^2-7+x=x^2-7x
<=> -3x^2+ 15x - 7 = -7x
<=> -3x^2 +23x - 7 =0
<=> \(x=\dfrac{23\pm\sqrt{445}}{6}\)
1.
\(\Leftrightarrow3x-1-2x=6\)
\(\Leftrightarrow x-1=6\)
\(\Leftrightarrow x=7\)
2.
\(\Leftrightarrow\left(2x-1\right)\left(7-x\right)+7x-x^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(7-x\right)+x\left(7-x\right)=0\)
\(\Leftrightarrow\left(7-x\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{1}{3}\end{matrix}\right.\)
a: =>x+3=x-2 hoặc x+3=2-x
=>2x=-1
=>x=-1/2
b: =>3x+7=x-2 hoặc 3x+7=-x+2
=>2x=-9 hoặc 4x=-5
=>x=-5/4 hoặc x=-9/2
c: =>|3x-4|=|2x-5|
=>3x-4=2x-5 hoặc 3x-4=-2x+5
=>x=-1 hoặc x=9/5
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
a) \(2x^3 + 6x^2 = x^2 +3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\)
\(\Leftrightarrow\left(x+3\right).x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
S = \(\left\{-3;0;\dfrac{1}{2}\right\}\)
b) \((3x-1) (x^2 +2 ) = (3x-1) (7x - 10)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
S = \(\left\{\dfrac{1}{3};3;4\right\}\)