Bài 1:
a) (3x + 2)2 - (3x - 2)2 = 5x + 38
<=> (3x + 2 + 3x - 2)(3x + 2 - 3x + 2) - 5x - 38 = 0
<=> 24x - 5x - 38 = 0
<=> 19x - 38 = 0
<=> 19x = 38
<=> x = 2
Vậy S ={2}
b) 3(x - 2)2 + 9(x - 1) = 3(x2 + x - 3)
<=> 3(x2 - 4x - 4) + 9x - 9 - 3x2 - 3x + 9 = 0
<=> 3x2 - 12x - 12 + 9x - 9 - 3x2 - 3x + 9 = 0
<=> -6x - 12 = 0
<=> 6x = -12
<=> x = -2
Vậy S ={-2}
c) (x + 3)2 - (x - 3)2 = 6x + 18
<=> (x + 3 + x - 3)(x + 3 - x + 3) - 6x - 18 = 0
<=> 12x - 6x - 18 = 0
<=> 6x - 18 = 0
<=> 6x = 18
<=> x = 3
Vậy S ={3}
d) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x - 10x + 5x2 + 11x + 22 = 0
<=> 3x + 21 = 0
<=> 3x = -21
<=> x = -7
Vậy S ={-7}
e) (x + 1)(x2 - x + 1) - 2x = x(x - 1)(x + 1)
<=> x3 + 1 - 2x - x3 + x = 0
<=> 1 - x = 0
<=> x = 1
Vậy S ={1}
f) (x - 2)3 + (3x - 1)(3x + 1) = (x + 1)3
<=> x3 - 6x2 + 12x - 8 + 9x2 - 1 - x3 - 3x2 - 3x - 1 = 0
<=> 9x - 10 = 0
<=> 9x = 10
<=> x = \(\frac{10}{9}\)
Vậy S = {\(\frac{10}{9}\)}
g) (2x - 3)(x + 1) = 4x2 - 9
<=> (2x - 3)(x + 1) - (2x - 3)(2x + 3) = 0
<=> (2x - 3)(x + 1 - 2x - 3) = 0
<=> (2x - 3)(-x - 2) = 0
<=> \(\left[{}\begin{matrix}2x-3=0\\-x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=-2\end{matrix}\right.\)
Vậy S ={\(\frac{3}{2}\); -2}
h) (x2 - 7x).\(\frac{2x-5}{7}=0\)
<=> \(\frac{x\left(x-7\right)\left(2x-5\right)}{7}=0\)
<=> x(x - 7)(2x - 5) = 0
<=> \(\left[{}\begin{matrix}x=0\\x-7=0\\2x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=7\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy S ={0; 7; \(\frac{5}{2}\)}
Bài 2:
a) \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}-\frac{10x}{30}+\frac{15\left(x-2\right)}{30}=0\)
\(\Leftrightarrow6x+24-30x+120-10x+15x-30=0\)
\(\Leftrightarrow114-19x=0\)
\(\Leftrightarrow19x=114\)
\(\Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
b) \(x+\frac{5}{2}=\frac{4x+3}{4}-\frac{2-3x}{8}\)
\(\Leftrightarrow\frac{8x}{8}+\frac{20}{8}-\frac{2\left(4x+3\right)}{8}+\frac{2-3x}{8}=0\)
\(\Leftrightarrow8x+20-8x-6+2-3x=0\)
\(\Leftrightarrow16-3x=0\)
\(\Leftrightarrow3x=16\)
\(\Leftrightarrow x=\frac{16}{3}\)
Vậy \(S=\left\{\frac{16}{3}\right\}\)
c) \(1-\frac{x-1}{3}=\frac{4x+3}{6}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{6}{6}-\frac{2\left(x-1\right)}{6}-\frac{4x+3}{6}+\frac{3\left(x-2\right)}{6}=0\)
\(\Leftrightarrow6-2x+2-4x-3+3x-6=0\)
\(\Leftrightarrow-3x-1=0\)
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy \(S=\left\{-\frac{1}{3}\right\}\)
d) \(\frac{3x-5}{4}-\frac{5}{3}=\frac{6x+7}{6}+\frac{x}{2}\)
\(\Leftrightarrow\frac{3\left(3x-5\right)}{12}-\frac{20}{12}-\frac{2\left(6x+7\right)}{12}-\frac{6x}{12}=0\)
\(\Leftrightarrow9x-15-20-12x-14-6x=0\)
\(\Leftrightarrow-9x-49=0\)
\(\Leftrightarrow9x=-49\)
\(\Leftrightarrow x=-\frac{49}{9}\)
Vậy \(S=\left\{-\frac{49}{9}\right\}\)
e) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=x+\frac{7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)}{12}-\frac{2\left(5x+3\right)}{12}+\frac{4\left(x+1\right)}{12}-\frac{12x}{12}-\frac{7}{12}=0\)
\(\Leftrightarrow18x+9-10x-6+4x+4-12x-7=0\)
\(\Leftrightarrow0x=0\)
Vậy phương trình trên vô số nghiệm
#Giờ có việc bận, khi nào on đc, làm tiếp sau nhé! :)
Bài 2:
f)
\(\frac{\left(x+2\right)^2}{8}-2\cdot\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\\ \Leftrightarrow\frac{x^2+4x+4}{8}-\frac{32x+16}{8}=\frac{200}{8}+\frac{x^2-4x+4}{8}\\ \Leftrightarrow x^2+4x+4-32x-16-200-x^2+4x-4=0\\ \Leftrightarrow-24x-216=0\\ \Rightarrow x=-9\)
g)
\(\frac{\left(2x-3\right)\cdot\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\frac{12x^2-27}{24}=\frac{4x^2-32x+64}{24}+\frac{8x^2-32x+32}{24}\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27-4x^2+32x-64-8x^2+32x-32=0\\ \Leftrightarrow-123+64x=0\\ \Rightarrow x=\frac{123}{64}\)
h)
\(\frac{\left(7x+1\right)\cdot\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\cdot\left(x-3\right)}{2}\\ \Leftrightarrow\frac{7x^2-13x-2}{10}+\frac{4}{10}=\frac{2x^2-8x+8}{10}+\frac{5x^2-20x+15}{10}\\ \Leftrightarrow7x^2-13x-2+4=2x^2-8x+8+5x^2-20x+15\\ \Leftrightarrow7x^2-13x+2=7x^2-28x+23\\ \Leftrightarrow7x^2-13x+2-7x^2+28x-23=0\\ \Leftrightarrow15x-21=0\\ \Rightarrow x=\frac{21}{15}=\frac{7}{5}\)
Bài 3:
a)
\(\left(x-2\right)\cdot\left(3x+5\right)=\left(2x-4\right)\cdot\left(x+1\right)\\ \Leftrightarrow3x^2-x-10=2x^2-2x-4\\ \Leftrightarrow3x^2-x-10-2x^2+2x+4=0\\ \Leftrightarrow x^2+x-6=0\\ \Leftrightarrow x^2+3x-2x-6=0\\ \Leftrightarrow x\cdot\left(x+3\right)-2\cdot\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
b)
\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\cdot\left(x+3\right)\\ \Leftrightarrow2x^2-2x=3-2x-x^2\\ \Leftrightarrow2x^2-2x-3+2x+x^2=0\\ \Leftrightarrow3x^2-3=0\\ \Leftrightarrow3\cdot\left(x^2-1\right)=0\\ \Leftrightarrow3\cdot\left(x-1\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c)
\(x^3+2x^2-2x-12=0\\ \Leftrightarrow x^3+4x^2+6x-2x^2-8x-12=0\\ \Leftrightarrow x^2\cdot\left(x-2\right)+4x\cdot\left(x-2\right)+6\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x^2+4x+6\right)=0\\ \Rightarrow x-2=0\\ \Rightarrow x=2\)
d)
\(\left(x^2+3x+2\right)\cdot\left(x^2+5x+6\right)=72\\ \Leftrightarrow x^4+8x^3+23x^2+28x-60=0\\ \Leftrightarrow\left(x^3+9x^2+32x+60\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+4x+12\right)\cdot\left(x+5\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
e)
\(\left(x^2-1\right)\cdot\left(x^2+4x+3\right)=45\\ \Leftrightarrow x^4+4x^3+2x^2-4x-48=0\\ \Leftrightarrow\left(x^3+6x^2+14x+24\right)\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+2x+6\right)\cdot\left(x+4\right)\cdot\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
