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c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)
\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)
Tương tự với a ; b
a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)
\(\Leftrightarrow3x+2x-10=6-5x+1\)
\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm
b, \(x^3-3x^2-x+3=0\)
\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)
Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }
c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3+x^2-3x-2=0\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn
Vậy ...
\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)
\(4x^2-40x+100=18x+9\)
\(4x^2-58x+91=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)
\(c,x^3+3x^2-6x-8=0\)
\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)
\(Th1:x+4=0\Leftrightarrow x=-4\)
\(Th2:x-2=0\Leftrightarrow x=2\)
\(Th3:x+1=0\Leftrightarrow x=-1\)
\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)
\(< =>4x^2-40x+100=18x+9\)
\(< =>4x^2+58x+91=0\)
\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)
a. (x + 3).(x2 - 1)
= x.x2 - x.1 + 3.x2 - 3.1
= x3 - x + 3x2 - 3
= x3 + 3x2 - x - 3
b. (3x + 2).(4x - 1)
= 3x.4x - 3x + 2.4x - 2
= 12x2 - 3x + 8x - 2
= 12x2 + 5x - 2
c. (2x - 3).(3x + 2)
= 2x.3x + 2x.2 - 3.3x - 3.2
= 6x2 + 4x - 9x - 6
= 6x2 - 5x - 6
d. (12x - 5).(4x + 1)
= 12x.4x + 12x - 5.4x - 5
= 48x2 + 12x - 20x - 5
= 48x2 - 8x - 5
e. (x - 3).(x2 + 3x + 9)
= x.x2 + x.3x + x.9 - 3x2 - 3.3x - 3.9
= x3 + 3x2 + 9x - 3x2 - 9x - 27
= x3 - 27 (Đây là dạng HĐT x3 - 33)
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
Đề:.........
<=> (3x - 9) - 2x(x - 3)
<=> 3(x - 3) - 2x(x - 3)
<=> (x - 3)(3 - 2x)