a)(3x-1)(4x-8)=0

⇔3x-1=0 hoặc 4x-8=0

1.3x-1=0⇔3x=1⇔x=1/3

2.4x-8=0⇔4x=8⇔x=2

phương trình có 2 nghiệm:x=1/3 và x=2

b)(x-2)(1-3x)=0

⇔x-2=0 hoặc 1-3x=0

1.x-2=0⇔x=2

2.1-3x=0⇔-3x=1⇔x=-1/3

phương trình có 2 nghiệm:x=2 và x=-1/3

c)(x-3)(x+4)-(x-3)(2x-1)=0

⇔(x+4)(2x-1)=0

⇔x+4=0 hoặc 2x-1=0

1.x+4=0⇔x=-4

2.2x-1=0⇔2x=1⇔x=1/2

phương trình có hai nghiệm:x=-4 và x=1/2

d)(x+1)(x+2)=2x(x+2)

⇔(x+1)(x+2)-2x(x+2)=0

⇔2x(x+1)=0

⇔2x=0 hoặc x+1=0

1.2x=0⇔x=0

2.x+1=0⇔x=-1

phương trình có 2 nghiệm:x=0 và x=-1

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)

d: \(\Leftrightarrow3x^2-6x-2x+4=0\)

=>(x-2)(3x-2)=0

=>x=2 hoặc x=2/3

e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)

=>x(x-3)(x+1)=0

hay \(x\in\left\{0;3;-1\right\}\)

f: \(\Leftrightarrow x^2-5x-2+x=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow\left(x-2\right)^2=6\)

hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)

a,\(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

\(\Rightarrow\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}\)

\(\Rightarrow\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}=\dfrac{x+3+63}{63}+\dfrac{x+4+62}{62}\)

\(\Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

\(\Rightarrow x+66=0\) ( vì \(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}>0\) )

\(\Rightarrow x=-66\)

\(b,\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)

\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)

\(\Rightarrow\dfrac{x-12-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)

\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)

\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Rightarrow x-89=0\)

\(\Rightarrow x=89\)

b.

\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\\ \Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}=\dfrac{x-89}{15}+\dfrac{x-89}{16}\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\\ \\ \Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\\ \Leftrightarrow x-89=0\\ \Leftrightarrow x=89\)

\(\frac{12}{x-1}-\frac{8}{x+1}=1\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow\frac{12\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{8\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\) \(\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow\left(12x+12\right)-\left(8x-8\right)=x^2-1\)

\(\Leftrightarrow12x+12-8x+8=x^2-1\)

\(\Leftrightarrow12x+12-8x+8-x^2+1=0\)

\(\Leftrightarrow-x^2+4x+21=0\)

\(\Leftrightarrow x^2-4x-21=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)-25=0\)

\(\Leftrightarrow\left(x-2\right)^2-5^2=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{7;-3\right\}\)

a thiếu

chỗ x phải có chữ thỏa mãn nữa nha

sorry sora cưng

\(\frac{x^3+7x^2+6x-30}{x^3-1}=\frac{x^2-x+16}{x^2+x+1}\) \(\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow\frac{x^3+7x^2+6x-30}{\left(x-1\right)\left(x^2+x+1\right)}=\) \(\frac{\left(x^2-x+16\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)

\(\Rightarrow x^3+7x^2+6x-30=x^3-2x^2+17x-16\)

\(\Leftrightarrow9x^2-11x-14=0\)

\(\Leftrightarrow\left(9x^2-11x+\frac{121}{36}\right)-\frac{625}{36}=0\)

\(\Leftrightarrow\left(3x-\frac{11}{6}\right)^2-\left(\frac{25}{6}\right)^2=0\)

\(\Leftrightarrow\left(3x-6\right)\left(3x+\frac{7}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=\frac{-7}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-\frac{7}{9}\left(tm\right)\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{2;\frac{-7}{9}\right\}\)