3: =>x(x+3)+3=x-3
=>x^2+3x+3=x-3
=>3x+3=x-3
=>2x=-6
=>x=-3(nhận)
4: Sửa đề: \(\dfrac{2}{x+2}-\dfrac{1}{x-2}=\dfrac{14}{\left(x+2\right)\left(x-2\right)}\)
=>2x-4-x-2=14
=>x-6=14
=>x=20(nhận)
Giải phương trình sau:
\(2x\left(8x+1\right)\left(8x^2-x+2\right)-126=0\)
Sửa đề: 8x-1
=>2(8x^2-x)(8x^2-x+2)-126=0
=>2[(8x^2-x)^2+2(8x^2-x)]-126=0
=>(8x^2-x)^2+2(8x^2-x)-63=0
=>(8x^2-x+9)(8x^2-x-7)=0
=>8x^2-x-7=0
=>x=1 hoặc x=-7/8
Giải phương trình sau:
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\left(x\ne-2;x\ne3\right)\\ < =>\dfrac{9x-2}{x^2-3x+2x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{x\left(x-3\right)+2\left(x-3\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{\left(x-3\right)\left(x+2\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
suy ra: \(9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(< =>9x-2+2x^2-6x-\left(x^2+2x-x-2\right)=x^2+2x-3x-6\)
\(< =>9x-2+2x^2-6x-x^2-2x+x+2=x^2-x-6\)
\(< =>2x^2-x^2-x^2+9x-6x-2x+x+x=6+2-2\)
\(< =>3x=6\\ < =>x=2\left(tm\right)\)
ĐKXĐ: \(x\ne\left\{-2;3\right\}\)
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
\(\Leftrightarrow\dfrac{9x-2}{\left(x+2\right)\left(x-3\right)}+\dfrac{2x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)
\(\Leftrightarrow9x-2+2x^2-6x-x^2-x+2=x^2-x-6\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\left(loại\right)\)
Vậy: PT vô nghiệm.
Giải các phương trình sau:
a) \(x^3-3x^2-4x=0\)
b) \(3x^2-5x-2=0\)
a) \(x^3-3x^2-4x=0\)
\(\Leftrightarrow x\left(x^2-3x-4\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{0;4;-1\right\}\).
b) \(3x^2-5x-2=0\)
\(\Leftrightarrow3x^2+x-6x-2=0\)
\(\Leftrightarrow x\left(3x+1\right)-2\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3};2\right\}\).
\(5,\left(3x-6\right)\left(7-10x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-6=0\\7-10x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{10}\end{matrix}\right.\)
\(6,\left(6x+7\right)\left(3x-2\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+7=0\\3x-2=0\\2x-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=\dfrac{2}{3}\\x=\dfrac{9}{2}\end{matrix}\right.\)
\(7,\left(3,4x-6,8\right)\left(15x+2,5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3,4x-6,8=0\\15x+2,5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(5)\left(3x-6\right)\left(7-10x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-6=0\\7-10x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\10x=7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{10}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2;\dfrac{7}{10}\right\}\)
\(6)\left(6x+7\right)\left(3x-2\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+7=0\\3x-2=0\\2x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-7\\3x=2\\2x=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{6}\\x=\dfrac{2}{3}\\x=\dfrac{9}{2}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-7}{6};\dfrac{2}{3};\dfrac{9}{2}\right\}\)
\(7)\left(3,4x-6,8\right)\left(15x+2,5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3,4x-6,8=0\\15x+2,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3,4x=6,8\\15x=-2,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{6}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2;\dfrac{1}{6}\right\}\)
\(8)\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(\text{nhận}\right)\\x^2=-1\left(\text{loại}\right)\end{matrix}\right.\)
\(\text{Vậy phương trình trên có tập nghiệm là }S=\left\{1\right\}\)
11: \(\Leftrightarrow12x+3-9x+5+4x-8=0\)
=>7x=0
=>x=0
12: \(\Leftrightarrow9x+27+3-9x=-2x+2\)
=>-2x+2=30
=>-2x=28
=>x=-14
13: \(\Leftrightarrow3\left(2x-1\right)-2\left(3-x\right)=-1\)
=>6x-3-6+2x=-1
=>8x-9=-1
=>8x=8
=>x=1
14: \(\Leftrightarrow3x+3-2x+5+2x-4=0\)
=>3x+4=0
=>x=-4/3
15: \(\Leftrightarrow18x-6x+8=3-6x\)
=>18x+8=3
=>x=-5/18
\(7,\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)-4\left(2x+3\right)-x-6}{12}=0\)
\(\Leftrightarrow3x-6-8x-12-x-6=0\)
\(\Leftrightarrow-6x=24\)
\(\Leftrightarrow x=-4\)
\(8,\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x+2\right)-2\left(5x\right)-4\left(1-x\right)}{12}=0\)
\(\Leftrightarrow3x+6-10x-4+4x=0\)
\(\Leftrightarrow-3x=-2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
\(9,\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow\dfrac{5\left(x-3\right)-x-1-2\left(x-2\right)}{10}=0\)
\(\Leftrightarrow5x-15-x-1-2x+4=0\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(10,\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)+15\left(3x-2\right)-12\left(4x-3\right)}{60}=0\)
\(\Leftrightarrow40x-20+45x-30-48x+36=0\)
\(\Leftrightarrow37x=14\)
\(\Leftrightarrow x=\dfrac{14}{37}\)
\(7,\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\\ \Rightarrow3.\left(x-2\right)-4.\left(2x+3\right)=x+6\\ \Rightarrow3x-6-8x-12-x-6=0\\ \Rightarrow-6x-24=0\\ \Rightarrow x=-4\)
\(8,\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\\ \Rightarrow3.\left(x+2\right)-5x.2=4\left(1-x\right)\\ \Rightarrow3x+6-10x-4+4x=0\\ \Rightarrow-3x+2=0\\ \Rightarrow x=\dfrac{2}{3}\)
\(9,\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\\ \Rightarrow5.\left(x-3\right)-\left(x+1\right)=2.\left(x-2\right)\\ \Rightarrow5x-15-x-1-2x+4=0\\ \Rightarrow2x-12=0\\ \Rightarrow x=6\)
\(10,\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\\ \Rightarrow20.\left(2x-1\right)+15.\left(3x-2\right)=12.\left(4x-3\right)\\\Rightarrow40x-20+45x-30-48x+36=0\\ \Rightarrow37x-14=0\\ \Rightarrow x=\dfrac{14}{37}\)
\(a,\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\\ \Rightarrow3x^2-x+9x-3=10-5x-6x+3x^2\\ \Rightarrow8x-3=10-11x\\ \Rightarrow8x+11x=10+3\\ \Rightarrow19x=13\\ \Rightarrow x=\dfrac{13}{19}\\ ------------\\ b,\left(x+5\right)\left(2x-1\right)=\left(2x+3\right)\left(x+1\right)\\ \Rightarrow2x^2+10x-x-5=2x^2+3x+2x+3\\ \Rightarrow9x-5=5x+3\\ \Rightarrow9x-5x=3+5\\ \Rightarrow4x=8\\ \Rightarrow x=8:4\\ \Rightarrow x=2\)
\(---------\\ c,\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\\ \Rightarrow x^2+x+9x+9=x^2+3x+5x+15\\ \Rightarrow10x+9=8x+15\\ \Rightarrow10x-8x=15-9\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
\(-----------\\ d,\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\\ \Rightarrow6x^2+10x+3x+5=6x^2-2x-18x+6\\ \Rightarrow13x+5=-20x+6\\ \Rightarrow13x+20x=6-5\\ \Rightarrow33x=1\\ \Rightarrow x=\dfrac{1}{33}\)
a: =>(3x-1)(x+3)=(3x-5)(x-2)
=>3x^2+9x-x-3=3x^2-6x-5x+10
=>8x-3=-11x+10
=>19x=13
=>x=13/19
b: \(\Leftrightarrow2x^2-x+10x-5=2x^2+2x-3x-3\)
=>9x-5=-x-3
=>10x=2
=>x=1/5
c: \(\Leftrightarrow x^2+10x+9=x^2+8x+15\)
=>2x=6
=>x=3
d: \(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
=>13x+5=-20x+6
=>33x=1
=>x=1/33
e: =>x^2+4x+4+2x-8=x^2-6x+8
=>6x-4=-6x+8
=>12x=12
=>x=1
f: =>2x^2-3x+2x-3-3x+6=2x^2-4x+2
=>-4x+3=-4x+2
=>3=2(loại)
1.21:
a: \(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
=>15x=30
=>x=2
b: \(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
=>83x-2=81
=>x=1
c: \(\Leftrightarrow5\left(x-3\right)^2+2\left(x-6\right)^2=3\left(x+9\right)^2-10\left(13x-1\right)\)
=>5x^2-30x+45+2x^2-24x+72=3x^2+54x+243-130x+10
=>7x^2-54x+117=3x^2-76x+253
=>4x^2+22x-136=0
=>\(x=\dfrac{-11\pm\sqrt{665}}{4}\)
chứng minh rằng : trong hình thang có 2 đáy không bằng nhau, đoạn thẳng nối trung điểm 2 đường chéo bằng nửa hiệu độ dài 2 đáy