Bài 1:
a)
\(\frac{x}{3}+\frac{x}{8}-\frac{1}{2}=\frac{x+3}{8}+\frac{9}{24}\\ \Leftrightarrow\frac{8x}{24}+\frac{3x}{24}-\frac{12}{24}=\frac{3x+9}{24}+\frac{9}{24}\\ \Leftrightarrow8x+3x-12-3x-9-9=0\\ \Leftrightarrow8x-30=0\\ \Rightarrow x=\frac{30}{8}=\frac{15}{4}\)
b)
\(\frac{3x-2}{5}=\frac{4-7x}{3}\\ \Leftrightarrow\frac{9x-6}{15}-\frac{20-35x}{15}=0\\ \Leftrightarrow9x-6-20+35x=0\\ \Leftrightarrow44x-26=0\\ \Rightarrow x=\frac{26}{44}=\frac{13}{22}\)
c)
\(2x\cdot\left(x-5\right)+21=x\cdot\left(2x+1\right)-12\\ \Leftrightarrow2x^2-10x+21-2x^2-x+12=0\\ \Leftrightarrow33-11x=0\\ \Rightarrow x=3\)
d)
\(5-\left(x-6\right)=4\cdot\left(3-2x\right)\\ \Leftrightarrow5-x+6-12+8x=0\\ \Leftrightarrow7x-1=0\\ \Rightarrow x=\frac{1}{7}\)
e)
\(3-4x\cdot\left(25-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2-8x^2-x+300=0\\ \Leftrightarrow303-101x=0\\ \Rightarrow x=3\)
f)
\(\left(x-6\right)\cdot\left(2x-5\right)\cdot\left(3x^2+27\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\3x^2+27=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=\pm\sqrt{-9}\end{matrix}\right.\)
2x.(x-3)+5.(3-x)=0
\(\Leftrightarrow\)2x.(x-3)-5.(x-3)=0
\(\Leftrightarrow\)(2x-5).(x-3)=0
\(\Rightarrow\) 2x-5=0 \(\Rightarrow x=\frac{5}{2}\)
\(\Rightarrow\) x-3=0 \(\Rightarrow x=3\)
Bài làm
a) \(\frac{x}{3}+\frac{x}{8}-\frac{1}{2}=\frac{x+3}{8}+\frac{9}{24}\)
\(\Leftrightarrow\frac{8x}{24}+\frac{3x}{24}-\frac{12}{24}=\frac{3x+9}{24}+\frac{9}{24}\)
\(\Rightarrow8x+3x-12=3x+9+9\)
\(\Leftrightarrow8x=30\)
\(\Leftrightarrow x=3,75\)
Vậy x = 3,75 là nghiệm phương trình.
b) \(\frac{3x-2}{5}=\frac{4-7x}{3}\)
\(\Rightarrow3\left(3x-2\right)=5\left(4-7x\right)\)
\(\Leftrightarrow9x-6=20-35x\)
\(\Leftrightarrow42x=26\)
\(\Leftrightarrow x=\frac{13}{21}\)
Vậy x = 13/21 là nghiệm phương trình.
c) \(2x\left(x-5\right)+21=x\left(2x+1\right)-12\)
\(\Leftrightarrow2x^2-10x+21=2x^2+x-12\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3 là nghiệm ptrình.
d) \(5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\frac{1}{7}\)
Vậy x = 1/7 là nghiệm phương trình
e) \(3-4x\left(25-2x\right)=8x^2+x-300\)
\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Leftrightarrow x=3\)
Vậy x = 3 là nghiệm phương trình.
f) \(\left(x-6\right)\left(2x-5\right)\left(3x^2+27\right)=0\)
\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)
\(\Leftrightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\)
\(\Leftrightarrow3x^3+27=0\Leftrightarrow3x^3=-27\Leftrightarrow x^3=-9\Leftrightarrow\sqrt[3]{9}\)
Vậy S = { \(6;\frac{5}{2};\sqrt[3]{9}\)}
g) \(2x\left(x-3\right)+5\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)-5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy S = { 3; 5/2 }
h) \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy S = { 0; 1/3 }
i) \(\left(2x-1\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x-1-2x-1\right)\left(2x-1+2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}0x-2=0\\4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{0}\left(vo-li\right)\\x=0\end{matrix}\right.\)
Vậy x = 0 là nghiệm
k) \(\left(x^2+3x+2\right)=\left(x^2-x-2\right)^2\)
\(\Leftrightarrow\left(x^2+3x+2-x^2+x+2\right)\left(x^2+3x+2+x^2-x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=0\\2x^2+2x=0\Leftrightarrow2x\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-1\end{matrix}\right.\)
Vậy S = { -1 ; 0 }
h)
\(\left(x^2-4\right)-\left(x-2\right)\cdot\left(3-2x\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+2\right)-\left(x-2\right)\cdot\left(3-2x\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+2-3+2x\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
i)
\(\left(2x-1\right)^2=\left(2x+1\right)^2\\ \Leftrightarrow\left(2x-1\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow\left(2x-1+2x+1\right)\cdot\left(2x-1-2x-1\right)=0\\ \Leftrightarrow4x\cdot\left(-2\right)=0\\ \Rightarrow x=0\)
k)
\(\left(x^2+3x+2\right)^2=\left(x^2-x-2\right)^2\\ \Leftrightarrow\left(x^2+3x+2\right)^2-\left(x^2-x-2\right)^2=0\\ \Leftrightarrow\left(x^2+3x+2+x^2-x-2\right)\cdot\left(x^2+3x+2-x^2+x+2\right)=0\\ \Leftrightarrow\left(2x^2+2x\right)\cdot\left(4x+4\right)=0\\ \Leftrightarrow2x\cdot\left(x+1\right)\cdot4\cdot\left(x+1\right)=0\\ \Leftrightarrow8x\cdot\left(x+1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}8x=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Bài 2:
a)
\(x^2-6x+9=49\\ \Leftrightarrow x^2-6x+9-49=0\\ \Leftrightarrow x^2-6x-40=0\\ \Leftrightarrow x^2+4x-10x-40=0\\ \Leftrightarrow x\cdot\left(x+4\right)-10\cdot\left(x+4\right)=0\\ \Leftrightarrow\left(x-10\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-10=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
b)
\(x^3-2x^2-x+2=0\\ \Leftrightarrow x^2\cdot\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-1\right)\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x+1\right)\cdot\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
c)
\(x^3-3x^2-6x+8=0\\ \Leftrightarrow x^3-2x^2-8x-x^2+2x+8=0\\ \Leftrightarrow x^2\cdot\left(x-1\right)-2x\cdot\left(x-1\right)-8\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-2x-8\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+2x-4x-8\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left[x\cdot\left(x+2\right)-4\cdot\left(x+2\right)\right]\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x-4\right)\cdot\left(x+2\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-2\\x=1\end{matrix}\right.\)
d)
\(x^4+2x^3+5x^2+4x-12=0\\ \Leftrightarrow x^4+3x^3+8x^2+12x-x^3-3x^2-8x-12=0\\ \Leftrightarrow x^3\cdot\left(x-1\right)+3x^2\cdot\left(x-1\right)+8x\cdot\left(x-1\right)+12\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^3+3x^2+8x+12\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^3+x^2+6x+2x^2+2x+12\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left[x^2\cdot\left(x+2\right)+x\cdot\left(x+2\right)+6\cdot\left(x+2\right)\right]\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\cdot\left(x+2\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Bài 3:
a)
\(\left(x^2-5x\right)^2+10\cdot\left(x^2-5x\right)+24=0\\ \Leftrightarrow x\cdot\left(x-5\right)^2+10x\cdot\left(x-5\right)+24=0\\ \Leftrightarrow x^4-10x^3+35x^2-50x+24=0\\ \Leftrightarrow x^4-x^3-9x^3+9x^2+26x^2-26x-24x+24=0\\ \Leftrightarrow x^3\cdot\left(x-1\right)-9x^2\cdot\left(x-1\right)+26x\cdot\left(x-1\right)-24\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x^3-9x^2+26x-24\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x^3-2x^2-7x^2+14x+12x-24\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left[x^2\cdot\left(x-2\right)-7x\cdot\left(x-2\right)+12\cdot\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x^2-7x+12\right)=0\\ \)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x^2-3x-4x+12\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left[x\cdot\left(x-3\right)-4\cdot\left(x-3\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
b)
\(x\cdot\left(x+1\right)\cdot\left(x^2+x+1\right)=42\\ \Leftrightarrow x^4+2x^3+2x^2+x-42=0\\ \Leftrightarrow x^4-2x^3+4x^3-8x^2+10x^2-20x+21x-42=0\\ \Leftrightarrow x^3\cdot\left(x-2\right)+4x^2\cdot\left(x-2\right)+10x\cdot\left(x-2\right)+21\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x^3+4x^2+10x+21\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x^3+3x^2+x^2+3x+7x+21\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left[x^2\cdot\left(x+3\right)+x\cdot\left(x+3\right)+7\cdot\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+3\right)\cdot\left(x^2+x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c)
\(\left(5x^2+3x-2\right)^2-\left(4x^2-x-5\right)^2=0\\ \Leftrightarrow\left(5x^2+3x-2+4x^2-x-5\right)\cdot\left(5x^2+3x-2-4x^2+x+5\right)=0\\ \Leftrightarrow\left(9x^2+2x-7\right)\cdot\left(x^2+4x+3\right)=0\\ \Leftrightarrow\left(9x^2+9x-7x-7\right)\cdot\left(x^2+3x+x+3\right)=0\\ \Leftrightarrow\left[9x\cdot\left(x+1\right)-7\cdot\left(x+1\right)\right]\cdot\left[x\cdot\left(x+3\right)+\left(x+3\right)\right]=0\\ \Leftrightarrow\left(9x-7\right)\cdot\left(x+1\right)\cdot\left(x+3\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\\9x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-3\\x=\frac{7}{9}\end{matrix}\right.\)
