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a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
\((3x-2)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(\Leftrightarrow3x-2=0\) hoặc \(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\)
- \(3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\) ;
- \(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\Leftrightarrow\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\Leftrightarrow10\left(x+3\right)=7\left(4x-3\right)\Leftrightarrow x=\frac{17}{6}\).
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{2}{3};\frac{7}{16}\right\}\).
\(\left(3x-2\right)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=2\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\10\left(x+3\right)=7\left(4x-3\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
vậy x=2/3 hoặc x=17/6
\(ĐK:x\ne\frac{-1}{3}\)
\(PT\Leftrightarrow\left(\frac{4x-3}{3x+1}+2\right)\left(x^2+3x+1-4x-7\right)=0\)
\(\Leftrightarrow\left(\frac{10x-1}{3x+1}\right).\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\)\(x=\frac{1}{10}\)hoặc x=3 hoặc x=-2
Vậy...........
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
\(b,\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne2;\ne-3\)
\(\Leftrightarrow\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x^2-9=5\)
\(\Leftrightarrow x^2=14\)
\(x=\sqrt{14}\)
.....
a) \(\left(x+3\right)^2-\left(x-3\right)^2=6x\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=6x\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x\Leftrightarrow12x=6x\)\(\Leftrightarrow12x-6x=0\Leftrightarrow6x=0\Leftrightarrow x=0\)
Vậy phương trình có tập nghiệm S = { 0 }
b)\(-ĐKXĐ:\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-3\end{cases}}\)
- Ta có : \(\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\Leftrightarrow\frac{x-3}{x-2}-\frac{5}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-5}{\left(x-2\right)\left(x+3\right)}=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(thoaman\right)\\x=-3\left(kothoaman\right)\end{cases}}\)
Vậy phương trình có tập nghiệm S = { 3 }
a) ta có :x2+2x+2=(x+1)2+1>0,với mọi x
x2+2x+3=(x+1)2+2>0,với mọi x
ĐKXĐ:x\(\in\)R.Đặt x2+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
=>6(a2-1)+6a2=7a2+7a<=>6a2-6+6a2=7a2+7a<=>12a2-7a2-7a-6=0
<=>5a2-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)
<=>a=2=>x2+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)
Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)
\(\left(3x-2\right)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}\)
Giải \(\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\)
\(\Leftrightarrow5.2\left(x+3\right)=7\left(4x-3\right)\)
\(\Leftrightarrow10x+30=28x-21\)
\(\Leftrightarrow10x-28x=-21-30\)
\(\Leftrightarrow-18x=-51\)
\(\Leftrightarrow x=\frac{17}{6}\)
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