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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)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
X\(\Leftrightarrow-96x^2+505x+396=0\)ét vế trái : \(\frac{\left(4x+7\right)^2}{7}-\frac{\left(5x-1\right)^2}{7}=\frac{\left(4x+7-5x+1\right)\left(4x+7+5x-1\right)}{7}=\frac{\left(8-x\right)\left(9x+6\right)}{7}\)
pt <=> \(\frac{\left(8-x\right)\left(9x+6\right)}{7}=\frac{\left(8x-3\right)\left(3x+4\right)}{56}\)
\(\Leftrightarrow24\left(8-x\right)\left(3x+2\right)-\left(8x-3\right)\left(3x+4\right)=0\)
\(\Leftrightarrow-96x^2+505x+396=0\)
Giải phương trình trên được : \(x_1=\frac{505}{192}-\frac{\sqrt{407089}}{192}\); \(x_2=\frac{505}{192}+\frac{\sqrt{407089}}{192}\)
Vậy ..............................
\((3x-2)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(\Leftrightarrow3x-2=0\) 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
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!
Bài 1:
a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)
\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)
\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)
Suy ra: \(12x-45-12x^2+45x=0\)
\(\Leftrightarrow-12x^2+57x-45=0\)
\(\Leftrightarrow-12x^2+12x+45x-45=0\)
\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)
\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)
mà \(-3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)
b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)
Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)
\(\Leftrightarrow-x^2+16x-39=0\)
\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)
\(\Leftrightarrow x^2-13x-3x+39=0\)
\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)
Vậy: Tập nghiệm S={3;13}
c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)
\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)
\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)
Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)
\(\Leftrightarrow-21x^2+26x+11=0\)
\(\Leftrightarrow-21x^2-7x+33x+11=0\)
\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
\(\left(3x-2\right)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}\)
Giải \(\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\)
\(\Leftrightarrow5.2\left(x+3\right)=7\left(4x-3\right)\)
\(\Leftrightarrow10x+30=28x-21\)
\(\Leftrightarrow10x-28x=-21-30\)
\(\Leftrightarrow-18x=-51\)
\(\Leftrightarrow x=\frac{17}{6}\)
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)
\(=>\frac{2}{x+1}=\frac{1}{x+1}\)
\(=>2x+2=x+1\)
\(=>2x-x=1-2=-1\)
\(=>x=-1\)
vậy nghiệm của phương trình trên là {-1}
À quên ĐKXĐ của câu a là \(x\ne-1\)
Nên \(x\in\varnothing\)nhé :v
\(ĐK:x\ne\frac{-1}{3}\)
\(PT\Leftrightarrow\left(\frac{4x-3}{3x+1}+2\right)\left(x^2+3x+1-4x-7\right)=0\)
\(\Leftrightarrow\left(\frac{10x-1}{3x+1}\right).\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\)\(x=\frac{1}{10}\)hoặc x=3 hoặc x=-2
Vậy...........