giải phương trình
a) \(\frac{4x-17}{2x^2+5}=0\)
b) \(\frac{\left(x^2-2x\right)-\left(3x+6\right)}{x+2}=0\)
c) \(\frac{x^2-x-6}{x-3}=0\)
d) \(\frac{2x-5}{x+5}=3\)
e) \(\frac{5}{3x+2}=2x-1\)
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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!
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\(=\left\{0;-2\right\}\)
a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)
\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)
\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)
\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)
a)Ta có: \(\frac{4x-17}{2x^2+5}=0\)
\(\Leftrightarrow4x-17=0\)
\(\Leftrightarrow4x=17\)
\(\Leftrightarrow x=\frac{17}{4}\)
Vậy: \(x=\frac{17}{4}\)
b) ĐKXĐ: x≠-2
Ta có: \(\frac{\left(x^2-2x\right)-\left(3x+6\right)}{x+2}=0\)
\(\Leftrightarrow x^2-2x-3x-6=0\)
\(\Leftrightarrow x^2-5x-6=0\)
\(\Leftrightarrow x^2+x-6x-6=0\)
\(\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)(tm)
Vậy: x∈{-1;6}
c) ĐKXĐ: x≠3
Ta có: \(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-2\end{matrix}\right.\)
Vậy: x=-2
d) ĐKXĐ: x≠-5
Ta có: \(\frac{2x-5}{x+5}=3\)
⇔\(\frac{2x-5}{x+5}-3=0\)
⇔\(\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0\)
\(\Leftrightarrow2x-5-3\left(x+5\right)=0\)
\(\Leftrightarrow2x-5-3x-15=0\)
\(\Leftrightarrow-x-20=0\)
\(\Leftrightarrow-\left(x+20\right)=0\)
\(\Leftrightarrow x=-20\)(tm)
Vậy: x=-20