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ĐKXĐ : \(x\ne\left\{2;3;4;5;6\right\}\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right).\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right).\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{loại}\right)\\x=-10\end{matrix}\right.\Leftrightarrow x=-10\)
Vậy tập nghiệm phương trình S = {10}
a) x - 5 = 7 - x
<=> 2x = 12
<=> x = 6
Vậy tập nghiệm phương trình S = {6}
b) 3x - 15 = 2x(x - 5)
<=> 3(x - 5) = 2x(x - 5)
<=> (2x - 3)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\)
Tập nghiệm phương trình \(S=\left\{\dfrac{3}{2};5\right\}\)
a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>3x+5<10x-30
=>-7x<-35
hay x>5
b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
hay x>16/5
Mấy này bạn quy đồng lên cùng mẫu xong khử mẫu rồi giải. Dễ mà.
a)
\(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\\ < =>3x-9+5-10x=90\)
\(< =>3x-10x=90+9-5\\ < =>-7x=94\\ < =>x=-\dfrac{94}{7}\)
b)
\(\left(2x-3\right)\left(x^2+1\right)=0\\ < =>\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-1\left(voli\right)\end{matrix}\right.\\ < =>x=\dfrac{3}{2}\)
c)
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(x\ne-1;x\ne2\right)\)
suy ra: \(2\left(x-2\right)-x-1=3x-11\)
\(< =>2x-4-x-1-3x+11=0\)
\(< =>2x-x-3x=4+1-11\\ < =>-2x=-6\\ < =>x=3\left(tm\right)\)
a) \(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\)
\(\Leftrightarrow3\left(x-3\right)+5\left(1-2x\right)=90\)
\(\Leftrightarrow-4-7x=90\)
\(\Leftrightarrow x=-\dfrac{94}{7}\)
b) \(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-3=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow x=\dfrac{3}{2}\)
c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(Đk:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow x-5=3x-11\)
\(\Leftrightarrow x=3\)
\(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}=2\)
<=>\(\dfrac{1}{x^2+2x+x+2}+\dfrac{1}{x^2+2x+3x+6}=2\)
<=>\(\dfrac{1}{x\left(x+2\right)+\left(x+2\right)}+\dfrac{1}{x\left(x+2\right)+3\left(x+2\right)}=2\)
<=>\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=2\)
ADCT:\(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) ta đc:
\(\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=2\)
<=>\(\dfrac{1}{x+1}-\dfrac{1}{x+3}=2\)
<=>\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}=2\)
<=>\(x^2+4x+3=4< =>x^2+4x-1=0< =>x^2+4x+4-3=0< =>\left(x+2\right)^2=3< =>x+2=\sqrt{3}< =>\left[{}\begin{matrix}x=-2-\sqrt{3}\\x=\sqrt{3}-2\end{matrix}\right.\)
số k đẹp nhỉ :) , k biết tớ có sai ở đâu k, nếu sai thì xin lỗi nha
chúc bạn học tốt ^ ^
\(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}=2\)
ĐKXĐ:\(x\ne-1;x\ne-2;x\ne-3\)
\(\Leftrightarrow\dfrac{1}{x^2+x+2x+2}+\dfrac{1}{x^2+3x+2x+6}=2\)
\(\Leftrightarrow\dfrac{1}{x\left(x+1\right)+2\left(x+1\right)}+\dfrac{1}{x\left(x+3\right)+2\left(x+3\right)}=2\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+1\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=2\)
\(\Leftrightarrow\dfrac{x+3+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2\)
\(\Leftrightarrow\dfrac{2x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2\)
\(\Leftrightarrow\dfrac{2\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2\)
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}=2\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=1\\x+3=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-1\\x+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\left(KTM\right)\\x=-4\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;-4\right\}\)
ĐKXĐ: \(x\ne-2;-3;-1\)
PT \(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+3\right)}-\dfrac{2}{\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+x-2x-6}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\left(loại\right)\end{matrix}\right.\)
Vậy \(x=3\)
ĐKXĐ của phương trình là: \(x\ne-2;x\ne-3;x\ne-1\)
Ta có: \(\dfrac{x}{x^2+5x+6}=\dfrac{x}{x^2+3x+2}\)
<=> \(\dfrac{x}{\left(x+2\right)\left(x+3\right)}=\dfrac{x}{\left(x+2\right)\left(x+1\right)}\)
<=> \(\dfrac{x\left(x+1\right)}{\left(x+2\right)\left(x+3\right)\left(x+1\right)}-\dfrac{x\left(x+3\right)}{\left(x+2\right)\left(x+1\right)\left(x+3\right)}=0\)
<=> \(\dfrac{x^2+x-x^2-3x}{\left(x+2\right)\left(x+3\right)\left(x+1\right)}=0\)
<=> \(-2x=0\Leftrightarrow x=0\)
Ta thấy x=0 thỏa mãn ĐKXĐ
Vậy tập nghiệm của pt là S={0}