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\(1+\dfrac{1}{x+2}=\dfrac{12}{x^3+8}\Leftrightarrow\dfrac{\left(x^3+8\right)\left(x+2\right)}{\left(x^3+8\right)\left(x+2\right)}+\dfrac{\left(x^3+8\right)}{\left(x^3+8\right)\left(x+2\right)}=\dfrac{12\left(x+2\right)}{\left(x^3+8\right)\left(x+2\right)}\)
\(\Rightarrow x^4+2x^3+8x+16+x^3+8=12x+24\)
\(\Leftrightarrow x^4+3x^3-4x=0\\ \Leftrightarrow x\left(x^3+3x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x^3+3x^2-4=0\end{matrix}\right.\)
\(x^3+3x^2-4=0\Leftrightarrow\left(x^3+4x^2+4x\right)-\left(x^2+4x+4 \right)=0\)
\(\left(x-1\right)\left(x^2+4x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x^2+4x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+2\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\left(loại\right)\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={1}
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
(x-5)(x-9)>0\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\Leftrightarrow x>5\\x-9>0\Leftrightarrow x>9\end{matrix}\right.\)
Vậy x>9 thì (x-5)(x-9)>0
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
Câu 1:
a: ĐKXĐ: \(x\notin\left\{0;1;\dfrac{1}{2}\right\}\)
\(B=\dfrac{x^2+x}{x^2+x+1}-\dfrac{2x^3+x^2-x-2x^3+2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{x\left(x+1\right)}{x^2+x+1}-\dfrac{-2x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{x\left(x+1\right)}{x^2+x+1}+\dfrac{2x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{-x\left(x-1\right)}{2x-1}\)
\(=\dfrac{x\left(x+1\right)}{x^2+x+1}+\dfrac{-x}{x^2+x+1}=\dfrac{x^2}{x^2+x+1}\)
b: Để \(B=\dfrac{4}{3}\) thì \(\dfrac{x^2}{x^2+x+1}=\dfrac{4}{3}\)
\(\Leftrightarrow4x^2+4x+4-3x^2=0\)
=>x=-2(nhận)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=X\\\frac{1}{y}=Y\end{matrix}\right.\) hệ pt đã cho trở thành:
\(\left\{{}\begin{matrix}2X-Y=5\\X+2Y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4X-2Y=10\\X+2Y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5X=15\\X+2Y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}X=3\\Y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=3\\\frac{1}{y}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\y=1\end{matrix}\right.\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(\frac{1}{3};1\right)\)
với x>0 thì pt luôn xác định.
\(\Rightarrow\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}=\dfrac{12}{x^3+8}\)
\(\Leftrightarrow x^3+8+x^2-2x+4=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(x\left(x^2+x-2\right)=0\Rightarrow x=0\) hoặc \(x^2+x-2=0\)
x=0 hoac (x\(^2\)-1) +(x-1) =0
x=0 hoặc (x-1)(x+2)=0
x=0 hoax x=1 hoặc x=2 vỉ x>0 nên pt có 2 nghiệm là x=1 , x=2.
x = 0