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\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)
\(\Rightarrow32x+32=31x+62\)
\(\Rightarrow x=30\)
Vậy x=30
Chúc bn học tốt
1/
\(y\left(x+1\right)-x^2\left(x+1\right)=7\Leftrightarrow\left(x+1\right)\left(y-x^2\right)=7\)
TH1: \(\left\{{}\begin{matrix}x+1=1\\y-x^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=7\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+1=7\\y-x^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=37\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x+1=-1\\y-x^2=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x+1=-7\\y-x^2=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=63\end{matrix}\right.\)
2/
\(\left(1+\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right)...\left(1+\dfrac{1}{\left(x+1-1\right)\left(x+1+1\right)}\right)=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{\left(x+1\right)^2}{x\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{2.3.4...\left(x+1\right)}{3.4.5...\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow2012\left(x+1\right)=2011\left(x+2\right)\)
\(\Leftrightarrow x=2010\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
1)
\(A=\dfrac{1}{2}.\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}......\dfrac{4064256}{2015.2017}\\ =\dfrac{1.2.2.3.3.....2016.2016}{2.1.3.2.4.3.5....2015.2017}\\ =\dfrac{\left(2.3.4.....2016\right)}{\left(1.2.3.4....2015\right)}.\dfrac{\left(2.3.4....2016\right)}{\left(2.3.4.5....2017\right)}\\ =2016.\dfrac{1}{2017}=\dfrac{2016}{2017}\)
2) a)
Ta có : \(\left(2x-\dfrac{1}{6}\right)^2+\left|3y+12\right|\ge0\) \(\forall x,y\)
Mà \(\left(2x-\dfrac{1}{6}\right)^2+\left|3y+12\right|=0\) ( theo đề ra)
\(\)\(\Rightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{1}{6}\right)^2=0\\\left|3y+12\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{12}\\y=-4\end{matrix}\right.\)