HOC24
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\(pt\Leftrightarrow\dfrac{1}{\left(x-1\right)^2+1}+\dfrac{2}{\left(x-1\right)^2+2}=\dfrac{6}{\left(x-1\right)^2+3}\)
Đặt: \(\left(x-1\right)^2=t\ge0\)
\(pt\Leftrightarrow\dfrac{1}{t+1}+\dfrac{2}{t+2}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{t+2+2\left(t+1\right)}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{t+2+2t+2}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{3t+4}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\left(3t+4\right)\left(t+3\right)=6\left(t+1\right)\left(t+2\right)\)
Phân tích ra:v
\(C=2-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2+\sqrt{16}\)
\(C=2-1+\dfrac{1}{4}+4\)
\(C=5+\dfrac{1}{4}=\dfrac{21}{4}\)
Đặt: \(x+2=t\)
\(pt\Leftrightarrow\left(t+1\right)^3-\left(t-1\right)^3=56\)
\(\Rightarrow\left(t^3+3t^2+3t+1\right)-\left(t^3-3t^23t-1\right)=56\)
\(\Rightarrow t^3+3t^2+3t+1-t^3+3t^2-3t+1=56\)
\(\Rightarrow\left(t^3-t^3\right)+\left(3t^2+3t^2\right)+\left(3t-3t\right)+\left(1+1\right)=56\)
\(\Rightarrow6t^2+2=56\Leftrightarrow6t^2=54\Leftrightarrow t^2=9\)
\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
\(A=\left|x\right|+3\ge3\)
Dấu "=" xảy ra khi: \(x=0\)
gọi phân số đó là \(\dfrac{a}{b}\)
theo đề bài ta có:\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{32}{60}\\a+b=115\end{matrix}\right.\)
Từ \(\dfrac{a}{b}=\dfrac{32}{60}\Leftrightarrow60a=32b\Leftrightarrow15a=8b\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{8}=\dfrac{b}{15}=\dfrac{a+b}{8+15}=\dfrac{115}{23}=5\)
\(\Rightarrow\left\{{}\begin{matrix}a=40\\b=75\end{matrix}\right.\)
Vậy \(\dfrac{a}{b}=\dfrac{40}{75}\)
a) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
Vì \(x^2-5>x^2-25\) nên \(\left\{{}\begin{matrix}x^2-5>0\\x^2-25< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2>5\\x^2< 25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{5}< x< -\sqrt{5}\left(vl\right)\\-5< x< 5\end{matrix}\right.\)
b) \(\left(x+5\right)\left(9+x^2\right)< 0\)
Vì \(9+x^2>0\) nên \(x+5< 0\Leftrightarrow x< -5\)
c) \(\left(x+3\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\) nên \(x+3=0\Leftrightarrow x=-3\)
d) \(\left(x+5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-2\\x=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-28a=b\\35c=d\end{matrix}\right.\)
Vì \(b;d\in Z^-\) nên \(\left\{{}\begin{matrix}-28a< 0\\35c< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\c< 0\end{matrix}\right.\)
Vậy \(a>c\)
\(3x-2⋮2x+1\)
\(\Rightarrow2\left(3x-2\right)⋮2x+1\)
\(\Rightarrow6x-4⋮2x+1\)
\(\Rightarrow6x+3-7⋮2x+1\)
\(\Rightarrow3\left(2x+1\right)-7⋮2x+1\)
\(\Rightarrow7⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=1\\2x+1=-1\\2x+1=7\\2x+1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\\x=-4\end{matrix}\right.\)
\(xy-y-2x-2=5\)
\(\Rightarrow xy-y-2x+2=9\)
\(\Rightarrow y\left(x-1\right)-2\left(x-1\right)=9\)
\(\Rightarrow\left(y-2\right)\left(x-1\right)=9\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\left(đpcm\right)\)