Mn giúp em với ạ :(( em cảm ơn
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2. get lucky money
3. talent show
4. always
5. food stands
6. fireworks
7. never
8. eat traditional foods
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\(h'\left(x\right)=f'\left(x\right)-g'\left(x\right)=0\Rightarrow x=\left\{a;b;c\right\}\)
Ta thấy \(h'\left(x\right)>0\) trên \(\left(b;c\right)\) và \(h'\left(x\right)< 0\) trên \(\left(a;b\right)\)
\(\Rightarrow x=b\) là điểm cực tiểu trên \(\left[a;c\right]\) hay \(\min\limits_{\left[a;c\right]}h\left(x\right)=h\left(b\right)\)
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\(=>Qthu1=0,2.340000=68000J\)
\(=>Qthu2=2100.0,2.20=8400J\)
\(=>Qtoa=2.4200.25=210000J\)
\(=>Qthu1+Qthu2< Qtoa\)=>đá nóng chảy hoàn toàn
\(=>0,2.2100.20+0,2.340000+0,2.4200.tcb=2.4200\left(25-tcb\right)\)
\(=>tcb=14,5^oC\)
Cho em hỏi ngu tí ạ vậy tcb ở nhưng phép tính trên vứt đi đâu ạ
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Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x-1}=a\\\sqrt[3]{x+1}=b\\\sqrt[3]{-2x}=c\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+b=c\\a^3+b^3=-2c^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c\\\left(a+b\right)^3-3ab\left(a+b\right)=-2c^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c\\c^3-3abc=-2c^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c\\c\left(c^2-ab\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c\\c\left[\left(a+b\right)^2-ab\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c\\c\left[\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\right]=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}c=0\\a=b=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt[3]{-2x}=0\Leftrightarrow x=0\)
:(