_ Mức đáp ứng nhu cầu:
........• Năng lượng: 98,2%
........• Protein: 88,96%
........• Canxi: 70,4%
........• Sắt: 134,05%
........• Vitamin A: 180,38%
........• Vitamin : 123%
........• Vitamin : 38,67%
........• Vitamin PP: 223,78%
........• Vitamin C: 63,07%

sorry hơi muộn

\(1,\\ a,=3x\left(1-3y\right)\\ b,=9xy\left(2xy-x^2+4y\right)\\ c,=\left(x-y\right)\left(15x-5y\right)=5\left(x-y\right)\left(3x-y\right)\\ 2,\\ a,\Rightarrow2x^2\left(x^2-4\right)=0\Rightarrow2x^2\left(x-2\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\\ b,\Rightarrow\dfrac{2}{5}x\left(x+10\right)-\left(x+10\right)=0\\ \Rightarrow\left(x+10\right)\left(\dfrac{2}{5}x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-10\\\dfrac{2}{5}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-10\\x=\dfrac{5}{2}\end{matrix}\right.\)

\(3,\)

\(a,\left\{{}\begin{matrix}AK=KD\\BI=IC\end{matrix}\right.\Rightarrow KI\) là đtb hình thang ABCD

\(b,\) Vì KI là đtb hình thang ABCD nên \(KI=\dfrac{AB+CD}{2}=\dfrac{17}{2}=8,5\left(cm\right)\)

\(c,\) \(\left\{{}\begin{matrix}AK=KD\\KE//AB\end{matrix}\right.\Rightarrow BE=ED\Rightarrow KE\) là đtb tam giác ABD

\(\Rightarrow KE=\dfrac{1}{2}AB=2,5\left(cm\right)\)

\(\left\{{}\begin{matrix}BI=IC\\IF//AB\end{matrix}\right.\Rightarrow AF=FC\Rightarrow IF\) là đtb tam giác ABC

\(\Rightarrow IF=\dfrac{1}{2}AB=2,5\left(cm\right)\)

Ta có \(EF=KI-KE-IF=8,5-2,5-2,5=3,5\left(cm\right)\)

giúp bài cuối hả?

Lời giải:

\(P.\frac{1}{\sqrt{2}}=\frac{\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}}{\sqrt{(2x-1)+2\sqrt{2x-1}+1}-\sqrt{(2x-1)-2\sqrt{2x-1}+1}}\)

\(=\frac{\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}}{\sqrt{(\sqrt{2x-1}+1)^2}-\sqrt{(\sqrt{2x-1}-1)^2}}\)

\(=\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-(\sqrt{2x-1}-1)}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\)

\(2^x+2^{x+1}+2^{x+2}+...+2^{x+15}=2^{2019}-8\)
<=>\(2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-2^3\)
<=>\(2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)
<=>\(2^x=2^3\)
<=>\(x=3\)