\(MA^4+MB^4+MC^4+MD^4\)

\(=\left(MA^2+MC^2\right)^2+\left(MB^2+MD^2\right)^2-2MA^2.MC^2-2MB^2.MD^2\)

\(=32R^4-8S_{MAC}^2-8S_{MBD}^2\)

\(=32R^4-8R^2\left(MH^2+MK^2\right)\) với H,K lần lượt là hình chiếu vuông góc của M trên AC,BD

\(=32R^4-8R^2.R^2=24R^4\)

17 nha

\(N=\frac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{y}}\)

\(N=\frac{1+\sqrt{x}+\sqrt{y}\left(\sqrt{x}+1\right)}{1+\sqrt{y}}\)

\(N=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)}{1+\sqrt{y}}\)

\(N=\sqrt{x}+1\)

\(P=\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(P=\frac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(P=\frac{\sqrt{y}\left(4\sqrt{y}+7\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(P=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(P=\sqrt{y}-1\)

\(C=\sqrt{x}+\sqrt{y}+\sqrt{x^2y}+\sqrt{xy^2}\)

\(C=\sqrt{x}\left(\sqrt{xy}+1\right)+\sqrt{y}\left(\sqrt{xy}+1\right)\)

\(C=\left(\sqrt{xy}+1\right)\left(\sqrt{x}+\sqrt{y}\right)\)

\(D=x+2\sqrt{xy}+y-4\)

\(D=\left(\sqrt{x}+\sqrt{y}\right)^2-4\)

\(D=\left(\sqrt{x}+\sqrt{y}-4\right)\left(\sqrt{x}+\sqrt{y}+4\right)\)

\(E=x+\sqrt{x}+\frac{1}{4}-\frac{49}{4}\)

\(E=\left(\sqrt{x}+\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

\(E=\left(\sqrt{x}+\frac{1}{2}-\frac{7}{2}\right)\left(\sqrt{x}+\frac{1}{2}+\frac{7}{2}\right)\)

\(E=\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)\)

\(F=2a-5\sqrt{ab}+3b\)

\(F=2a-2\sqrt{ab}-3\sqrt{ab}+3b\)

\(F=2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

\(F=\left(2\sqrt{a}-3\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

\(\hept{\begin{cases}\left(x+3\right)\left(y-5\right)=xy\\\left(x-2\right)\left(y+5\right)=xy\end{cases}}\Leftrightarrow\hept{\begin{cases}xy-5x+3y-15=xy\\xy+5x-2y-10=xy\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-5x+3y=15\\5x-2y=10\end{cases}}\Leftrightarrow\hept{\begin{cases}y=25\\5x-2y=10\end{cases}}\Leftrightarrow\hept{\begin{cases}x=12\\y=25\end{cases}}\)

ĐK : \(\hept{\begin{cases}x\ne1\\y\ne-1\end{cases}}\)

\(\hept{\begin{cases}\frac{x+1}{x-1}=\frac{y+3}{y+1}\\3x+2y=0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=\left(x-1\right)\left(y+3\right)\\3x+2y=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}xy+x+y+1=xy+3x-y-3\\3x+2y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-2y=4\\3x+2y=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5x=4\\3x+2y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{4}{5}\\y=-\frac{6}{5}\end{cases}\left(tm\right)}\)

\(cos^4a-sin^4a+2sin^2a\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+2sin^2a\)

\(=cos^2a\left(cos^2a+sin^2a\right)+2sin^2a\)

Bài làm này chắc ổn hơn bài làm trước ✔

\(cos^4a-sin^4a+2sin^2a\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+2sin^2a\)

\(=\left(cos^2a-sin^2a\right)1+2sin^2a\)

\(=cos^2a-sin^2a+2sin^2a\)

\(=cos^2a+sin^2a\)

\(=1\)

Ta có : 

\(\hept{\begin{cases}AC^2=AD^2+DC^2\\BD^2=AD^2+AB^2\end{cases}}\) vì \(CD>AB\Rightarrow AC>BD\)

b. ta có \(AC^2-BD^2=\left(AD^2+DC^2\right)-\left(AD^2+AB^2\right)=CD^2-AB^2\)

Ta có : \(\hept{\begin{cases}BD^2=BA^2+AD^2\\BC^2=\left(DC-AB\right)^2+AD^2\end{cases}\Leftrightarrow\hept{\begin{cases}225=BA^2+AD^2\\169=\left(14-AB\right)^2+AD^2\end{cases}}}\)

lấy hiệu ta có : \(56=28AB-196\Leftrightarrow AB=9\Rightarrow AD=\sqrt{225-9^2}=12\)

Diện tích hình thang là :\(\frac{1}{2}\left(AB+CD\right)\times AD=\frac{1}{2}\left(9+14\right)\times12=138cm^2\)

Mik có lớp 4 thôi seo giúp đc . Seo bn zô zuyên hế . ^~^

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