\(\sqrt{\left(x-1\right)\left(x-5\right)}-3\sqrt{1-x}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-5\right)\ge0\\1-x\ge0\\\sqrt{\text{​​}\left(x-1\right)\left(x-5\right)}=3\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-5\ge0\\x-1\le0\end{matrix}\right.\\x\le1\\\sqrt{\text{​​}\left(x-1\right)\left(x-5\right)}=3\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\\x\le1\\\left(x-1\right)\left(x-5\right)=9\left(1-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-5\right)+9\left(x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-5+9\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x+4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
- Vậy \(S=\left\{1;-4\right\}\)

\(\sqrt{1-x}(\sqrt{5-x}-3)=0\)

x = 1 hay \(\sqrt{5-x}=3 \)

x = 1 hay x= - 4

a, Xét tam giác BKC vuông tại K

\(sinKCB=\dfrac{BK}{BC}\Rightarrow BK=sinKCB.BC\approx3,75cm\)

\(cosKCB=\dfrac{KC}{BC}\Rightarrow KC=BC.cosKCB\approx7cm\)

Xét tam giác BKA vuông tại K

\(sinKAB=\dfrac{BK}{AB}\Rightarrow AB=BK.sinKAB\approx3,34cm\)

bạn ktra lại đoạn AK giúp mình nhé 

D\(=\dfrac{a^2\left(a+1\right)+b^2\left(b-1\right)+a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(a+1\right)}=\dfrac{a^3+a^2+b^3-b^2+a^3b^2+a^2b^3}{\left(a+b\right)\left(1-b\right)\left(a+1\right)}\)

\(ĐKXĐ:a\ne-b;a\ne-1;b\ne1\)

\(D=\dfrac{a^2}{\left(a+b\right)\left(1-b\right)}-\dfrac{b^2}{\left(a+b\right)\left(1+a\right)}-\dfrac{a^2b^2}{\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{a^3+a^2-b^2+b^3-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{a\left(a-b\right)+\left(a-b\right)+b^2\left(1-a^2\right)}{\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{\left(a-b\right)\left(a+1\right)-b^2\left(a-1\right)\left(a+1\right)}{\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{\left(1+a\right)\left(a-b-ab^2+b^2\right)}{\left(1+a\right)\left(1-b\right)}\)

\(=\dfrac{a-b-ab^2+b^2}{1-b}\)

\(=\dfrac{b\left(b-1\right)-a\left(b^2-1\right)}{1-b}=\dfrac{a\left(1-b\right)\left(1+b\right)-b\left(1-b\right)}{1-b}=\dfrac{\left(1-b\right)\left(a+ab-b\right)}{1-b}=a+ab-b\)

\(N=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{x-1}.\dfrac{x-1}{2x\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4x\sqrt{x}-4\sqrt{x}}{2x\sqrt{x}}\\ =\dfrac{4x\sqrt{x}}{2x\sqrt{x}}=2\left(x>0;x\ne1\right)\)

\(C=\dfrac{1}{\sqrt{x}\left(x\sqrt{x}-1\right)}.\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{x-1}\left(x>0;x\ne1\right)\)