Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=-6\sqrt{b}-\sqrt{6}.\sqrt{b}+\sqrt{6}b=-6\sqrt{b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Hoành độ giao điểm d1 ; d2 thỏa mãn phương trình
\(3x+1=-x\Leftrightarrow4x+1=0\Leftrightarrow x=-\frac{1}{4}\)
\(\Rightarrow y=-\frac{3}{4}+1=\frac{1}{4}\)
Vậy d1 cắt d2 tại A(-1/4;1/4)
Để 3 điểm đồng quy khi d3 cắt A(-1/4;1/4) <=> \(\frac{1}{4}=-\frac{1}{4}+\frac{1}{2}\)( đúng )
Vậy 3 điểm đồng quy
b, d1 : \(y=1-x\)
Hoành độ giao điểm d1 ; d2 thỏa mãn phương trình
\(1-x=3x+5\Leftrightarrow4x=-4\Leftrightarrow x=-1\)
\(\Rightarrow y=-3+5=2\)
Vậy d1 cắt d2 tại T(-1;2)
Để 3 điểm đồng quy khi d3 cắt T(-1;2) <=> \(-1-\frac{2}{3}+\frac{5}{3}=0\)( luôn đúng )
Vậy 3 điểm đồng quy
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)
b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)
c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)
d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(2\sqrt{3x}-\sqrt{48x}+\sqrt{108x}+\sqrt{3x}\\ =3\sqrt{3x}-\sqrt{4^2\cdot3x}+\sqrt{6^2\cdot3x}\\ =3\sqrt{3x}-4\sqrt{3x}+6\sqrt{3x}=5\sqrt{3x}\)
b,
\(2\sqrt{25xy}+\sqrt{5}\cdot\sqrt{45x^3y^3}-3y\sqrt{16x^3y}\\ =2\sqrt{5^2xy}+\sqrt{5\cdot45}\cdot\sqrt{\left(xy\right)^2\cdot xy}-3y\sqrt{\left(4x\right)^2\cdot xy}\\ =2\cdot5\sqrt{xy}+\sqrt{225}\cdot xy\sqrt{xy}-3y\cdot4x\sqrt{xy}\\ =10\sqrt{xy}+15xy\sqrt{xy}-12xy\sqrt{xy}=\sqrt{xy}\left(3xy+10\right)\)
c,
\(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{12}{3-\sqrt{13}}\\ =\frac{2\left(\sqrt{3}+1\right)}{3-1}+\frac{3\left(\sqrt{3}+2\right)}{3-4}+\frac{12\left(3+\sqrt{13}\right)}{9-13}\\ =\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{3\left(\sqrt{3}+2\right)}{-1}+\frac{12\left(3+\sqrt{13}\right)}{-4}\\ =\sqrt{3}+1-3\sqrt{3}-6-9-3\sqrt{13}\\ =-14-2\sqrt{3}-3\sqrt{13}\)
d,
\(\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{7}+\sqrt{3}}\\ =\frac{\sqrt{3}+\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}\\ =\sqrt{3}+\sqrt{2}-\sqrt{5}+\sqrt{3}+\sqrt{5}+\sqrt{2}+\sqrt{7}-\sqrt{3}=\sqrt{7}+\sqrt{3}\)
Chúc bạn học tốt nha.
\(1a.2\sqrt{25xy}+\sqrt{225x^3y^3}-3y\sqrt{16x^3y}=10\sqrt{xy}+15xy\sqrt{xy}-12xy\sqrt{xy}=10\sqrt{xy}+3xy\sqrt{xy}=\sqrt{xy}\left(10+3xy\right)\left(x,y\ge0\right)\)
\(b.-\sqrt{36b}-\dfrac{1}{3}\sqrt{54b}+\dfrac{1}{5}\sqrt{150b}=-6\sqrt{b}-\sqrt{6b}+\sqrt{6b}=-6\sqrt{b}\left(b\ge0\right)\)
\(2.\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)
\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}-\sqrt{3x}-3\sqrt{1-2x}=0\)
\(\Leftrightarrow\sqrt{1-2x}=4\sqrt{3x}\left(x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow1-2x=48x\)
\(\Leftrightarrow x=\dfrac{1}{50}\left(KTM\right)\)
KL....