b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

b) Áp dụng bđt Svac-xơ:

\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)

=> hpt vô nghiệm

c) Ở đây x,y,z là các số thực dương

Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)

Dấu = xảy ra khi \(x=y=z=\dfrac{3}{3}=1\)

(Pt trên là pt (1), pt dưới là pt (2))

Đk: \(x;y\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3=2x^3+x^2y\\3=2y^3+xy^2\end{matrix}\right.\)

\(\Rightarrow2\left(x^3-y^3\right)+\left(x^2y-xy^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+2xy+y^2\right)=0\)\(\Leftrightarrow\left(x-y\right)\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

TH1: \(x=y\) thay vào pt (1) \(\Rightarrow\dfrac{3}{y^2}=2y+y\)

\(\Leftrightarrow3=3y^3\) \(\Leftrightarrow y=1\) \(\Rightarrow x=y=1\)  (TM)

TH2:\(x=-y\) thay vào pt (1) \(\Rightarrow\dfrac{3}{y^2}=-2y+y\)

\(\Leftrightarrow\dfrac{3}{y^2}=-1\left(L\right)\)

Vậy (x;y)=(1;1)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy+\left(x+y\right)=4\\\left(x+y+1\right)\left(5+2xy+x+y\right)=27\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}u=x+y\\v=xy\end{matrix}\right.\left(u^2\ge4v\right)\)

Khi đó hpt tt \(\left\{{}\begin{matrix}u^2-2v+u=4\\\left(u+1\right)\left(5+2v+u\right)=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2v=u^2+u-4\left(1\right)\\\left(u+1\right)\left(5+u^2+u-4+u\right)=27\end{matrix}\right.\)

Phương trình (1) \(\Leftrightarrow\left(u+1\right)\left(u^2+2u+1\right)=27\)

\(\Leftrightarrow u+1=\sqrt[3]{27}\) \(\Leftrightarrow u=2\)

\(\Rightarrow v=\dfrac{u^2+u-4}{2}=1\)

Khi đó\(\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Rightarrow\) x,y là nghiệm của pt: \(t^2-2t+1=0\) \(\Leftrightarrow t=1\) 

\(\Rightarrow x=1;y=1\)

Vì tỉ số giữa hai nghiệm khác 1 nên pt có hai nghiệm pb

\(\Rightarrow\Delta=4m^2-4\left(2m-1\right)>0\)

\(\Leftrightarrow m\ne1\)

Áp dụng viet có: \(\left\{{}\begin{matrix}y_1+y_2=2m\\y_1y_2=2m-1\end{matrix}\right.\) 

Giả sử \(y_1=2y_2\) 

Có hệ: \(\left\{{}\begin{matrix}y_1+y_2=2m\\y_1=2y_2\\y_1y_2=2m-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y_1=\dfrac{4m}{3}\\y_2=\dfrac{2m}{3}\\y_1y_2=2m-1\end{matrix}\right.\)\(\Rightarrow\dfrac{4m}{3}.\dfrac{2m}{3}=2m-1\)

\(\Leftrightarrow8m^2-18m+9=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{3}{2}\\m=\dfrac{3}{4}\end{matrix}\right.\)(tm)

13)\(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}\)\(=\dfrac{\left(\sqrt{10}-\sqrt{2}\right)\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}=\dfrac{2+\sqrt{3}}{2}\)

14)sai đề? phải là \(\sqrt{3-\sqrt{5}}\) 

\(=\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{10}-2\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}=\dfrac{\left|\sqrt{5}-1\right|\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}\)

\(=\dfrac{3+\sqrt{5}}{4}\)

15)\(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}=\left|1-\sqrt{2016}\right|\sqrt{1+2\sqrt{2016}+2016}\)

\(=\left(\sqrt{2016}-1\right)\sqrt{\left(1+\sqrt{2016}\right)^2}=\left(\sqrt{2016}-1\right)\left(1+\sqrt{2016}\right)\)

\(=2015\)

Bài trên sai r
( l10) 

(d): \(\left(m-2\right)x-y+m=0\)
\(d_{\left(M;d\right)}=\dfrac{\left|m-2-0+m\right|}{\sqrt{\left(m-2\right)^2+1}}\)\(=\dfrac{\left|2m-2\right|}{\sqrt{\left(m-2\right)^2+1}}=\sqrt{\dfrac{\left(2m-2\right)^2}{\left(m-2\right)^2+1}}\)

Đặt \(A=\dfrac{\left(2m-2\right)^2}{\left(m-2\right)^2+1}\)   

\(\Leftrightarrow m^2\left(A-4\right)-4m\left(A-2\right)+5A-4=0\) (*)
Tại A=4 thì pt(*) \(\Leftrightarrow-8m+16=0\) \(\Leftrightarrow m=2\)

Tại \(A\ne4\) .Coi pt (*) là pt bậc 2 => Pt có nghiệm

\(\Leftrightarrow\Delta\ge0\)\(\Leftrightarrow-A^2+8A\ge0\) \(\Leftrightarrow A\in\left[0;8\right]\)

\(\Rightarrow maxA=8\) \(\Leftrightarrow\) m=3

=>\(\sqrt{\dfrac{\left(2m-2\right)^2}{\left(m-2\right)^2+1}}\le\sqrt{8}\) khi m=3
=> Khoảng cách lớn nhất từ m đến d là \(\sqrt{8}\) khi m=3