a) \(\Delta'=m^2-\left(m-4\right)=m^2-m+4=m^2-2.m.\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\)

\(=\left(m-\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0;\forall m\)

=> phương trình (1) luôn có hai nghiệm phân biệt với mọi m

b) Áp dụng định lí Viet ta có: 

\(x_1.x_2=m-4\)

\(x_1+x_2=-2m\)

=> \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1.x_2=\left(-2m\right)^2-2\left(m-4\right)=4m^2-2m+8\)

=> \(x_1^3+x_2^3=\left(x_1+x_2\right)\left(x_1^2-x_1x_2+x_2^2\right)=\left(-2m\right)\left(4m^2-2m+8-\left(m-4\right)\right)\)

\(=-2m\left(4m^2-3m+12\right)\)

Theo bài ra ta có:

 \(x_1+x_2=\frac{x_1^2}{x_2}+\frac{x_2^2}{x_1}\)

 \(\Leftrightarrow x_1+x_2=\frac{x_1^3+x_2^3}{x_1.x_2}\)

Thay vào ta có:

\(-2m=\frac{-2m\left(4m^2-3m+12\right)}{m-4}\)( đk m khác 4)

\(\Leftrightarrow\orbr{\begin{cases}m=0\\m-4=4m^2-3m+12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}m=0\left(tm\right)\\4m^2-4m+16=0\left(l\right)\end{cases}\Leftrightarrow m=0}\)

Vì \(4m^2-4m+16=\left(2m-1\right)^2+15>0\) với mọi m

Vậy m =0

cảm ơn nhìu

\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)

\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)

\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)

\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)

\(S=\frac{\left(2019-1\right)}{2}=1009\)

\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)

\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)

\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)

\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)

ĐK: \(x^2-1\ge0\) (1)

\(pt\Leftrightarrow\left(x^2+3\sqrt{x^2-1}\right)^2=x^4-x^2+1\)

\(\Leftrightarrow6\sqrt{x^2-1}+9\left(x^2-1\right)=-x^2+1\)

\(\Leftrightarrow6\sqrt{x^2-1}+10\left(x^2-1\right)=0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(6+10\sqrt{x^2-1}\right)=0\)

\(\Leftrightarrow\sqrt{x^2-1}=0\)

\(\Leftrightarrow x^2-1=0\Leftrightarrow x=\pm1\)Thỏa mãn đk (1)

Vậy ...

\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)

\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)

\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\

\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

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

\(=10\)

\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)

\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)

\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)

\(=\sqrt{3}-1\)

G A B C N M E F

a) Gọi F' là giao điểm của AE và BC

MN//BC => \(\frac{MN}{BC}=\frac{AN}{AC}\)

NE//F'C => \(\frac{EN}{FC}=\frac{AN}{AC}\)

=> \(\frac{EN}{F'C}=\frac{MN}{BC}=\frac{2EN}{2FC}=\frac{EN}{FC}\Rightarrow F'C=FC\)

mà F', F cùn thuộc cạnh BC

=> F' trùng F

=> A, E, F thẳng hàng

b) Xét tam giác BNC có: Flaf trung điểm BC; G là trung điểm BN

=> FG là đường trung bình tam giác BNC

=> FG//=1/2 NC

=> FG=9:2=4,5 cm

Xét tam giác BNM tương tự

có: EG//=1/2 BM 

=> EG=12:2=6 cm

Ta lại có: EG//BM => EG//AB

FG //NC => FG//AC

Mà AB vuông AC

=> EG vuông FG

=> Tam giác EGF vuông tại G có: FG=4,5 cm và EG=6 cm

Áp dụng định lí pitago: 

=> \(EF^2=GE^2+GF^2=4,5^2+6^2=7,5^2\)

=> EF=7,5

\(\widehat{EGF}=90^o\)

\(\cos\widehat{GEF}=\frac{GE}{EF}=\frac{6}{7,5}=\frac{4}{5}\Rightarrow\widehat{GEF}=arcos\frac{4}{5}\)

\(\cos\widehat{GFE}=\frac{GF}{EF}=\frac{4,5}{7,5}=\frac{3}{5}\Rightarrow\widehat{GFE}=arcos\frac{3}{5}\)

c) Ta có: MN//BC 

=> \(\frac{BM}{AB}=\frac{CN}{AC}\Rightarrow\frac{AB}{AC}=\frac{BM}{CN}=\frac{2GE}{2GF}=\frac{GE}{GF}\)

Xét tam giác vuông GEF và tam giác vuông ABC 

có: \(\frac{AB}{AC}=\frac{GE}{GF}\)

=> tam giác GEF đồng dạng với tam giác ABC

Đặt: \(P=\left(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\right)^2\)

=> \(2P=2\left(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\right)^2\)

\(2P=\left(\sqrt{2}.\sqrt{2+\sqrt{3}}-\sqrt{2}.\sqrt{3+\sqrt{5}}\right)^2\)

\(2P=\left(\sqrt{4+2\sqrt{3}}-\sqrt{6+2\sqrt{5}}\right)^2\)

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

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

\(2P=\left(\sqrt{3}-\sqrt{5}\right)^2=3+5-2\sqrt{15}=8-2\sqrt{15}\)

=> \(P=4-\sqrt{15}\)

\(\frac{2}{3}\sqrt{3}\)- \(\frac{1}{4}\sqrt{18}\)+ \(\frac{2}{5}\sqrt{2}-\frac{1}{4}\sqrt{12}\)

= \(\frac{2}{3}\sqrt{3}-\frac{3}{4}\sqrt{2}+\frac{2}{5}\sqrt{2}-\frac{2}{4}\sqrt{3}\)

= \(\sqrt{3}\left(\frac{2}{3}-\frac{1}{2}\right)\)- \(\sqrt{2}\left(\frac{3}{4}-\frac{2}{5}\right)\)

= \(\frac{\sqrt{3}}{6}\)- \(\frac{7}{20}\sqrt{2}\)

kq ra hơi kì

#mã mã#