Bạn ơi,phương trình của bạn như nào vậy?

\(\frac{x}{3\sqrt{x-1}}=\frac{6}{5}\left(1\right)\)

\(\frac{x}{3\sqrt{x}-1}=\frac{6}{5}\left(2\right)\)

Phương trình bạn muốn hỏi là 1 hay 2?

Phương trình 2 bạn ơi

a,(a+b)^2-4ab=(a-b)^2

\(\Leftrightarrow a^2+2ab+b^2-4ab=a^2-2ab+b^2\)

\(\Leftrightarrow a^2-a^2+2ab+2ab-4ab+b^2-b^2=0\)

\(\Leftrightarrow0=0\)

=> Với giá trị nào của a,b luôn tồn tại.(Đề là gì vậy?)

b,(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc               

=>Hằng đẳng thức                                                                                                                                       

c,(a-b)^3=a^3-b^3-3ab-(a-b)   

\(\Leftrightarrow a^3-3a^2b+3ab^2-b^3-a^3+b^3-3ab\left(a-b\right)=0\)

\(\Leftrightarrow-3a^2b+3ab^2-3a^2b+3ab^2=0\)

\(\Leftrightarrow-6a^2b+6ab^2=0\)

\(\Leftrightarrow-6ab\left(a-b\right)=0\)

\(a,\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{2.3}+2}\)

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

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

\(=3-2\)

\(=1\)

\(b,\sqrt{11+2\sqrt{6}}-3+\sqrt{2}\)

==>Đề sai???

\(a,\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{3+2\sqrt{2.3}+2}-\sqrt{3-2\sqrt{2.3}+2}\)

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

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

\(=2\sqrt{2}\)

\(b,\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{5-2\sqrt{2.5}+2}-\sqrt{5+2\sqrt{5.2}+2}\)

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

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)

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

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

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

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

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

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

=\(2\sqrt{2}\) 

b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\) 

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

=/\(\sqrt{5}-\sqrt{2}\) / \(-\) /\(\sqrt{5}+\sqrt{2}\)/

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

=\(\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) 

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

Chép đề đúng chưa bạn? 2 phân số đầu có ngoặc không vậy?

Nguyễn Công Tỉnh đúng r bạn, mình sửa lại r

\(D=x^2+2x\left(y+2\right)+2y^2+6y+10\)

\(=x^2+2x\left(y+2\right)+\left(y^2+4y+4\right)+\left(y^2+2y+1\right)+5\)

\(=x^2+2x\left(y+2\right)+\left(y+2\right)^2+\left(y+1\right)^2+5\)

\(=\left(x+y+2\right)^2+\left(y+1\right)^2+5\ge5\forall x\)

\(\Rightarrow\)Min D = 5 tại \(\hept{\begin{cases}x+y+2=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}}\)

=.= hk tốt!!

\(E=x^2+4xy+5y^2=x^2+4xy+4y^2+y^2=\left(x+2y\right)^2+y^2\ge0\forall x,y\)

=> Min E = 0 tại \(\hept{\begin{cases}x+2y=0\\y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}}\)

\(a,\left(\sqrt{27}-2\sqrt{17}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=3\sqrt{21}-2\sqrt{119}+7+7\sqrt{8}\)

Đề sai chăng???

\(b,\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2\sqrt{2}+1}\)

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

\(=\sqrt{2}-1+\sqrt{2}+1\)

\(=2\sqrt{2}\)

\(c,9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)

\(=9\sqrt{2}-8\sqrt{2}-5\sqrt{2}+8\sqrt{2}\)

\(=\sqrt{2}\left(9-8-5+8\right)\)

\(=4\sqrt{2}\)

\(d,\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)

\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4+2.2\sqrt{2}+2}\)

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

\(=\sqrt{2}-1-2-\sqrt{2}\)

\(=-3\)