1)ý 1) [tex]A=\frac{(a^2-b^2)^2}{a^4b^4}+\frac{4}{ab}=\frac{1}{a^ 4}-\frac{2}{a^2b^2}+\frac{1}{b^4}+\frac{4}{ab}=\left ( \frac{1}{a^2}-\frac{1}{b^2} \right )^2+\frac{4}{ab}=\left ( \frac{1}{a}-\frac{1}{b} \right )^2+\frac{4}{ab}=\left ( \frac{1}{a}+\frac{1}{b} \right )^2=1[/tex]
ý 2) [tex]a+b=ab[/tex]
[tex](a+b-2)^3-(a-1)^3-(b-1)^3-3(a+b)+6=0<=>(ab-2)^3-(a^3-3a^2+3a-1)-(b^3-3b^2+3b-1)-3ab+6=0<=>(ab-2)^3-ab\left [ (ab)^2-3ab \right ]+3(ab)^2-6ab+8-3ab-3ab=0[/tex] (đúng )
câu 2 ) ý 1)
[tex]\frac{x+2-4(x-1)}{\sqrt{x+2}+2\sqrt{x-1}}=\frac{3.(-x+2)}{x+\sqrt{(x+2)(x-1)}}=> -x+2=0 \vee \frac{1}{\sqrt{x+2}+2\sqrt{x-1}}=\frac{1}{x+\sqrt{(x+2)(x-1)}}<=> x+\sqrt{(x+2)(x-1)}=\sqrt{x+2}+2\sqrt{x-1}[/tex]
với pt : [tex]x+\sqrt{(x+2)(x-1)}=\sqrt{x+2}+2\sqrt{x-1}[/tex]
[tex]\sqrt{x+2}\frac{x-1-1}{\sqrt{x-1}+1}=\frac{4x-4-x^2}{2\sqrt{x-1}+x}=>x-2=0 \vee \frac{\sqrt{x+2}}{\sqrt{x-1}+1}+\frac{x-2}{2\sqrt{x-1}+x}=0[/tex]
pt [tex]\frac{\sqrt{x+2}}{\sqrt{x-1}+1}+\frac{x-2}{2\sqrt{x-1}+x}=0[/tex] vô no
ý 2)
[tex]y-2x-1+2(\sqrt{x^2+y}-x-1)=0<=>y-2x-1+2.\frac{y-2x-1}{\sqrt{x^2+y}+x+1}=0=> y-2x-1=0[/tex]
thay xuống (2) : [tex](x-3)\sqrt{2x+5}+(2x-3)\sqrt{x-1}+2=0[/tex]
có no x=2 liên hợp
câu 4)
[tex]\sqrt{x}+\sqrt{ y}=2\sqrt{505}[/tex]
[tex]x=y=505[/tex]
[tex]x=0;y=2020[/tex]
[tex]x=2020;y=0[/tex]
câu 5 ) ý 1 )[tex]P=\frac{2}{a(1-2a)}+\frac{3}{b(1-2b)}+\frac{4}{c(1-2c)}\geq \frac{(2+3+4)^2}{(2a+3b+4c)-2(2a^2+3b^2+4c^2)}=\frac{9^2}{3-2(2a^2+3b^2+4c^2)}[/tex]
ta lại có : [tex](2a^2+3b^2+4c^2)\geq \frac{(2a+3b+4c)^2}{2+3+4}=\frac{9}{9}=1[/tex]
=> [tex]P\geq 81[/tex]
dấu "=" xảy ra khi [tex]a=b=c=\frac{1}{3}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
