$\sqrt{10x^2-24x+16}+\sqrt{10x^2-4x+4}=\sqrt{20}$
$\Leftrightarrow \sqrt{(3x-4)^2+x^2}+\sqrt{(3x)^2+(x-2)^2}=\sqrt{20}$
$VT=\sqrt{(4-3x)^2+x^2}+\sqrt{(3x)^2+(2-x)^2}\geq\sqrt{(4-3x+3x)^2+(2-x+x)^2}=\sqrt{20}$$=VP$
( BDT vector)
= khi $(4-3x)(2-x)=x.3x$ hay $x=\frac 45$
độ dài vector a độ dài vecto b lớn hơn độ dài vector (a b) cx cần chúng minh ý? –  mquang12a1tqk 30-05-17 02:55 AM
lm cái này phải c/m bđt nữa –  ๖ۣۜTQT☾♋☽ 30-05-17 12:05 AM
cách giải khác để tham khảo thôi, dễ nhất vẫn là chuyển vế rồi bình phương hai lần, ra nghiệm thay lại pt đầu –  mquang12a1tqk 29-05-17 10:13 PM
$\Leftrightarrow \sqrt{(10x^2-24x+16)(10x^2-4x+4)}=14x-10x^2\Leftrightarrow (10x^2-24x+16)(10x^2-4x+4)=(14x-10x^2)^2\Leftrightarrow 4(4-5x)^2=0$
ủa. mk đăng câu hỏi lúc nào nhỉ ? –  свинья 01-06-17 03:41 AM
dấu = cơ mak,gõ ra đi,xem tk nào dài hơn –  ๖ۣۜTQT☾♋☽ 29-05-17 08:02 PM
:v nhanh hơn ấy 1 dòng là ra –  ๖ۣۜJinღ๖ۣۜKaido 29-05-17 07:59 PM
định làm cách anh nhưng cmn tìm dấu = gõ lâu hơn cái này –  ๖ۣۜTQT☾♋☽ 29-05-17 07:51 PM
:v cạn lời bình phương đại pháp –  ๖ۣۜJinღ๖ۣۜKaido 29-05-17 07:50 PM

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