\(2\left(a^2+b^2\right)=5ab\)

\(\Leftrightarrow2a^2+2b^2-5ab=0\)

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

\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)

\(a>b\Rightarrow2a>b\Leftrightarrow2a-b>0\)

\(\Rightarrow a=2b\)

\(Q=\frac{3\left(2b-b\right)}{2\left(2b+b\right)}=\frac{3b}{6b}=\frac{1}{2}\)

thank bạn nha

4 ( x + y ) = xy + 11

\(\Leftrightarrow\)4x + 4y - xy = 11

\(\Leftrightarrow\)x ( 4 - y ) - 16 + 4y = -5

\(\Leftrightarrow\)x ( 4 - y ) - 4 ( 4 - y ) = -5

\(\Leftrightarrow\)( x - 4 ) ( 4 - y ) = -5

lập bảng giá trị, tìm được x,y

Câu hỏi của NOO PHƯỚC THỊNH - Toán lớp 8 - Học toán với OnlineMath

>: nhấn vô coi câu tl rồi bt( câu b ý) 

0 nha bạn hihi

^{ssssssssssssssss_ssssss}s

đáp án

12+91-12-91=9

học tốt

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pascal bạn oi

Theo mk nghĩ thôi nhé, mk viết đáp số thôi nha

\(a,b,c=0\)

Trong 3 số a,b,c luôn tồn tại hai số cùng \(\ge\frac{1}{2}\) hoặc \(\le\frac{1}{2}\)Giả sử hai số đó là a và b

Ta có:\(c\left(2a-1\right)\left(2b-1\right)\ge0\Leftrightarrow c\left(4ab-2a-2b+1\right)\ge0\)

\(\Leftrightarrow4abc-2ac-2bc+c\ge0\Leftrightarrow4abc+c\ge2ac+2bc\)

Ta lại có:\(1=a^2+b^2+c^2+2abc\ge2ab+2abc+c^2\)

\(\Leftrightarrow1-c^2\ge2ab\left(c+1\right)\Leftrightarrow1-c\ge2ab\Leftrightarrow1\ge2ab+c\)\(\ge2\sqrt{2abc}\)

\(\Rightarrow1\ge8abc\Rightarrow abc\le\frac{1}{8}\).Từ \(a^2+b^2+c^2+2abc=1\Rightarrow\)

\(2+c=2a^2+2b^2+2c^2+4abc+c\)\(\ge2a^2+2b^2+2c^2+2ac+2bc\)

\(\Leftrightarrow1+1+c-a^2-b^2-c^2+2ab\ge a^2+b^2+c^2+2ab+2ac+2bc\)

\(\Leftrightarrow\left(a+b+c\right)^2\le1+2abc+c+2ab\le1+\frac{1}{4}+1=\frac{9}{4}\)

\(\Rightarrow a+b+c\le\frac{3}{2}\).Nên GTLN của M là \(\frac{3}{2}\) khi \(a=b=c=\frac{1}{2}\)