Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{a-b}{a^2+ab}+\frac{a+b}{a^2-ab}=\frac{3a-b}{a^2-b^2}\)
\(\Leftrightarrow\frac{a-b}{a\left(a+b\right)}+\frac{a+b}{a\left(a-b\right)}=\frac{3a-b}{\left(a-b\right)\left(a+b\right)}\)
\(\Leftrightarrow\frac{\left(a-b\right)^2+\left(a+b\right)^2}{a\left(a-b\right)\left(a+b\right)}=\frac{3a^2-ab}{a\left(a-b\right)\left(a+b\right)}\)
\(\Leftrightarrow a^2-2ab+b^2+a^2+2ab+b^2=3a^2-ab\)
\(\Leftrightarrow2a^2+2b^2=3a^2-ab\)
\(\Leftrightarrow a^2-ab=2b^2\)
\(\Leftrightarrow\left(a^2+ab\right)-\left(2ab+2b^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-2b\right)=0\Rightarrow\orbr{\begin{cases}a=-b\left(l\text{do }\left|a\right|\ne\left|b\right|\right)\\a=2b\left(TM\right)\end{cases}}\)
Thay a = 2b vào B tự tính
B sai đề
\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
\(\Leftrightarrow\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)
\(\Leftrightarrow\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)=64c^2\)
\(\Leftrightarrow\left(2a+2b\right)\left(8a-8b\right)=16\left(a^2-b^2\right)\)
\(\Leftrightarrow16\left(a^2-b^2\right)=16\left(a^2-b^2\right)\left(true\right)\)
Vậy \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)khi \(a^2-b^2=4c^2\)
(5a−3b+8c)(5a−3b−8c)(5a-3b+8c)(5a-3b-8c)
=(5a−3b)2−(8c)2=(5a-3b)2-(8c)2
=(5a−3b)2−16.4c2=(5a-3b)2-16.4c2
Thay a2−b2=4c2a2-b2=4c2 ta có :
=25a2−30ab+9b2−16(a2−b2)=25a2-30ab+9b2-16(a2-b2)
=25a2−30ab+9b2−16a2+16b2=25a2-30ab+9b2-16a2+16b2
=9a2−30ab+25b2=9a2-30ab+25b2
=(3a−5b)2(đpcm)=(3a-5b)2(dpcm)
Có bđt x2 + y2 \(\ge\)( x + y) /2 ( * )
( * ) \(\Leftrightarrow\)2x2 + 2y2\(\ge\)x2 + 2xy + y2 \(\Leftrightarrow\)x2 - 2xy +y2 \(\ge\)0 \(\Leftrightarrow\)( x- y)2 \(\ge\)0
Dấu "=" xảy ra khi x = y =1
Thay bđt ( * ) vào bài toán ta có:
a4 + b4 \(\ge\)(a2 + b2)2 / 2 \(\Leftrightarrow\)a4 + b4 \(\ge\)[(a + b)2 /2]2 /2 = 2 ( đpcm)
Dấu "=" xảy ra khi a = b = 1
Thay a = b = 1 vào bt ta có:
\(\frac{5a^2}{b}\)+ \(\frac{3b^2}{a^2}\)\(\ge\)8