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1: \(\Leftrightarrow a^5-a^4b+b^5-ab^4>=0\)
\(\Leftrightarrow a^4\left(a-b\right)-b^4\left(a-b\right)>=0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\left(a+b\right)\cdot\left(a^2+b^2\right)>=0\)(luôn đúng khi a,b dương)
Ta có: \(P=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}-\frac{1}{\left(a-b\right)\left(b-c\right)}-\frac{1}{\left(a-c\right)\left(c-b\right)}\)
\(=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{a-c}{\left(a-c\right)\left(a-b\right)\left(b-c\right)}+\frac{a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}=\frac{0}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}=0\)(đpcm)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1< 3^{32}\)
Gợi ý: Sử dụng liên tục tính chất \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
2(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 + 1)
= (38 - 1)(38 + 1)(316 + 1)
= (316 - 1)(316 + 1)
= 332 - 1 < 332
Ta có: ab(a+b)-\(\frac{ab\left(a^3+b^3\right)}{a^2+2ab+b^2}\)
=\(ab\left(a+b\right)\)-\(\frac{ab\left(a^3+b^3\right)}{\left(a+b\right)^2}\)
=\(\frac{ab\left(a+b\right)^3}{\left(a+b\right)^2}\)-\(\frac{ab\left(a^3+b^3\right)}{\left(a+b\right)^2}\)
=\(\frac{ab\left[\left(a+b\right)^3-\left(a^3+b^3\right)\right]}{\left(a+b\right)^2}\)
=\(\frac{ab.3ab\left(a+b\right)}{\left(a+b\right)^2}\)
=\(\frac{3\left(ab\right)^2}{a+b}\)
Mình năm nay lớp 7 nên chưa chắc đúng đâu nha :
\(a\left(b+1\right)+b\left(a+1\right)=\left(a+b\right)\left(b+1\right)\left(1\right)\)
=) \(ab+a+ab+b=\left(a+b\right)\left(b+1\right)\)
=) \(1+a+1+b=\left(a+b\right)\left(b+1\right)\)
=) \(2+a+b=\left(a+b\right)\left(b+1\right)\)
=) \(2=\left(a+b\right)\left(b+1\right)-\left(a+b\right)\)
=) \(2=\left(a+b\right).\left(b+1-1\right)\)=) \(2=\left(a+b\right).b=ab+b^2\)
=) \(2=1+b^2\)=) \(b^2=2-1=1\)=) \(b=1\)
=) \(a=1:b=1:1=1\)
Thay vào \(\left(1\right)\):
\(1.\left(1+1\right)+1.\left(1+1\right)=\left(1+1\right).\left(1+1\right)\)
=) \(1.2+1.2=2.2\)
=) \(4=4\)( Đúng )
Vậy nếu \(ab=1\Leftrightarrow a\left(b+1\right)+b\left(a+1\right)=\left(a+b\right)\left(b+1\right)\left(ĐPCM\right)\)