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a) (−2).3....> hoặc \(\ge\).....(−2).5(−2).3.........(−2).5
b) 4.(−2)...< hoặc \(\le\)....(−7).(−2)4.(−2).......(−7).(−2)
c) (−6)2+2....\(\le\) hoặc \(\ge\)....36+2(−6)2+2........36+2
d) 5.(−8).....> hoặc \(\ge\).....135.(−8)
a)ta có:(-2).3=-6 ; (-2).5=-10
Vì -6>-10 nên (-2).3>(-2).5
b)Ta có:4.(-2)=-8 ; (-7).(-2)=14
vì -8<14 nên 4.(-2)<(-7).(-2)
c)Ta có:(-6)2+2=36+2=38 ; 36+2=38
Vì 38=38 nên (-6)2+2=36+2
d)Ta có:5.(-8)=-40 ; 135.(-8)=-1080
Vì -40>-1080 nên 5.(-8) > 135.(-8)
\(\left(a+b\right)\left(a^3+b^3\right)\le2\left(a^4+b^4\right)\)
\(\Leftrightarrow a^4+b^4+a^3b+ab^3\le2\left(a^4+b^4\right)\)
\(\Leftrightarrow a^4+b^4-a^3b-ab^3\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right]\ge0\) * đúng *
b
Hiển hiên
\(\left(a+b\right)\left(a^3+b^3\right)\le2\left(a^4+b^4\right)\)
\(\Leftrightarrow a^4+b^4-a^3b-ab^3\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
Dấu "=" xảy ra <=> a=b
a) 12 + (-8) > 9 + (-8)
b) 13 - 19 < 15 - 19
c) (-4)2 + 7 ≥ 16 + 7
d) 452 + 12 > 450 + 12
a: m<n nên m-n<0
a>b nên a(m-n)<b(m-n)
b: a>b nên a-b>0
m(a-b)<n(a-b)
Đặt \(A=\frac{\left(a+b+c+d\right)\left(a+b+c\right)\left(a+b\right)}{abcde}\)
\(\Rightarrow16A=\frac{\left(a+b+c+d+e\right)^2\left(a+b+c+d\right)\left(a+b+c\right)\left(a+b\right)}{abcde}\)
Áp dụng AM-GM ta có:
\(\Rightarrow16A\ge\frac{4e\left(a+b+c+d\right)^2\left(a+b+c\right)\left(a+b\right)}{abcde}\)
\(\Rightarrow16A\ge\frac{4e.4d\left(a+b+c\right)^2\left(a+b\right)}{abcde}\)
\(\Rightarrow16A\ge\frac{4e.4d.4c\left(a+b\right)^2}{abcde}\)
\(\Rightarrow16A\ge\frac{4e.4d.4c.4ab}{abcde}\)
\(\Rightarrow A\ge16\)
Dấu "=" xảy ra khi đồng thời:
\(\text{a+b+c+d+e=4, a+b+c+d=e, a+b+c=d, a+b=c, a=b}\)
\(\Rightarrow e=2,d=1,c=\frac{1}{2},a=\frac{1}{4},b=\frac{1}{4}\)
e)\(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=1+\frac{b}{a}+\frac{a}{b}+1\)
\(=\left(1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\)
\(=2+\left(\frac{a.a}{b.a}+\frac{b.b}{a.b}\right)\)
\(=2+\frac{a.a+b.b}{b.a}\)
Vì \(\frac{a.a+b.b}{a.b}>=2\)
Nên \(2+\frac{a.a+b.b}{a.b}>=2+2=4\)
Hay \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)>=4\)
a) \(a^2+b^2-2ab\)
\(=\left(a-b\right)^2\)
Vì \(\left(a-b\right)^2\) là binh phương của một số nên \(\left(a-b\right)^2>=0\)
Hay \(a^2+b^2-2ab>=0\)
c) \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow\)\(\left(ax\right)^2+2axby+\left(by\right)^2\le\left(ax\right)^2+\left(ay\right)^2+\left(bx\right)^2+\left(by\right)^2\)
\(\Leftrightarrow\)\(2axby\le\left(ay\right)^2+\left(bx\right)^2\)
\(\Leftrightarrow\)\(\left(ay\right)^2-2axby+\left(bx\right)^2\ge0\)
\(\Leftrightarrow\)\(\left(ay-bx\right)^2\ge0\) luôn đúng
Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{a}{x}=\frac{b}{y}\)