chun to ran neu \(\frac{a}{b}\)<\(\frac{c}{d}\)(b.0,d>0) thi \(\frac{a}{b}\)<\(\frac{a+c}{b+d}\)<\(\frac{c}{d}\)
Những câu hỏi liên quan
chung to neu a,b thuoc n* \(\frac{1}{a}+\frac{1}{b}=\frac{4}{a+b}\) thi a=b
cmr
a] neu\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\) trong do b khac 0 thi c=0
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-\left(a-b+c\right)}{a+b-c-\left(a-b-c\right)}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=1\)
\(\Rightarrow\frac{a+b+c}{a+b-c}=1\Rightarrow a+b+c=a+b-c\Rightarrow2c=0\Rightarrow c=0\)
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1. Chmr neu a, b>0 thi
\(\sqrt{a}+\sqrt{b}\le\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\)
\(\frac{\sqrt{a}^2}{\sqrt{b}}+\frac{\sqrt{b}^2}{\sqrt{a}}\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{b}+\sqrt{a}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" xảy ra khi \(a=b\)
Chung minh rang neu : \(0<\frac{a}{b}<1;b>0;m>0\) thi \(\frac{a}{b}<\frac{a+m}{b+m}\)
cac ban khong lam thi minh lam nhe
sang tien cho ****
he he he he!
Vi :\(0<\frac{a}{b}<1\left(b>0\right)\) nen a<b ma m>0, do do am<bm , them ab vao 2 ve :
ab+am<ab+bm hay a(b+m)<b(a+m) ma b>0 va b+m>0 nen suy ra :
\(\frac{a}{b}<\frac{a+m}{b+m}\)
**** nhe moi ng
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Gia su \(x=\frac{a}{m},y=\frac{b}{m}\)va x<y.Hay chung to rang neu chon \(z=\frac{a+b}{2m}\)thi ta co x<y<z
Su dung tinh chat neu a,b,c thuoc zva a<b thi a+c<b+c
giup mik voi nha tik cho cam on
Do x < y
=> \(\frac{a}{m}< \frac{b}{m}\)
=> \(\frac{a}{m}+\frac{a}{m}< \frac{a}{m}+\frac{b}{m}< \frac{b}{m}+\frac{b}{m}\)
=> \(\frac{2a}{m}< \frac{a+b}{m}< \frac{2b}{m}\)
=> \(\frac{a}{m}< \frac{a+b}{m}:2< \frac{b}{m}\)
=> \(\frac{a}{m}< \frac{a+b}{2m}< \frac{b}{m}\)
=> x < z < y
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x. (x^2)^3 = x^5
x^7 ≠ x^5
Nếu,
x^7 - x^5 = 0
mủ lẻ nên phương trình có 3 nghiệm
Đáp số:
x = -1
hoặc
x = 0
hoặc
x = 1
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cmr neu \(\sqrt{a.a'}+\sqrt{b.b'}+\sqrt{c.c'}=\sqrt{\left(a+b+c\right)\left(a'+b'+c'\right)}\)
voi a,a',b,b',c,c'>0 thi \(\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}\)
G/s x = \(\frac{a}{m}\) , y = \(\frac{b}{m}\) ( a, b, m \(\in Z\) , m > 0 ) , x < y . Hay chung to rang neu chon z= \(\frac{a+b}{2m}\) thi ta co x < z < y .
chung minh rang :
Neu a,b trai dau thi \(\frac{b-a}{b\sqrt{\frac{-a}{b}}}=\frac{a-b}{a\sqrt{\frac{-b}{a}}}\)
CMR neu a+b=1 thi: \(\frac{b}{a^3-1}-\frac{a}{b^3-1}=\frac{2\left(a-b\right)}{a^2b^2+3}\)