trong 3 số a,bc có 1 số âm ,1so duong,1so 0
Tim a,b,c thoa man biet
|a|=b^2(b-c)
Những câu hỏi liên quan
a) cho 2 stn a va b voi a<b thoa man 3(a+b) =5(a-b)
tim thuong cua 2 so
b) Tim cac so Nguyen Duong a, b, c biet rang : a3-b3-c3 =3abc va a2=2(b+c)
cho 3 so thuc duong a, b, c thoa man 1/a+1/c=2/b. tim GTNN cua (a+b)/(2a-b)+(b+c)(/2c-b)
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2ac}{a+c}\)
\(P=\frac{a+b}{2a-b}+\frac{b+c}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{\frac{2ac}{a+c}+c}{2c-\frac{2ac}{a+c}}=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\ge4\)
Dấu "=" xảy ra khi \(a=b=c\)
Tim a,b,c thoa man khac 0 thoa man:
a+b-2/c= b+c+1/a= c+a+1/b= a+b+c/ 2
moi nguoi giup toi voi........cho ket qua thoi nha1. gia tri x+y+z biet :lx-5l+ly-4l+lz-4l02.so cac uoc nguyen duong cu 2^6.3^33.so cac uoc nguyen duong cua 184.so duong thang tao thanh boi 8 biem phan biet doi mot khong thang hang5.so tu nhien n thoa man 2^n+1:4^210246.UCLL(12,18,24)7.so tu nhien n thoa man 2+4+6.......+n1108.tim x 576.15:x5409.gia tri a+b+c bieta+b10,a+c25,b+c25so phan tu cua tap hop A cac so le nho hon 23
Đọc tiếp
moi nguoi giup toi voi........cho ket qua thoi nha
1. gia tri x+y+z biet :
lx-5l+ly-4l+lz-4l=0
2.so cac uoc nguyen duong cu 2^6.3^3
3.so cac uoc nguyen duong cua 18
4.so duong thang tao thanh boi 8 biem phan biet doi mot khong thang hang
5.so tu nhien n thoa man 2^n+1:4^2=1024
6.UCLL(12,18,24)
7.so tu nhien n thoa man 2+4+6.......+n=110
8.tim x 576.15:x=540
9.gia tri a+b+c biet
a+b=10,a+c=25,b+c=25
so phan tu cua tap hop A cac so le nho hon 23
Cho a,b,c duong thoa man a +b+c=1
Tim GTLN P=\(\sqrt{\dfrac{ab}{ab+c}}+\sqrt{\dfrac{bc}{bc+a}}+\sqrt{\dfrac{ac}{ac+b}}\)
Sử dụng AM-GM, ta có
\(P=\sum\sqrt{\dfrac{ab}{ab+c}}=\sum\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}=\sum\sqrt{\dfrac{ab}{\left(c+b\right)\left(c+a\right)}}\le\dfrac{1}{2}\sum\dfrac{a}{c+b}+\dfrac{b}{c+a}=\dfrac{3}{2}\)
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tim a,b,c khac 0 thoa man :a+b+c=(ab+ac)/2=(bc+ba)/3=(ca+cb)/4
cho cac so duong a,b,c thoa man : ab+a+b=3
tim GTNN cua bieu thuc C=a^2+b^2
Cho a, b, c > 0 thoa man a + b + c = 3.
Tim GTNN : \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac+ab+bc+ac+a^2+b^2+c^2}+\dfrac{7}{ab+bc+ac}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương:
\(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)
Ta có: \(\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
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Cho a, b, c > 0 thoa man a + b + c = 3.
Tim GTNN : \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\)
Áp dụng BĐT Cauchy-Schwarz ta có
BT\(\ge\)\(\frac{\left(1+1+1\right)^2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}=\frac{9}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}\)
\(=\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}+\frac{7}{ab+bc+ac}\)
\(\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}+\frac{7}{ab+bc+ac}\)\(=1+\frac{7}{ab+bc+ac}\)
Ta lại có ab+bc+ac =< (a+b+c)^2/3 =3
\(\Rightarrow BT\ge1+\frac{7}{3}=\frac{10}{3}\)
Vậy GTNN là \(\frac{10}{3}\)khi a=b=c=1
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