help!!!!! cho 3 so a,b,c thoa man a3>36;abc=1 cmr:
a3+3(b2+c2)>3(ab+bc+ca)
cho 3 so thuc a,b,c thoa man a+b+c=3
chung minh rang: a4+b4+c4≥a3+b3+c3
a)cho biet : a+b/a-b=c+a/c-a
chứng minh rằng : a^2=bc
b) cho 4 so kha c0 : a1;a2;a3;a4 thoa man a2^2= a1.a3
a3^2= a2.a4
cmr :
\(\frac{a1^3+a2^3+a3^3}{a2^3+a3^3+a4^3}=\frac{a1}{a4}\)
\(a,\frac{a+b}{a-b}=\frac{c+a}{c-a}\Rightarrow\frac{a+b}{c+a}=\frac{a-b}{c-a}=\frac{a+b+a-b}{c+a+c-a}=\frac{2a}{2c}=\frac{a}{c}\)
\(\text{Suy ra: }\frac{a+b}{c+a}=\frac{a}{c}\Rightarrow c.\left(a+b\right)=a.\left(c+a\right)\Rightarrow ac+bc=ac+a^2\)
=>a2=bc
b)Viết đề rõ lại giúp
bÀI TẬP :a)cho biet : a+b/a-b=c+a/c-a
chứng minh rằng : a^2=bc
b) cho 4 so kha c0 : a1;a2;a3;a4 thoa man a2^2= a1.a3
a3^2= a2.a4
cmr : \(\frac{a1^3+a2^3+a3^3}{a2^3+a3^3+a4^3}=\frac{a1}{a4}\)
nếu thấy đề sai chỗ nào thì sửa lại giúp mik nhé !
Cho 4 so khac o a1,a2,a3,a4 thoa man dieu kien a2^2=a1a3 va a3^2=a2a4. chung minh : a2^3+a3^2+a3^3/a3^2+a3^3+a4^3=a1/a4
Cho a b c khac 0 thoa man a/1=b/2=c/3.
CMR (a+b+c).(1/a+4/b+9/c)=36
cho 2 so thuc a,b thoa man a>1va b>1 chung minh rang\(\frac{a^3+b^3-\left(a^2+b^2\right)}{\left(a-1\right)\left(b-1\right)}\)
\(A=\frac{a^3+b^3-\left(a^2+b^2\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2\left(a-1\right)+b^2\left(b-1\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2}{b-1}+\frac{b^2}{a-1}\)
(chơi 3 cách luôn cho máu :3)
Cách 1, Áp dụng Svacxơ đc
\(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge\frac{\left(a+b\right)^2}{a+b-2}=\frac{t^2}{t-2}\left(t=a+b>2\right)\)
Ta luôn có \(\frac{t^2}{t-2}\ge8\left(1\right)\)thật vậy
\(\left(1\right)\Leftrightarrow t^2\ge8t-16\Leftrightarrow t^2-8t+16\ge0\Leftrightarrow\left(t-4\right)^2\ge0\left(True\right)\)
=> Đpcm
Cách 2, \(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge2\sqrt{\frac{a^2.b^2}{\left(b-1\right)\left(a-1\right)}}=2.\frac{a}{\sqrt{a-1}}.\frac{b}{\sqrt{b-1}}\)
Ta đi c/m \(\frac{a}{\sqrt{a-1}}\ge2\left(#\right)\)thật vậy
\(\left(#\right)\Leftrightarrow a\ge2\sqrt{a-1}\Leftrightarrow a^2\ge4a-4\Leftrightarrow a^2-4a+4\ge0\Leftrightarrow\left(a-2\right)^2\ge0\left(true\right)\)
=> (#) đúng
tương tự\(\frac{b}{\sqrt{b-1}}\ge2\)
\(\Rightarrow A\ge2.2.2=8\)(Đpcm)
Cách 3 , \(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}=\frac{\left(a-1+1\right)^2}{b-1}+\frac{\left(b-1+1\right)^2}{a-1}\)
\(=\frac{\left(a-1\right)^2+2\left(a-1\right)+1}{b-1}+\frac{\left(b-1\right)^2+2\left(b-1\right)+1}{a-1}\)
\(=\frac{\left(a-1\right)^2}{b-1}+\frac{2\left(a-1\right)}{b-1}+\frac{1}{b-1}+\frac{\left(b-1\right)^2}{a-1}+\frac{2\left(b-1\right)}{a-1}+\frac{1}{a-1}\)
\(=\left[\frac{\left(a-1\right)^2}{b-1}+\frac{\left(b-1\right)^2}{a-1}\right]+2\left(\frac{a-1}{b-1}+\frac{b-1}{a-1}\right)+\left(\frac{1}{b-1}+\frac{1}{a-1}\right)\)
\(\ge2\sqrt{\frac{\left(a-1\right)^2.\left(b-1\right)^2}{\left(b-1\right)\left(a-1\right)}}+2.2\sqrt{\frac{a-1}{b-1}.\frac{b-1}{a-1}}+\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}\)
\(=2\sqrt{\left(a-1\right)\left(b-1\right)}+\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}+4\)
\(\ge2\sqrt{2\sqrt{\left(a-1\right)\left(b-1\right)}.\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}}+4\)
\(=2.2+4=8\)
Dấu "=" xảy ra tại a = b = 2
Cho 4 so khac 0:a1,a2,a3,a4 thoa man a2^2=a1xa3 va a3^2=a2xa4
Chung minh rang a1^3+a2^3+a3^3/a2^3+a4^3=a1/a4
tim 3 so nguyen a, b, c thoa man : a+b = -4; b+c = -6; c+a = 12
Số nguyên b là :
[ (-4) + (-6) - 12 ] : 2 = -11
Số nguyên a là:
( - 4) - ( - 11 ) = 7
Số nguyên c là :
( - 6 ) - ( - 11 ) = 5
cho 3 so a b c thoa man 3 dieu kien sau a<b<c