cho a,b,c,d la cac so thuc thoa ma dang thuc a+b+c+d=0.chung minh rang:
\(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
Những câu hỏi liên quan
cho a , b, c la cac so thuc duong thoa man he thuc a+b+c=6abc
Chung minh rang \(\dfrac{bc}{a^3\left(c+2b\right)}+\dfrac{ac}{b^3\left(a+2c\right)}+\dfrac{ab}{c^3\left(b+2a\right)}\ge2\)
Cho cac so nguyen a, b, c. d thoa man \(a^3+b^3=2\left(c^3-8d^3\right)\). Chung minh rang \(a+b+c+d⋮3\)
Ta có: \(a^3+b^3=2\left(c^3-8d^3\right)\)
\(\Leftrightarrow a^3+b^3=2c^3-16d^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3c^3-15d^3=3\left(c^3-5d^3\right)\)
\(VP⋮3\Rightarrow a^3+b^3+c^3+d^3⋮3\)(1)
Ta có: \(a^3-a+b^3-b+c^3-c+d^3-d\)
\(=\left(a-1\right)a\left(a+1\right)+\left(b-1\right)b\left(b+1\right)\)
\(+\left(c-1\right)c\left(c+1\right)+\left(d-1\right)d\left(d+1\right)\)
Vì tích 3 số tự nhiên liên tiếp chia hết cho 3 nên \(\left(a-1\right)a\left(a+1\right)+\left(b-1\right)b\left(b+1\right)\)
\(+\left(c-1\right)c\left(c+1\right)+\left(d-1\right)d\left(d+1\right)\)chia hết cho 3 (2)
Từ (1) và (2) suy ra \(a+b+c+d⋮3\left(đpcm\right)\)
cho a, b, c la cac so thuc duong thoa man a + b + c =abc chung minh rang :
\(\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ac\right)}+\frac{1}{c^2\left(1+ab\right)}\le\frac{1}{4}\)
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
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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
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voi a,b,c,d, la cac so duong thoa man a*b = c*d =1 chung minh bat dang thuc : ( a+b )*( c+d ) +4 >= 2*( a+b+c+d ) cac ban oi giup minh voi OK
voi a,b,c,d la cac so duong thoa man a*b = c*d = 1. Chung minh bat dang thuc ( a+b )*( c+d ) + 4 >= 2( a+b+c+d )
14 tháng 10 2019 lúc 18:46
cho cac so thuc duong a b c thoa a^2+b^2+c^2>=3 chung minh
\(\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}+\frac{\left(b+1\right)\left(c+2\right)}{\left(c+1\right)\left(c+5\right)}+\frac{\left(c+1\right)\left(a+2\right)}{\left(a+1\right)\left(a+5\right)}\ge\frac{3}{2}\)
Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)
Thật vậy, BĐT trên tương đương:
\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)
\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)
\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)
Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)
\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)
\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)
\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)
\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
1.tìm các nghiem nguyen cua phuong trinh: 54x^3+1=y^3
2.cho x+y=1 và xy khac 0.chung mih \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
3.cho a,b,c la cac so thuc duong.chung minh :\(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)^2+\frac{14abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
Câu 2 thế y = 1 - x rồi quy đồng như bình thường là ra bn nhé
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Cho a,b,c la cac so thuc >0
Cmr \(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(c+a\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}>=1\)