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Định dùng Abel mà quên là ko có điều kiện vs lại thường dùng cho BĐT:v
Đặt \(\frac{x}{a}=m;\frac{y}{b}=n\)
Khi đó \(m+n=1;mn=-2\).Ta cần chứng minh:\(m^3+n^3=7\).Thật vậy !
Ta có:
\(m^3+n^3=\left(m+n\right)^3-3mn\left(m+n\right)=1^3-3\cdot\left(-2\right)\cdot1=1+6=7\)
=> đpcm
a, Chứng minh \(x^3+y^3+z^3=\left(x+y\right)^3-3xy.\left(x+y\right)+z^3\)
Biến đổi vế phải thì ta phải suy ra điều phải chứng minh
b, Ta có: \(a+b+c=0\)thì
\(a^3+b^3+c^3==\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)
( Vì \(a+b+c=0\)nên \(a+b=-c\))
Theo giả thuyết \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)
\(=xyz.\frac{3}{xyz}=3\)
Câu 3:
a: \(G=\dfrac{a^2}{b\left(a+b\right)}-\dfrac{b^2}{a\left(a-b\right)}+\dfrac{-\left(a^2+b^2\right)}{ab}\)
\(=\dfrac{a^3\left(a-b\right)-b^3\left(a+b\right)-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a^4-a^3b-ab^3-b^4-a^4+b^4}{ab\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{-ab\left(a^2+b^2\right)}{ab\left(a-b\right)\left(a+b\right)}=\dfrac{-a^2-b^2}{a^2-b^2}\)
b: \(\dfrac{a}{b}=\dfrac{a+1}{b+5}\)
nên ab+5a=ab+b
=>5a=b
\(G=\dfrac{-a^2-\left(5a\right)^2}{a^2-\left(5a\right)^2}=\dfrac{-a^2-25a^2}{a^2-25a^2}=\dfrac{-26}{-24}=\dfrac{13}{12}\)
câu 1 là :từ a/x + b/y + c/z =0 suy ra (ayz+bxz+cxy)/xyz =0 suy ra ayz+bxz+cxy=0 (1)
vì x/a + y/b + z/c =1 (gt) suy ra (x/a + y/b + z/c )^2 = 1^2 . suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(xy/ab + yz/bc + xz/ac) =1
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2[(ayz+bxz+cxy)/abc = 1 (2)
Từ (1) và (2) suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =1 (đpcm)
\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)
\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)
\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)
\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)
\(=\frac{2005\times2010-6}{2005\times2011}\)
\(=\frac{2004}{2005}\)
Bài 1:
a) Ta có: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2}{x+2y}+\frac{y}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}+\frac{y\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x-4y+xy+2y^2+4}{\left(x-2y\right)\cdot\left(x+2y\right)}\)
b) Ta có: \(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
\(=\frac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\frac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{2x-2y}{x^2+xy+y^2}\)
c) Ta có: \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}\)
\(=\frac{xy}{2x-y}+\frac{x^2-1}{2x-y}\)
\(=\frac{x^2+xy-1}{2x-y}\)
d) Ta có: \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)
\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)
\(=\frac{2x^2-2y^2+2y^2}{x}\)
\(=\frac{2x^2}{x}=2x\)
Bài 2:
a) Ta có: \(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{3\left(4x+1\right)}{6}-\frac{2\left(3x+2\right)}{6}\)
\(=\frac{12x+3-6x-4}{6}\)
\(=\frac{6x-1}{6}\)
b) Ta có: \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)
c) Ta có: \(\frac{x+3}{x^2+1}-\frac{1}{x^2+2}\)
\(=\frac{\left(x+3\right)\left(x^2+2\right)}{\left(x^2+1\right)\left(x^2+2\right)}-\frac{x^2+1}{\left(x^2+2\right)\left(x^2+1\right)}\)
\(=\frac{x^3+2x+3x^2+6-x^2-1}{\left(x^2+1\right)\left(x^2+2\right)}\)
\(=\frac{x^3+2x^2+2x+5}{\left(x^2+1\right)\left(x^2+2\right)}\)
e) Ta có: \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
\(=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x+1\right)\left(x-1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x+1\right)\left(x-1\right)}-\frac{2\cdot2\cdot\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x-3+4x^2-2x-4\left(x^2-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{4x^2+x-3-4x^2+4}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x+1}{2x\left(x+1\right)\left(x-1\right)}=\frac{1}{2x\left(x-1\right)}\)
d) Ta có: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
\(=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{x+2}{\left(3x-2\right)\left(3x+2\right)}\)
f) Ta có: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)
1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
Ta có : \(\frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+2.\frac{xy}{ab}=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-2.\frac{xy}{ab}=1-2\left(-2\right)=5\)
\(\Rightarrow\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)\left(\frac{x^2}{a^2}-2.\frac{xy}{ab}+\frac{y^2}{b^2}\right)\)
\(=1\left(5+4\right)=9\)
Vậy \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=9\)