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Đặt \(\left(\frac{a}{a-1};\frac{b}{b-1};\frac{c}{c-1}\right)\rightarrow\left(x;y;z\right)\) thì có
\(xyz=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)
\(\Leftrightarrow xy+yz+xz+1-\left(x+y+z\right)=0\)
Và BĐT cần chứng minh là \(x^2+y^2+z^2\ge1\)
Đúng vì \(x^2+y^2+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)-2\left(x+y+z\right)+2\)
\(=\left(x+y+z-1\right)^2+1\ge1\)
Vậy ta có ĐPCM
Bài làm :
Ta có :
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ac=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Leftrightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\left(1\right)\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\left(2\right)\)
Thay (1) vào (2) ; ta được :
\(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
=> Điều phải chứng minh
Ta có \(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2ac+2bc=0\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
Ta lại có giả sử
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=3\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3=3.a^2b^2c^2\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3-3.a^2b^2c^2=0\)
\(\Leftrightarrow\left(ab+bc+ac\right)^3-3ca\left(ab+bc\right)\left(ab+bc+ac\right)-3ab^3c\left(-ac\right)-3a^2b^2c^2=0\)
\(\Leftrightarrow0+3a^2b^2c^2-3a^2b^2c^2+0=0\)
\(\Leftrightarrow0=0\left(lđ\right)\)
Vậy bất đẳng thức được chứng minh
Có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)(do a,b,c khác 0)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Suy ra: \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{3}{abc}\)(vì \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\))
Vậy...........
1a)
Đặt \(a^2+a+1=t\Rightarrow a^2+a+2=t+1\)
\(\Rightarrow A=t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(a^2+a-2\right)\left(a^2+a+5\right)\)
Mà \(a>1\Rightarrow\hept{\begin{cases}a^2+a-2>0\\a^2+a+5>0\end{cases}}\forall a>1\)
Vậy A là hợp số
1b)
Ta có :
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1=....=\left(2^{1006}-1\right)\left(2^{1006}+1\right)+1\)
\(=2^{2012}-1+1=2^{2012}\)
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
Ta đặt: \(\hept{\begin{cases}ab=x\\bc=y\\ca=z\end{cases}}\)
\(\Rightarrow x+y+z=0\)
Ta cần chứng minh
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\left(bc\right)^3+\left(ca\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x^3+3xy\left(x+y\right)+y^3\right)+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(z^2+y^2+2xy-yz-zx+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)(đúng)
\(\RightarrowĐPCM\)