Question

B1; Cho biết \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)và \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\).Chứng minh rằng

a+b+c=abc

B2; Cho 3 số x,y,z thỏa mãn x,y,z=1. CMR:

\(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)

B3: Giải phương trình

\(\left(x^2-x+1\right)^2+5x^4=6x^2\left(x^2-x+1\right)\)

Làm Ơn Giúp với

Answer 1

\(Dat:\left\{{}\begin{matrix}x^2-x+1=a\\x^2=h\end{matrix}\right.\Rightarrow a^2+5h^2=6ah\Rightarrow a^2-6ah+5h^2=\left(a^2-ah\right)+5\left(h^2-ah\right)=a\left(a-h\right)+5h\left(h-a\right)=\left(a-5h\right)\left(a-h\right)=0\Leftrightarrow\left[{}\begin{matrix}a=h\\a=5h\end{matrix}\right.\) giai tiep

Answer 2

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\Rightarrow a+b+c=abc\left(nhan:abc\right)\)

Answer 3

B1

Có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=4\)

<=> \(2+\frac{2.\left(a+b+c\right)}{abc}=4\) (vì \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\))

<=> \(\frac{2\left(a+b+c\right)}{abc}=2\)

<=> \(a+b+c=abc\)(đpcm)

Answer 4

viết lại đề bài 2 đi

Answer 5

Có \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)

=\(\frac{1}{xyz+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+\frac{1}{y}}\) (vì xyz=1 )

=\(\frac{1}{x\left(1+y+yz\right)}+\frac{1}{1+y+yz}+\frac{1}{\frac{y+zy+1}{y}}\)

=\(\frac{1}{\frac{1}{zx}\left(1+y+yz\right)}+\frac{1}{y+1+yz}+\frac{y}{y+zy+1}\)

=\(\frac{zx+1+y}{1+y+zx}=1\)

Vậy \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=1\)