mũ 2 từng vế ta có

\(7-x^2=x^2-2x+\)\(1\)

\(\Rightarrow x^2-2x+1-7+x^2=0\)

\(\Rightarrow2x^2-2x-6=0\)

\(\Rightarrow x^2-x-3=0\)

\(\Rightarrow x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{13}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{13}{4}\)

\(\Rightarrow x-\frac{1}{2}=\sqrt{\frac{13}{4}}hay-\sqrt{\frac{13}{4}}\)

\(\Rightarrow x=\frac{1}{2}+\sqrt{\frac{13}{4}}hay=\frac{1}{2}-\sqrt{\frac{13}{4}}\)

\(dk\hept{\begin{cases}x\ge-\sqrt{7}\\x\le\sqrt{7}\\x\ge1\end{cases}\Rightarrow1\le x\le\sqrt{7}}\)

\(\Leftrightarrow7-x^2=x^2-2x+1\Leftrightarrow x^2-x=3\)

\(\left(x-\frac{1}{2}\right)^2=\frac{1}{4}+3=\frac{13}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1-\sqrt{13}}{2}\left(loai\right)\\x=\frac{1+\sqrt{13}}{2}\left(nhan\right)\end{cases}}\)

no i can't

gọi 2 loại vải lần lượt là :a1 ; b1

gọi số mét vải lần lươt là :a2 ; b2

Vì số tiền và số mét là 2 đại lượng TLN nên :

Ta có : \(\frac{a1}{a2}=\frac{b2}{b1}=\frac{126}{a2}=\frac{90}{100}\)

\(\Rightarrow126.100=a2.90=12600\)

\(\Rightarrow a2=12600:90=140\)

Vậy số mét loại 2 mua được là 140 m

kick mình nha :3

Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) suy ra \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

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

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\\\frac{1}{c}+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}\end{cases}}\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)

Khi đó \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

\(\frac{1}{2^2}< \frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}\right)\)

\(\frac{1}{3^2}< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)\)

\(\frac{1}{4^2}< \frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)\)

\(\frac{1}{5^2}< \frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)\)

\(\frac{1}{\left(n-1\right)^2}< \frac{1}{2}\left(\frac{1}{n-2}-\frac{1}{n}\right)\)

\(\frac{1}{n^2}< \frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\)

\(A< 1\)

Ta có

\(A=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)...\left(\frac{3^6}{9}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)...\left(\frac{729}{9}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)=0\)

vì 729/9=81

Vậy A=0

k me đi

\(\left(\frac{3}{4}-81\right).\left(\frac{3^2}{5}-81\right).\left(\frac{3^3}{6}-81\right).\left(\frac{3^4}{7}-81\right).\left(\frac{3^5}{8}-81\right).\left(\frac{729}{9}-81\right)....\left(\frac{3^{2013}}{2016}-81\right)\)

=>....................................................................................................................(81-81)..............................................

=>.....................................................................................................................0.....................................................

=>A=0