sua de \(\frac{3}{x^4-x^3+x-1}\) \(-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}\) (dk \(x\ne+-1\) )

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

=\(\frac{2}{x^4+x^2+1}>0\)

 P\(< \frac{32}{9}\Leftrightarrow\frac{2}{x^4+x^2+1}< \frac{32}{9}\)

\(\Leftrightarrow16x^4+16x^2+7>0\)

\(\Rightarrow\)\(0< P< \frac{32}{9}\) VOI X KHAC 1;-1

Tiếng gj cơ ban

TIẾNG GÌ

\(a^4+b^4+c^4=\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\ge a^2b^2+a^2c^2+b^2c^2\)

\(=\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge ab.bc+bc.ca+ab.ca=abc\left(a+b+c\right)\)(đpcm)

Dấu = xảy ra khi \(a=b=c\)

thôi, mk ra rồi . cảm ơn các bạn nhiều !  :)

\(\sqrt{2-\sqrt{3}}=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{2.\left(2-\sqrt{3}\right)}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{\sqrt{2}.\left(\sqrt{3}-1\right)}{\sqrt{2}.\sqrt{2}}=\frac{\sqrt{2}.\sqrt{3}-\sqrt{2}.1}{\sqrt{2.2}}=\frac{\sqrt{2.3}-\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{6}-\sqrt{2}}{2}\)

a, \(MS\perp BC;MR\perp AC\) ( gt ) nên \(\widehat{MSC}=\widehat{MRC}=90^o\)

Tam giác ABC có \(\widehat{C}=90^o\)( gt ) do đó \(\widehat{MSC}=\widehat{MRC}=\widehat{SCR}=90^o\)

Vậy tam giác cân ABC là hình tam giác ( vì có 3 góc )

P/s: Tham khảo nhé