mình chỉ học lop5 thôi thông cảm

ta có P/Q = R/S => PS= RQ (1)

P/Q-P = R/S-R => P( S-R) = R(Q-P)

                        => PS -PR = RQ-RP

từ (1) => P/Q-P= R/S-R (bn tự kết luận nhé

còn người ta cho Q khác P để Q-P khác 0 vì Q-P là mẫu số và R-S cũng vậy nên S khác R

a)

\(\dfrac{P}{Q}=\dfrac{R}{S}\Rightarrow PS=QR\)

\(\Leftrightarrow PS+QS=QR+QS\)

\(\Leftrightarrow S\left(P+Q\right)=Q\left(R+S\right)\)

điều kiện Q,s khác 0 => chia hau vế cho QS

\(\Leftrightarrow\dfrac{S\left(P+Q\right)}{QS}=\dfrac{Q\left(R+S\right)}{QS}\Leftrightarrow\dfrac{\left(P+Q\right)}{Q}=\dfrac{\left(R+S\right)}{S}\) đpcm

ĐKXĐ: \(x\ge0;x\ne1\)

\(S=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(\frac{1}{3}-S=\frac{1}{3}-\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{x-2\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}>0;\forall x>0;x\ne1\)

\(\Rightarrow S< \frac{1}{3}\)

a: \(\dfrac{P}{Q}=\dfrac{R}{S}\)

nên \(\dfrac{P}{Q}+1=\dfrac{R}{S}+1\)

hay \(\dfrac{P+Q}{Q}=\dfrac{R+S}{S}\)

b: P/Q=R/S=k

=>P=Qk; R=Sk

\(\dfrac{P}{Q-P}=\dfrac{Qk}{Q-Q\cdot k}=\dfrac{k}{1-k}\)

\(\dfrac{R}{S-R}=\dfrac{S\cdot k}{S-S\cdot k}=\dfrac{k}{1-k}\)

Do đó: \(\dfrac{P}{Q-P}=\dfrac{R}{S-R}\)

Áp dụng Viet với lưu ý \(tanA+tanB+tanC=tanA.tanB.tanC\) ta có:

\(x_4+tanA+tanB+tanC=p\) (1)

\(x_4\left(tanA+tanB+tanC\right)+tanA.tanB+tanB.tanC+tanC.tanA=q\) (2)

\(x_4\left(tanA.tanB+tanB.tanC+tanC.tanA\right)+tanA.tanB.tanC=r\)(3)

\(x_4.tanA.tanB.tanC=s\) (4)

\(\left(1\right)\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC=p-x_4\)

\(\left(4\right)\Rightarrow x_4\left(p-x_4\right)=s\)

Thế vào (2):

\(x_4\left(p-x_4\right)+tanA.tanB+tanB.tanC+tanC.tanA=q\)

\(\Rightarrow tanA.tanB+tanB.tanC+tanC.tanA=q-x_4\left(p-x_4\right)=q-s\)

Thế vào (3):

\(x_4\left(q-s\right)+p-x_4=r\)

\(\Rightarrow p-r=x_4\left(1-q+s\right)\Rightarrow x_4=\frac{p-r}{1-q+s}\)

*ba góc