Bài này lớp 7 thôi mà !

a) Cộng 1 vào 2 vế

b) Nghịch đảo 2 vế,trừ 1 ở 2 vế rồi lại nghịch đảo 2 vế

mình chỉ học lop5 thôi thông cảm

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

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: \(\dfrac{P}{Q-P}=\dfrac{R}{S-R}\)

\(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(B=\frac{x-\sqrt{x}+3\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\frac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)

b/ \(C=\left(\frac{\sqrt{x}-1}{\sqrt{x}-5}.\frac{\sqrt{x}+6}{\sqrt{x}-1}\right).\frac{\sqrt{x}-5}{\sqrt{x}}\)

\(C=\frac{\sqrt{x}+6}{\sqrt{x}-5}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}+6}{\sqrt{x}}=1+\frac{6}{\sqrt{x}}\)

Cai này thì so sánh \(\frac{6}{\sqrt{x}}\) vs 2

Nếu0< x<9\(\Rightarrow\frac{6}{\sqrt{x}}< 2\)

Nếu x=9\(\Rightarrow\frac{6}{\sqrt{x}}=2\)

Nếu x>9\(\Rightarrow\frac{6}{\sqrt{x}}>2\)

\(S=\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\\ =\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}\\ =\frac{\left(2a-1\right)\left(b+1\right)}{3\left(b+1\right)\left(2a-1\right)}\\=\frac{1}{3}\)

\(A=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)

\(B=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

b/ \(P=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

Có \(\sqrt{x}-5< \sqrt{x}+3\Rightarrow P< 1\)