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

Bài 1.

a) Do hai phân thức bằng nhau , ta có :

( x +2)P( x² - 2²) = ( x - 1)Q( x -2)

=( x + 2)P( x - 2)( x + 2) = ( x - 1)Q( x - 2)

Suy ra : P = x - 1 ; Q = ( x + 2)²

b) Do hai phân thức bằng nhau , ta có :

( x + 2)P(x² - 2x + 1) = ( x - 2)Q( x² - 1)

= ( x + 2)P( x - 1)² = ( x - 2)Q( x - 1)( x + 1)

Suy ra : P = ( x - 2)( x + 1) = x² - x - 2

Q = ( x + 2)( x - 1) = x² + x + 2

Bài 2. a) Do : \(\dfrac{P}{Q}=\dfrac{R}{S}=>PS=QR\)

Xét : ( P + Q)S= PS + QS = QR + QS = Q( R + S)

-> \(\dfrac{P+Q}{Q}=\dfrac{R+S}{S}\)

b) Do : \(\dfrac{P}{Q}=\dfrac{R}{S}=>PS=QR\)

Xét : ( S - R)P = PS - PR = QR - PR = R( Q - P)

-> \(\dfrac{R-S}{R}=\dfrac{Q-P}{P}\)

- > \(\dfrac{R}{R-S}=\dfrac{P}{Q-P}\)

Chọn B

\(R=\frac{1}{2.32}+\frac{1}{3.33}+......+\frac{1}{1976.2006}\Rightarrow30R=\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1976}-\frac{1}{32}-\frac{1}{33}-....-\frac{1}{2006}=\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{31}-\frac{1}{1977}-\frac{1}{1978}-....-\frac{1}{2006};S=\frac{1}{2.1977}+\frac{1}{3.1978}+....+\frac{1}{31.2006}=\Rightarrow1975S=\frac{1}{2}+\frac{1}{3}+....+\frac{1}{31}-\frac{1}{1977}-\frac{1}{1978}-....-\frac{1}{2006}=R\Rightarrow30R=1975S\Rightarrow R=\frac{1975}{30}S=\frac{395}{6}\Rightarrow\frac{R}{S}=\frac{395}{6}\)

Ta có :

\(\dfrac{31}{2}.\dfrac{32}{2}.\dfrac{33}{2}.....\dfrac{60}{2}=31.32.33.....\dfrac{60}{2^{30}}\)

(31.32.33....60)(1.2.3....30)/2³⁰(1.2.3....30)

= (1.3.5.....59)(2.4.6.....60 )/( 2.4.6....60 ) = 1.3.5....59

\(\Rightarrow P=Q\)

\(\dfrac{1}{R\left(x\right)}=\dfrac{1}{x\left(x+2\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)

\(\Rightarrow S=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2022}-\dfrac{1}{2024}+\dfrac{1}{2023}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2024}-\dfrac{1}{2025}\right)+\dfrac{1}{2.2023}\)

Một kết quả rất xấu

a, Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Rightarrow a=b=c\)

b, Ta có: \(a^2=bc\Rightarrow\dfrac{a}{c}=\dfrac{b}{a}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{c}=\dfrac{b}{a}=\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrowđpcm\)

a) $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1$

(tính chất dãy tỉ số bằng nhau)

$\dfrac{a}{b}=1=>a=b$

$\dfrac{b}{c}=1=>b=c$

$\dfrac{c}{a}=1=>c=a$

Vậy a = b = c.

b) Ta có : $a^2=bc=>\dfrac{a}{c}=\dfrac{b}{a}=\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}$(tính chất dãy tỉ số bằng nhau)

$=>\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}$

$=>\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}$

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

\(a^2=bc\Rightarrow\dfrac{a}{c}=\dfrac{b}{a}\)

Đặt:

\(\dfrac{a}{c}=\dfrac{b}{a}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=ak\end{matrix}\right.\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{ck+ak}{ck-ak}=\dfrac{k\left(c+a\right)}{k\left(c-a\right)}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

Ta có: \(\dfrac{9}{11}=\dfrac{36}{44};\dfrac{5}{4}=\dfrac{55}{44}\)

Khi đó giá trị 1 phần là: \(38:\left(55-36\right)=2\)

\(\Rightarrow\) Tử số: \(36.2=72\)

Mẫu: \(44.2=88\)

Vậy \(\dfrac{a}{b}=\dfrac{72}{88}.\)

a)S=\(\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

=\(\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(\left(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right):\dfrac{2x-6}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

=\(\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

=\(\dfrac{6\left(2x-6\right)x\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(2x-6\right)}+\dfrac{x}{6-x}\)

=\(\dfrac{6}{x-6}+\dfrac{x}{6-x}\)

=\(\dfrac{6}{x-6}-\dfrac{x}{x-6}=\dfrac{6-x}{x-6}=-1\)

b ) S khi rút gọn=-1 => mọi giá trị của x đều thỏa mãn S=-1