\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Rightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)

\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Dễ  thấy\(\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)>0\Rightarrow x-50=0\Rightarrow x=50\)

Vậy x = 50

Ta có 

\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà : \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\)

\(\Rightarrow x-50=0\)

\(\Rightarrow x=50\)

Vậy : \(x=50\)

\(\Leftrightarrow\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}-4=0\)

\(\Rightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)nên 

\(\Rightarrow x-50=0\Rightarrow x=50\)

Vậy x=50

ai làm ơn trả lời hộ mình câu này với

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)

a. \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Rightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

\(\Rightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Rightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Rightarrow x-105=0\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Rightarrow x=105\)

b. \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)

\(\Rightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+\frac{25-x}{25}+1+\frac{23-x}{27}+1+\frac{21-x}{29}+1=0\)

\(\Rightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\Rightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

\(\Rightarrow50-x=0\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)

\(\Rightarrow x=50\)

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Dễ dàng thấy nhân tử thứ hai luôn bé thua 0 nên \(x-105=0\)\(\Leftrightarrow x=105\)

b) Kĩ thuật làm tương tự câu a cộng mỗi phân số VT với 1 thì VP=0 và ta có nhân tử chung 50-x

\(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5.\)

\(\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)\)\(=0\)

\(\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\left(50-x\right).\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

=> 50 - x = 0 \(\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)

=> x = 50

\(pt\Leftrightarrow\frac{29}{21}-\frac{x}{21}+\frac{27}{23}-\frac{x}{23}+\frac{25}{25}-\frac{x}{25}+\frac{23}{27}-\frac{x}{27}+\frac{21}{29}-\frac{x}{29}=-5\Leftrightarrow-x\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}\Leftrightarrow-x=\frac{-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}}{\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}}=-50\Leftrightarrow x=50\\ \Rightarrow S=\left\{50\right\}\)

\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)

\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.

pt<=>29-x/21+1+27-x/23+1+...=0

<=>50-x/21+50-x/23+50-x/25+50-x/27+50-x/29=0

<=>(50-x).(1/21+1/23+1/25+1/27+1/29)=0

Do 1/21+1/23+1/25+1/27+1/29>0 nên 50-x=0 hay x=50

a) 3x - 2(5 + 2x) =45 - 2x

=> 3x - 10 - 4x = 45 - 2x

=> 3x - 4x + 2x = 45 + 10

=> x = 55

b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)

=> 5(2x + 17) = 3(x - 3)

=> 10x + 85 = 3x - 9

=> 7x = -94

=> x = -94/7

c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)

=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)

=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)

=> (-11x - 3).7 = (4x - 33).12

= -77x - 21 = 48x - 396

=> x = 3

d) (x - 1)(5x + 3) = (3x - 8)(x - 1)

=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0

=> (x - 1)(2x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\) 

e) (x - 1)(x² + 5x - 2) - (x³ - 1) = 0

=> (x - 1)(x² + 5x - 2) - (x - 1)(x² + x + 1) = 0

=> (x - 1)(4x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)

f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\) 

=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))

=> x = 50

b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)

\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)

f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)

\(\Leftrightarrow x=-66\)