bằng 10 nha mg

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)

\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)

\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)

\(\Leftrightarrow x=1999\)(thỏa mãn)

Vậy \(x=1999\)

chịu

\(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}=-4\\\)

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

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

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

Vì\(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}\ne0\)

\(\Rightarrow50-x=0\)

\(\Leftrightarrow x=50\)

29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 = -4

<=> (29-x/21 + 1) + (27-x/23 + 1) + (25-x/25 + 1) + (23-x/27 + 1) = -4 + 4

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

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

Mà 1/21 + 1/23 + 1/25 + 1/27 > 0

Nên 50-x=0 <=> x=50

Vậy ...

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

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

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

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

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

\(\Leftrightarrow x=105\)

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

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

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

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

\(\Leftrightarrow50-x=0;\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)\ne0\)

\(\Leftrightarrow x=50\)

a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)

mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)

nên \(x^2-10x-2000=0\)

\(\Leftrightarrow x^2+40x-50x-2000=0\)

\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)

\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)

Vậy: S={-40;50}

\(\Leftrightarrow\dfrac{x-5}{95}-1+\dfrac{x-132}{32}+1=\dfrac{x-131}{31}+1+\dfrac{x-10}{90}-1\)

=>x-100=0

hay x=100

\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10

⇔\(\dfrac{2625\left(x-130\right)}{52500}\)+\(\dfrac{2100\left(x-100\right)}{52500}\)+\(\dfrac{1750\left(x-60\right)}{52500}\)+\(\dfrac{1500\left(x-10\right)}{52500}\)=\(\dfrac{525000}{52500}\)

⇔2625\(x\)-341250+2100\(x\)-210000+1750\(x\)-105000+1500\(x\)-15000=525000

⇔ 7975\(x\) = 1196250

⇔ \(x\) = \(\dfrac{1196250}{7975}\)

⇔\(x \) = 150