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 ...

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 \(x=1999\)

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

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

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

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

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

\(\Leftrightarrow x=50\)

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

=>50-x=0

hay x=50

c: \(\Leftrightarrow\dfrac{x-2}{2001}+1=\dfrac{x-1}{2002}+\dfrac{x}{2003}\)

\(\Leftrightarrow\left(\dfrac{x-2}{2001}-1\right)=\left(\dfrac{x-1}{2002}-1\right)+\left(\dfrac{x}{2003}-1\right)\)

=>x-2003=0

hay x=2003

a, \(\Leftrightarrow3x^2-3+5=3x^2+2x-3x-2\)

\(\Leftrightarrow3x^2-3x-2x+3x=-2+3-5\)

<=>x=-4

b, \(\Leftrightarrow\dfrac{x+4}{5}-\dfrac{5x}{5}+\dfrac{20}{5}=\dfrac{2x}{6}-\dfrac{3\left(x-2\right)}{6}\)

\(\Leftrightarrow\dfrac{x+4-5x+20}{5}=\dfrac{2x-3x+6}{6}\)

\(\Leftrightarrow\dfrac{6\left(-4x+24\right)}{30}=\dfrac{5\left(-x+6\right)}{30}\)

<=>-24x+144=-5x+30

<=>-5x+24x=144-30

<=>19x=114

<=>x=6

a ) <=> 3x² - 3 + 5 = 3x² + 2x - 3x - 2

<=> 3x² - 3x² - 2x + 3x = -2 - 5 + 3

<=> x = - 4

Vậy s = \(\left\{-4\right\}\)

b)<=> \(\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

<=> 6x + 24 - 30x - 120 = 10x - 15x + 30

<=> 6x -30x - 10x + 15x = 30 - 24 + 120

<=> -19x = 126

<=> x =-6,6

Vậy s = \(\left\{-6,6\right\}\)

a ) <=> 3x² - 3 + 5 = 3x² + 2x - 3x - 2

<=> 3x² - 3x² - 2x + 3x = -2 - 5 + 3

<=> x = - 4

Vậy s = \(\left\{-4\right\}\)

b)<=> 6x + 24 - 30x - 120 = 10x - 15x + 30

<=> 6x + 24 - 30x - 120 = 10x - 15x + 30

<=> -19x = 126

<=> x =-6,6

d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)

\(\Leftrightarrow12x^2+16=12x^2-12x-8\)

=>-12x=24

hay x=-2

e: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(x-5\right)^2\)

\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)

=>-5x+19=-10x+25

=>5x=6

hay x=6/5

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

=>x-105=0

hay x=105

=>x-23=0

=>x=23

1)

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

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

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)

\(\Leftrightarrow x=105\)

b)

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

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

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)

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

\(\Leftrightarrow50-x=0\)

\(\Leftrightarrow x=50\)

2)

\(\left(5x+1\right)^2=\left(3x-2\right)^2\)

\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)

b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)

\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)

\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)

\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)

\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

c. \(\left(x+3\right)^4+\left(x+5\right)^4=2\)

Đặt: \(y=x+4\), ta có:

\(\left(y-1\right)^4+\left(y+1\right)^4=2\)

\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)

\(\Leftrightarrow2y^4+12y^2=0\)

\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)

\(\Leftrightarrow y=0\)

\(\Leftrightarrow x=-4\)

d) \(x^4-3x^3+4x^2-3x+1=0\)

\(\Leftrightarrow x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

\(\Leftrightarrow x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x=1\)