a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

Câu 1:

c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)

\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)

\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

=>x=10

Câu 2:

d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

Hẹ hẹ

Bài 3:

a: \(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\cdots\left|x+\frac{1}{2019\cdot2020}\right|=2020x\) (1)

=>2020x>=0

=>x>=0

Phương trình (1) sẽ trở thành:

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{2019\cdot2020}=2020x\)

=>\(2020x=2019x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\right)\)

=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2019\cdot2020}\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2019}-\frac{1}{2020}\)

=>\(x=1-\frac{1}{2020}=\frac{2019}{2020}\)

b: \(\left|x+\frac{1}{1\cdot3}\right|+\left|x+\frac{1}{3\cdot5}\right|+\cdots+\left|x+\frac{1}{197\cdot199}\right|=100x\) (2)

=>100x>=0

=>x>=0

(2) sẽ trở thành: \(x+\frac{1}{1\cdot3}+x+\frac{1}{3\cdot5}+\cdots+x+\frac{1}{197\cdot199}=100x\)

=>\(100x=99x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{197\cdot199}\right)\)

=>\(x=\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{197}-\frac{1}{199}\right)=\frac12\left(1-\frac{1}{199}\right)\)

=>\(x=\frac12\cdot\frac{198}{199}=\frac{99}{199}\)

c: \(\left|x+\frac12\right|+\left|x+\frac16\right|+\left|x+\frac{1}{12}\right|+\cdots+\left|x+\frac{1}{110}\right|=11x\left(3\right)\)

=>11x>=0

=>x>=0

(3) sẽ trở thành:

\(11x=x+\frac12+x+\frac16+\ldots+x+\frac{1}{110}\)

=>\(11x=10x+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)

=>\(x=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{10\cdot11}\)

=>\(x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\) (nhận)

Bài 2:

a: \(\left|5-\frac23x\right|\ge0\forall x;\left|\frac23y-4\right|\ge0\forall y\)

Do đó: \(\left|5-\frac23x\right|+\left|\frac23y-4\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}5-\frac23x=0\\ \frac23y-4=0\end{cases}\Rightarrow\begin{cases}\frac23x=5\\ \frac23y=4\end{cases}\Rightarrow\begin{cases}x=5:\frac23=\frac{15}{2}\\ y=4:\frac23=6\end{cases}\)

b: \(\left|\frac23-\frac12+\frac34x\right|=\left|\frac34x+\frac16\right|\ge0\forall x\)

\(\left|1,5-\frac34-\frac32y\right|=\left|\frac34-\frac32y\right|\ge0\forall y\)

Do đó: \(\left|\frac34x+\frac16\right|+\left|\frac34-\frac32y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}\frac34x+\frac16=0\\ \frac34-\frac32y=0\end{cases}\Rightarrow\begin{cases}\frac34x=-\frac16\\ \frac32y=\frac34\end{cases}\Rightarrow\begin{cases}x=-\frac16:\frac34=-\frac16\cdot\frac43=-\frac{4}{18}=-\frac29\\ y=\frac34:\frac32=\frac24=\frac12\end{cases}\)

c: \(\left|x-2020\right|\ge0\forall x;\left|y-2021\right|\ge0\forall y\)

Do đó: \(\left|x-2020\right|+\left|y-2021\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2020=0\\ y-2021=0\end{cases}\Rightarrow\begin{cases}x=2020\\ y=2021\end{cases}\)

d: \(\left|x-y\right|\ge0\forall x,y\)

\(\left|y+\frac{21}{10}\right|\ge0\forall y\)

Do đó: \(\left|x-y\right|+\left|y+\frac{21}{10}\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ y+\frac{21}{10}=0\end{cases}\Rightarrow x=y=-\frac{21}{10}\)

Bài 1:

a: \(\left|\frac32x+\frac12\right|=\left|4x-1\right|\)

=>\(\left[\begin{array}{l}4x-1=\frac32x+\frac12\\ 4x-1=-\frac32x-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}4x-\frac32x=\frac12+1\\ 4x+\frac32x=-\frac12+1\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac52x=\frac32\\ \frac{11}{2}x=\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32:\frac52=\frac35\\ x=\frac12:\frac{11}{2}=\frac{1}{11}\end{array}\right.\)

b: \(\left|\frac75x+\frac12\right|=\left|\frac43x-\frac14\right|\)

=>\(\left[\begin{array}{l}\frac75x+\frac12=\frac43x-\frac14\\ \frac75x+\frac12=\frac14-\frac43x\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac75x-\frac43x=-\frac14-\frac12\\ \frac75x+\frac43x=\frac14-\frac12\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac{1}{15}x=-\frac34\\ \frac{41}{15}x=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac34:\frac{1}{15}=-\frac34\cdot15=-\frac{45}{4}\\ x=-\frac14:\frac{41}{15}=-\frac14\cdot\frac{15}{41}=-\frac{15}{164}\end{array}\right.\)

c: \(\left|\frac54x-\frac72\right|-\left|\frac58x+\frac35\right|=0\)

=>\(\left|\frac54x-\frac72\right|=\left|\frac58x+\frac35\right|\)

=>\(\left[\begin{array}{l}\frac54x-\frac72=\frac58x+\frac35\\ \frac54x-\frac72=-\frac58x-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x-\frac58x=\frac35+\frac72\\ \frac54x+\frac58x=-\frac35+\frac72\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac58x=\frac{41}{10}\\ \frac{15}{8}x=\frac{29}{10}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{41}{10}:\frac58=\frac{41}{10}\cdot\frac85=\frac{164}{25}\\ x=\frac{29}{10}:\frac{15}{8}=\frac{29}{10}\cdot\frac{8}{15}=\frac{116}{75}\end{array}\right.\)

d: \(\left|\frac78x+\frac56\right|-\left|\frac12x+5\right|=0\)

=>\(\left|\frac78x+\frac56\right|=\left|\frac12x+5\right|\)

=>\(\left[\begin{array}{l}\frac78x+\frac56=\frac12x+5\\ \frac78x+\frac56=-\frac12x-5\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac78x-\frac12x=5-\frac56\\ \frac78x+\frac12x=-5-\frac56\end{array}\right.\)

=>\(\left[\begin{array}{l}\frac38x=\frac{25}{6}\\ \frac{11}{8}x=-\frac{35}{6}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{25}{6}:\frac38=\frac{25}{6}\cdot\frac83=\frac{200}{18}=\frac{100}{9}\\ x=-\frac{35}{6}:\frac{11}{8}=-\frac{35}{6}\cdot\frac{8}{11}=-\frac{140}{33}\end{array}\right.\)

lI dau la lI

Bài làm

\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)

\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}+2=\frac{x+50}{23}+\frac{x+49}{24}+2\)

\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)=\left(\frac{x+50}{23}+1\right)+\left(\frac{x+49}{24}+1\right)\)

\(\Leftrightarrow\left(\frac{x+52}{21}+\frac{21}{21}\right)+\left(\frac{x+51}{22}+\frac{22}{22}\right)=\left(\frac{x+50}{23}+\frac{23}{23}\right)+\left(\frac{x+49}{24}+\frac{24}{24}\right)\)

\(\Leftrightarrow\frac{x+52+21}{21}+\frac{x+51+22}{22}=\frac{x+50+23}{23}+\frac{x+49+24}{24}\)

\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}=\frac{x+73}{23}+\frac{x+73}{24}\)

\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)

\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)

\(\Leftrightarrow x+73=0\)

\(\Leftrightarrow x=-73\)

Vậy x = -73

 # Học tốt #

\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)

\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}-\frac{x+50}{23}-\frac{x+49}{24}=0\)

\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)-\left(\frac{x+50}{23}+1\right)-\left(\frac{x+49}{24}+1\right)=0\)

\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)

\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)

=> \(x+73=0\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\ne0\right)\)

<=> x=-73

\(\begin{cases}y=x+20\left(1\right)\\ \frac{y}{50}-\frac{x}{40}=\frac14\left(2\right)\end{cases}\)

thay (1) vào (2) ta được:

\(\frac{x+20}{50}-\frac{x}{40}=\frac14\)

\(\Leftrightarrow4\left(x+20\right)-5x=50\)

4x + 80 - 5x = 50

-x = -30

⇒ x = 30

⇒ y = 30 + 20 = 50

vậy (x; y) = (30; 50)

\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)

=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)

Hnay đi học, cô giáo có sửa cho bạn bài đó hong dọ, do cô mình giao cái bài về nhà  y sì dãy í, mà mai nộp ròi, nhưng mình k biết làm, nếu bạn biết , chỉ mình với :((

Chứng minh rằng (9^2n+1994^93)chia heets cho 5

a) \(x\cdot\frac{1}{2}+x\cdot\frac{1}{4}+x\cdot\frac{1}{8}=\frac{21}{24}\)

\(x\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)=\frac{7}{8}\)

\(x\cdot\frac{7}{8}=\frac{7}{8}\)

\(\Rightarrow x=\frac{7}{8}\div\frac{7}{8}=1\)

b) \(\left(x+4\right)+\left(x+9\right)+\left(x+14\right)+.....+\left(x+44\right)+\left(x+49\right)=1430\)

\(\left(x+x+x+....+x+x\right)+\left(4+9+14+...+44+49\right)=1430\)

\(10x+265=1430\)

\(10x=1430-265\)

\(10x=1165\)

\(\Rightarrow x=\frac{1165}{10}=116,5\)

c) \(x\cdot0,25-0,5=1\)

\(x\cdot0,25=1+0,5\)

\(x\cdot0,25=1,5\)

\(\Rightarrow x=1,5\div0,25=6\)

Tim x:

     x X 12  + x X 14  + x X 18   = \(\frac{21}{24}\)

x X (\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)) = \(\frac{21}{24}\)

x X (\(\frac{4}{8}+\frac{2}{8}+\frac{1}{8}\)) = \(\frac{21}{24}\)

x X \(\frac{9}{8}\)= \(\frac{21}{24}\)

x           = \(\frac{21}{24}\): \(\frac{9}{8}\)

x           = \(\frac{21}{24}\)X \(\frac{8}{9}\)

x           = \(\frac{7}{9}\)

\(x\times\frac{1}{2}+x\times\frac{1}{4}+x\times\frac{1}{8}=\frac{21}{24}\)

\(\Rightarrow x\times\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right]=\frac{21}{24}\)

\(\Rightarrow x\times\left[\frac{4+2+1}{8}\right]=\frac{21}{24}\)

\(\Rightarrow x\times\frac{7}{8}=\frac{21}{24}\)

\(\Rightarrow x\times\frac{7}{8}=\frac{7}{8}\)

\(\Rightarrow x=\frac{7}{8}:\frac{7}{8}=1\)