b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)

Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)

(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)

a./

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)

Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)

(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

c./ \(\Leftrightarrow2\cdot2^2\cdot2^3\cdot...\cdot2^x=2^{10}\Leftrightarrow2^{1+2+3+...+x}=2^{10}\Leftrightarrow1+2+3+...+x=10\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=10\Leftrightarrow x\left(x+1\right)=20=4\cdot5\Rightarrow x=4\)

a) Ta có: \(\frac{x}{9}=\frac{-12}{27}\)

=> \(27.x=-12.9\)

=> \(27x=-108\)

=> \(x=108:27\)

=>\(x=4\)

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

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

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

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

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

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

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

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

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> \(4\left(x+1\right)=3\left(x-2\right)\)

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)

Tự làm nốt và kết luận 

b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)

Vậy ....

c) \(\frac{x}{y}=\frac{10}{9}\Leftrightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{z}=\frac{3}{4}\Leftrightarrow\frac{y}{3}=\frac{z}{4}\Leftrightarrow\frac{y}{9}=\frac{x}{12}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\). Mà \(x-y+z=78\). Áp dụng t/c dãy tỉ số bằng nhau 

\(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Rightarrow x=6.10=60;y=6.9=54;z=6.12=72\)

Vậy..........

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

\(2\cdot2^2\cdot2^3\cdot2^4\cdot\cdot\cdot2^x=32768\)

\(\Leftrightarrow2^{1+2+3+4+\cdot\cdot\cdot+x}=2^{15}\)

\(\Leftrightarrow1+2+3+4+..+x=15\)

\(\Leftrightarrow\)\(\frac{\left(1+x\right)x}{2}=15\)

\(\Leftrightarrow x\left(x+1\right)=30=5\left(5+1\right)\)

Vậy x=5

Bài 2:

Bậc của đơn thức là 2+5+3=10

Bài 3:

\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=5\)

+)TH1: \(x\ge\frac{1}{4}\) thì bt trở thành

\(2x-\frac{1}{2}=5\Leftrightarrow2x=\frac{11}{2}\Leftrightarrow x=\frac{11}{4}\left(tm\right)\)

+)TH2: \(x< \frac{1}{4}\) thì pt trở thành

\(2x-\frac{1}{2}=-5\Leftrightarrow2x=-\frac{9}{2}\Leftrightarrow x=-\frac{9}{4}\left(tm\right)\)

Vậy x={-9/4;11/4}

\(b,B=\frac{4^2\cdot2^3}{2^6}\)

\(=\frac{2^4\cdot2^3}{2^6}\)

\(=\frac{2^7}{2^6}=2\)

\(b,\frac{x}{5}=\frac{y}{7}\Rightarrow\frac{x}{5}=\frac{2y}{14}\)

ap dụng tính chất DTSBN ta có

\(\frac{2y}{14}=\frac{x}{5}=\frac{2y-x}{14-5}=\frac{27}{9}=3\)

\(\hept{\begin{cases}\frac{x}{5}=3\Rightarrow x=15\\\frac{y}{7}=3\Rightarrow y=21\end{cases}}\)

1) 

\(B=\frac{4^2.2^3}{2^6}=\frac{\left(2^2\right)^2.2^3}{2^6}=\frac{2^4.2^3}{2^6}=\frac{2^7}{2^6}=2\)

2)

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

 \(\frac{x}{5}=\frac{y}{7}=\frac{2y}{14}=\frac{2y-x}{14-5}=\frac{27}{9}=3\)

\(\Rightarrow\hept{\begin{cases}x=3.5=15\\y=3.7=21\end{cases}}\)

Bài 1; B= \(\frac{4^2.2^3}{2^6}\) \(\Rightarrow B=\frac{4^2.4^2}{2^6}=\frac{\left(4.4\right)^2}{64}=\frac{16^2}{64}=\frac{256}{64}=4\)