\(x^4\cdot x^7\cdot...\cdot x^{100}\)

\(=x^{4+7+...+100}\)

\(=x^{52\cdot33}=x^{1716}\)

\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}\)

Ta có : \(x^1\cdot x^2=x^{1+2}=x^3\)

Tương tự : \(x^1\cdot x^2\cdot x^3=x^{1+2+3}=x^6\)

Áp dụng vào bài toán :

\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}=x^{1+2+3+...+2006}\)

\(\Rightarrow x^{1+2+3+...+2006}=x^{2013021}\)

Sửa câu a : \(x^{1+2+3+...+2006}=x^{2033136}\)

Bài 1:

a) 8² . 2² = 64 . 4 = 256

b) 2⁵. 4 = 2⁵. 2²= 2⁵⁺² = 2⁷

c) 9⁴. 243 = 9⁴. 3⁵=(3²)⁴.3⁵=3⁸.3⁵=3⁸⁺⁵=3¹³

Bài 1 :

\(a.8^2.2^2=\left(8.2\right)^2=16^2\)

\(b.2^5.4=2^5.2^2=2^7\)

\(c.9^4.243=\left(3^2\right)^4.3^5=3^8.3^5=3^{13}\)

Bài 2:

a. ( đề sai =.= )

\(b.12.\left(x-1\right):3=43+2^3\)

\(\Leftrightarrow12.\left(x-1\right)=43+2^3.3\)

\(\Leftrightarrow12.\left(x-1\right)=67\)

\(\Leftrightarrow x-1=\frac{67}{12}\)

\(\Leftrightarrow x=\frac{79}{12}\)

\(1,\\ a,\Rightarrow x=\dfrac{5\cdot4}{3}=\dfrac{20}{3}\\ b,\Rightarrow x=8\\ c,\Rightarrow x=\dfrac{27\cdot2}{3,6}=15\\ 2,\\ a,\dfrac{6}{9}=\dfrac{42}{63};\dfrac{6}{42}=\dfrac{9}{63};\dfrac{63}{9}=\dfrac{42}{6};\dfrac{63}{42}=\dfrac{9}{6}\\ b,\dfrac{4}{18}=\dfrac{2}{9};\dfrac{4}{2}=\dfrac{18}{9};\dfrac{9}{18}=\dfrac{2}{4};\dfrac{9}{2}=\dfrac{18}{4}\)

a) \(215+x=400\)

\(\Rightarrow x=400-215\)

\(\Rightarrow x=185\)

b) \(12,5-2x=\dfrac{1}{5}\)

\(\Rightarrow2x=12,5-\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{123}{10}\)

\(\Rightarrow x=\dfrac{123}{10}:2\)

\(\Rightarrow x=\dfrac{123}{20}\)

còn câu c d nhanh cậu ơi

Bạn xem lại đề phần e nha .

\(c)x.0,25-\dfrac{3}{25}:x=2\dfrac{1}{4}\)

\(x.0,25-\dfrac{25}{3}.x=\dfrac{9}{4}\)

\(x.\left(0,25-\dfrac{25}{3}\right)=\dfrac{9}{4}\)

\(x.\dfrac{-97}{12}=\dfrac{9}{4}\)

\(x=\dfrac{-27}{97}\)

\(d)\left(x-\dfrac{1}{3}\right).\left(x-\dfrac{1}{4}\right)=0\)

Do : 2 số đó nhân với nhau bằng 0 nên một trong 2 số là số 0.

Do đó :

\(Th1:x-\dfrac{1}{3}=0\)

\(x=\dfrac{1}{3}\)

\(Th2:x-\dfrac{1}{4}=0\)

\(x=\dfrac{1}{4}\)

3​.x + 2.x + x = 300​

​x.(3+2+1) = 300

​x.6 = 300

​x=300 : 6

​x = 50

3.x+2.x+x=300

x.(3+2)=300

x.5=300

x=300:5

x=60

Vậy x=60

x.(3+2)+x=300

x.5+x=300

2x.5=300

x=300:5

x=60

x=60:2

x=30

 (5.x+1)^2=36/49 => 5x + 1 = +- 6/7 

khi 5x + 1 = 6/7 => 5x = -1/7 => x = -1/35

khi 5x + 1 = -6/7 => 5x = -13/7 => x = -13/35

\(4\left(x+2\right)-8\left(x+3\right)=36\)

\(\Rightarrow4x+8-8x-24=36\)

\(\Rightarrow4x-8x=36+24-8\)

\(\Rightarrow-4x=52\)

\(\Rightarrow x=-13\)

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

=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)

=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)

=> \(x-\frac{11}{5}=2\)

=> \(x=2+\frac{11}{5}\)

=> \(x=\frac{21}{5}\)

thanks bn

1/15 +1/35 + 1/63 + ... + 1/3135

= 1/3*5 + 1/5*7 + 1/7*9 + ... + 1/55*57

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/55 - 1/57

= 1/3 - 1/57

= 18/57

1/3 + 1/6 + 1/10 +  ... + 1/x(x + 1)*2 = 2009/2011

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1)*2 = 2009/2011

=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1)  = 2009/2011

=> 2(1/2 - 1/x + 1) = 2009/2011

=> 1/2 - 1/x + 1 = 2009/4022

=> 1/x + 1 = 1/2011

=> x + 1 = 2011

=> x = 2010

vậy-

a) \(\left(x^2-9\right)\cdot\left(4^x-16\right)=0\)

\(\Rightarrow x^2-9=0\)hoặc \(4^x-16=0\)

               \(x^2=9\)                     \(4^x=16\)

               \(x^2=\left(\pm3\right)^2\)         \(4^x=4^2\)                

\(\Rightarrow x=\pm3\)hoặc \(x=2\)

b) \(5^x+5^{x+2}=650\)

\(\Rightarrow5^x+5^x\cdot25=650\)

\(\Rightarrow5^x\cdot\left(1+25\right)=650\)

\(\Rightarrow5^x\cdot26=650\)

\(\Rightarrow5^x=650\div26=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)Vậy \(x=2\)

c) \(2^{x+2}-2^x=96\)

\(2^x\cdot4-2^x=96\)

\(2^x\cdot\left(4-1\right)=96\)

\(2^x\cdot3=96\)

\(2^x=96\div3=32\)

\(2^x=2^5\)Vậy \(x=5\)