11 000 + X = 11 000

 X =11 000 - 11 000

X = 0

tk nhé

2000+1000+8000=11000

<=>11000+x=11000

=>X=11000-11000

=>x=0

X = 0. OK

x=-2 nha bạn Cá là có trong violympic

\(\Rightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

\(\Rightarrow\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)

Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)

=>x+2=0

=>x=-2

Vậy x=-2

\(\dfrac{5}{13}\times\dfrac{7}{11}+\dfrac{5}{13}\times\dfrac{2}{11}+\dfrac{8}{13}\times\dfrac{9}{11}\)

\(=\dfrac{5}{13}\times\left(\dfrac{7}{11}+\dfrac{2}{11}\right)+\dfrac{8}{13}\times\dfrac{9}{11}\)

\(=\dfrac{5}{13}\times\dfrac{9}{11}+\dfrac{8}{13}\times\dfrac{9}{11}\)

\(=\dfrac{9}{11}\times\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\)

\(=\dfrac{9}{11}\times1\)

\(=\dfrac{9}{11}\)

a) \(20,23\times21+20,23\times80-20,23\)

\(=20,23\times\left(21+80-1\right)\)

\(=20,23\times100\)

\(=2023\)

\(\dfrac{5}{13}\) \(\times\) \(\dfrac{7}{11}\) + \(\dfrac{5}{13}\) \(\times\) \(\dfrac{2}{11}\) + \(\dfrac{8}{13}\) \(\times\) \(\dfrac{9}{11}\)

= \(\left(\dfrac{5}{13}\times\dfrac{7}{11}+\dfrac{5}{13}\times\dfrac{2}{11}\right)\)+ \(\dfrac{8}{13}\) \(\times\) \(\dfrac{9}{11}\)

= \(\dfrac{5}{13}\)\(\times\) (\(\dfrac{7}{11}\) + \(\dfrac{2}{11}\)) + \(\dfrac{8}{13}\) \(\times\) \(\dfrac{9}{11}\)

= \(\dfrac{5}{13}\) \(\times\) \(\dfrac{9}{11}\) + \(\dfrac{8}{13}\) \(\times\) \(\dfrac{9}{11}\)

= \(\dfrac{9}{11}\) \(\times\)( \(\dfrac{5}{13}\) +\(\dfrac{8}{13}\))

= \(\dfrac{9}{11}\) \(\times\) 1

= \(\dfrac{9}{11}\)

=>

\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

=>\(\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}=0\right)\)

vì \(\frac{1}{10^{10}}>\frac{1}{11^{11}}>\frac{1}{12^{12}}>\frac{1}{13^{13}}=>\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)

=>x+2=0

=>x=-2

tick nhé

\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)

\(=>\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

\(=>\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)

Vì \(\frac{1}{10^{10}}>\frac{1}{11^{11}}>\frac{1}{12^{12}}>\frac{1}{13^{13}}\)

Nên \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)khác 0

Suy ra:  x+2=0

             x   =0 - 2

             x = -2

                              TICK NHA

nhiều thế

sfdhg 

Theo đầu bài ta có: \(\frac{x+2}{10^{10}}=\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}=\frac{x+2}{13^{13}}\)
Gọi kết quả của mỗi phân số là a. Khi đó x + 2 phải thỏa mãn điều kiện:
\(a\cdot10^{10}=a\cdot11^{11}=a\cdot12^{12}=a\cdot13^{13}\)
Mà 10¹⁰, 11¹¹, 12¹², 13¹³ không bao giờ bằng nhau nên a chỉ có thể bằng 0.
Mà 0 nhân với bất kì số nào cũng bằng 0 nên x + 2 = 0 --> x = -2
Vậy tập nghiệm S = { -2 }

X=-2

 Tick rồi mk ns cách làm cho,hứa

a) = 1.1204
b) = 0.5714

a) = 1.1204
b) = 0.5714

a) = 1.1204
b) = 0.5714