12 - ( x - 11 ) = ( -9 )
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\(=\dfrac{12}{15}\times\left(\dfrac{26}{11}-\dfrac{9}{11}-\dfrac{6}{11}\right)=\dfrac{12}{15}\times\dfrac{11}{11}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(x-\frac{6}{7}+x-\frac{7}{8}+x-\frac{8}{9}=x-\frac{9}{10}+x-\frac{10}{11}+x-\frac{11}{12}\)
\(x+x+x-x-x-x=\frac{6}{7}+\frac{7}{8}+\frac{8}{9}-\frac{9}{10}-\frac{10}{11}-\frac{11}{12}\)
\(0=\frac{6}{7}+\frac{7}{8}+\frac{8}{9}-\frac{9}{10}-\frac{10}{11}-\frac{11}{12}\)
X triệt tiêu hết ròi! Vậy đề bài yêu cầu tìm gì vậy. Nhưng mà...giá trị của 2 vế ko bằng nhau.
\(\Leftrightarrow\left(\frac{x+1}{7}-1\right)+\left(\frac{x+1}{8}-1\right)+\left(\frac{x+1}{9}-1\right)=\left(\frac{x+1}{10}-1\right)+\left(\frac{x+1}{11}-1\right)+\left(\frac{x+1}{12}-1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=0\)
\(\text{Vì}\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\ne\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\)\(\Rightarrow\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
\(\Leftrightarrow\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1\)\(+\frac{x-11}{12}+1\) ( cộng 2 vế với 3 )
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\right)\)
\(\Leftrightarrow x=-1\)
12+11+10+....+x=12
=>12+11+10+..+x-12=0
=>11+10+...+x=0
=> x = -11
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TK MÌNH NHÉ
Trả lời:
\(x=\frac{9^{11}+2}{9^{11}+3}=\frac{9^{11}+3-1}{9^{11}+3}=\frac{9^{11}+3}{9^{11}+3}-\frac{1}{9^{11}+3}=1-\frac{1}{9^{11}+3}\)
\(y=\frac{9^{12}+2}{9^{12}+3}=\frac{9^{12}+3-1}{9^{12}+3}=\frac{9^{12}+3}{9^{12}+3}-\frac{1}{9^{12}+3}=1-\frac{1}{9^{12}+3}\)
Ta có: \(9^{11}< 9^{12}\)
\(\Leftrightarrow9^{11}+3< 9^{12}+3\)
\(\Leftrightarrow\frac{1}{9^{11}+3}>\frac{1}{9^{12}+3}\)
\(\Leftrightarrow-\frac{1}{9^{11}+3}< -\frac{1}{9^{12}+3}\)
\(\Leftrightarrow1-\frac{1}{9^{11}+3}< 1-\frac{1}{9^{12}+3}\)
\(\Leftrightarrow x< y\)
Vậy x < y
\(\frac{1}{9x10}\)\(+\frac{1}{10x11}\)\(+\frac{1}{11x12}\)\(+.....\)\(+\frac{1}{805x806}\)
\(=\frac{1}{9}\)\(-\frac{1}{10}\)\(+\frac{1}{10}\)\(-\frac{1}{11}\)\(+\frac{1}{11}\)\(-\frac{1}{12}\)\(+.....\frac{1}{805}\)\(-\frac{1}{806}\)
\(=\frac{1}{9}\)\(-\frac{1}{806}\)
\(=\frac{797}{7254}\)
a) \(2\dfrac{4}{9}+6\dfrac{7}{11}+7\dfrac{5}{9}+13\dfrac{4}{11}\)
\(=2+\dfrac{4}{9}+6+\dfrac{7}{11}+7+\dfrac{5}{9}+13+\dfrac{4}{11}\)
\(=\left(2+6+7+13\right)+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\left(\dfrac{7}{11}+\dfrac{4}{11}\right)\)
\(=28+1+1\)
\(=30\)
b) \(\dfrac{3}{4}+\dfrac{1}{2}\times\dfrac{12}{5}-1\dfrac{1}{2}\)
\(=\dfrac{3}{4}+\dfrac{1}{2}-1-\dfrac{1}{2}\)
\(=\dfrac{3}{4}-1\)
\(=\dfrac{3}{4}-\dfrac{4}{4}\)
\(=-\dfrac{1}{4}\)
12 - ( x - 11 ) = -9
12 - x + 11 = -9
12 - x = -9 - 11
12 - x = -20
x = 12 - ( - 20 )
x = -32
12-(x-11)=(-9)
x-11=12-(-9)
x-11=21
x =21+11
x=32
Vậy x=32