A= (x+1)(x-1)+11
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Đặt:
\(X=\left(1+\dfrac{1}{9}\right)\left(1+\dfrac{1}{10}\right)\left(1+\dfrac{1}{11}\right).....\left(1+\dfrac{1}{200}\right)\)
\(X=\dfrac{10}{9}.\dfrac{11}{10}.\dfrac{12}{11}......\dfrac{201}{200}\)
\(X=\dfrac{10.11.12......201}{9.10.11......200}\)
\(X=\dfrac{201}{9}\)
\(Y=\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{12}\right).....\left(1-\dfrac{1}{99}\right)\)
\(Y=\dfrac{9}{10}.\dfrac{10}{11}.\dfrac{11}{12}.....\dfrac{98}{99}\)
\(Y=\dfrac{9.10.11......98}{10.11.12.....99}\)
\(Y=\dfrac{9}{99}=\dfrac{1}{11}\)
a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
\(a)\)
\(x+\frac{1}{10}+x+\frac{11}{11}=x+\frac{11}{12}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{1}{10}-\frac{11}{11}\)
\(\Leftrightarrow x=\frac{-11}{60}\)
\(b)\)
\(-\left|x-\frac{1}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|x-\frac{1}{2}\right|=\frac{-1}{2}\) (Vô lý)
Vậy \(x\in\varnothing\)
\(A=\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{55}{334}=\dfrac{11}{334}\)
\(B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{20}{21}\)
a/ 99x+(1+2+...+99)=0
99x+(1+99)x49,5=0
x=5
b/3x-1-2-3+1+2+3+4+...+10=0
3x+4+5+...+10=0
x=-49/3
Tìm x :
a) (x - 3) + (x - 2) + (x - 1) + .... + 10 + 11 = 11
(x - 3) + (x - 2) + (x - 1) + .... + 10 = 0
[(x - 3) + (x - 2) + (x - 1)] + (0 + 1 + 2 + ... + 10) = 0
[(x - 3) + (x - 2) + (x - 1)] + 55 = 0
x - 3 + x - 2 + x - 1 = -55
x + x + x - (3 + 2 + 1) = -55
x3 - 6 = -55
x3 = -55 + 6
x3 = -49
x = -49 : 3
x = -\(\frac{49}{3}\)
a)11(x-6)=4x+11
11x - 11*6=4x+11
11x-4x=11+11*6
7x=77
x=77/7
x=11
b)lx-3l+1=x
lx-3l=x-1
\(\Rightarrow x-3\in\left\{-\left(x-1\right);x-1\right\}\)
Ta có:
TH1:x-3=-(x-1)
x-3=-x+1
x-(-x)=1+3
2x=4
x=4/2
x=2
TH2:x-3=x-1
x-x=3-1
0=2
\(x\in rỗng\)
Vậy x=2