tìm x biết:3x + x3 = 11 x 11
dấu x ở giữa 2 số 11 là dấu nhân
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NV
8 tháng 2 2017
\(1.x^2+11x=0\)
\(\Leftrightarrow x\left(x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)
\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)
chia thành 4 TH :
\(TH1:X-1=0\)
\(\Leftrightarrow x=1\)
\(TH2:x+1=0\)
\(\Leftrightarrow x=-1\)
\(TH3:X+9=0\)
\(\Leftrightarrow X=-9\)
\(TH4:x-9=0\)
\(\Leftrightarrow x=9\)
Kết luận ....
\(3.\left(\left|x+1\right|-5\right)\left(x^2-9\right)\)
\(\Leftrightarrow\left(\left|x+1\right|-5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|-5=0\\x-3=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|=5\\x=3\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+1=+_-5\Leftrightarrow x+1=5,x+1=-5\Leftrightarrow x=4,x=-6\\x=3x\\x=-3\end{cases}}\)
kết luận x=.....
\(4.\left(3x-16\right)⋮\left(x+2\right)\)
\(\Leftrightarrow\left(3x+6\right)-22\)
\(\Leftrightarrow3\left(x+2\right)-22⋮\left(x+2\right)\)
Vì\(\left(x+2\right)⋮\left(x+2\right)\)
\(\Rightarrow\left(3x-16\right)⋮\left(x+2\right)\)
Kết luận x=.....
6 tháng 8 2016
\(a,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\times x+\frac{15}{16}=1\)
\(\Leftrightarrow4\times x=1-\frac{15}{16}\)
\(\Leftrightarrow4\times x=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{16}\div4\)
\(\Leftrightarrow x=\frac{1}{64}\)
\(b,x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Leftrightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-\left[715\times\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{13}+...+\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(\Leftrightarrow x-\left[715\times\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(\Leftrightarrow x-\left[715\times\frac{4}{55}\right]=\frac{3}{11}\)
\(\Leftrightarrow x-52=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+52\)
\(\Leftrightarrow x=\frac{575}{11}\)
LT
Lương Thị Vân Anh
CTVHS
18 tháng 7 2023
Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)
\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)
x + 4 = 113 ⇒ x = 109
TD
Thầy Đức Anh
Giáo viên
VIP
18 tháng 7 2023
\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x-4}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+4}\right)\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)
Với \(A=\dfrac{56}{113}\) thì
\(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)
\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=\dfrac{1}{113}\)
\(x=109\)
14 tháng 6 2016
\(3x.\left(x-\frac{11}{7}\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x-\frac{11}{7}=0\end{cases}}=>\orbr{\begin{cases}x=0\\x=\frac{11}{7}\end{cases}}\)
Chú ý :
Dấu \(.\)là dấu nhân đấy
5 tháng 11 2023
a: \(x+6\dfrac{1}{8}=8\)
=>\(x+\dfrac{49}{8}=\dfrac{64}{8}\)
=>\(x=\dfrac{64}{8}-\dfrac{49}{8}=\dfrac{15}{8}\)
b: \(\dfrac{11}{2}\cdot x=\dfrac{1}{5}:\dfrac{1}{3}\)
=>\(x\cdot\dfrac{11}{2}=\dfrac{1}{5}\cdot3=\dfrac{3}{5}\)
=>\(x=\dfrac{3}{5}:\dfrac{11}{2}=\dfrac{3}{5}\cdot\dfrac{2}{11}=\dfrac{6}{55}\)
c: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)
=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)
=>\(x\cdot1=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}\)
3x+x3=121 =>x{3+3}=121 =>x9=121 =>sai
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