3x + x3 = 11 x 11

3x + x3 = 121

Mà 121 = 38 + 83

Vậy x = 8

x = 8

vì 11 x 11 = 121

mà 1 = 8 + 3 nên x = 8

quý vị khán giả thấy đúng thì k nhá

3x + x3 = 11x11

3x + x3 = 121

Ta đặt phép tính như sau:

3x + x3 = 121 (cái này bạn đặt tính dọc nhé!)

- NX : Hàng đơn vị : x + 3 = 1 => x+3 có nhớ => x = 8

Vậy x =8

Thế thôi, k nhé?

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

Không tên

3x + x3 = 11 x 11 

=> 30 + x + x.10 + 3 = 121

=> 33 + x.11 = 121

=> x.11 = 121 - 33 = 88

=> x = 88 : 11 = 8 

\(1.x^2+11x=0\)

\(\Leftrightarrow x\left(x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)

\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)

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=.....

\(a,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(\Leftrightarrow4\times x+\frac{15}{16}=1\)

\(\Leftrightarrow4\times x=1-\frac{15}{16}\)

\(\Leftrightarrow4\times x=\frac{1}{16}\)

\(\Leftrightarrow x=\frac{1}{16}\div4\)

\(\Leftrightarrow x=\frac{1}{64}\)

\(b,x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(\Leftrightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-\left[715\times\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{13}+...+\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(\Leftrightarrow x-\left[715\times\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(\Leftrightarrow x-\left[715\times\frac{4}{55}\right]=\frac{3}{11}\)

\(\Leftrightarrow x-52=\frac{3}{11}\)

\(\Leftrightarrow x=\frac{3}{11}+52\)

\(\Leftrightarrow x=\frac{575}{11}\)

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

\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

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

 \(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)

\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=\dfrac{1}{113}\)

\(x=109\)

\(\dfrac{2}{1.5}\) + \(\dfrac{2}{5.9}\)  + \(\dfrac{2}{9.13}\)+...+  \(\dfrac{2}{x\left(x+4\right)}\)   = \(\dfrac{56}{113}\)

(\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+\dfrac{2}{9.13}+....+\dfrac{2}{x\left(x+4\right)}\) ) \(\times\) 2  = \(\dfrac{56}{113}\) \(\times\) 2

\(\dfrac{4}{1.5}\) + \(\dfrac{4}{5.9}\) +\(\dfrac{4}{9.13}\)+...+ \(\dfrac{4}{x\left(x+4\right)}\)                = \(\dfrac{112}{113}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{13}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

1  - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = 1 - \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = \(\dfrac{1}{113}\)

         \(x\) + 4 = 113

         \(x\)        = 113 - 4

         \(x\)        = 109

\(3x.\left(x-\frac{11}{7}\right)=0\)

=> \(\orbr{\begin{cases}3x=0\\x-\frac{11}{7}=0\end{cases}}=>\orbr{\begin{cases}x=0\\x=\frac{11}{7}\end{cases}}\)

Chú ý :

Dấu \(.\)là dấu nhân đấy 

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: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)

=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)

=>\(x\cdot1=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}\)