\(=321-21\cdot\left[2\cdot27+256:32-52\right]\)

\(=321-21\cdot10=321-210=111\)

quá chuẩn luôn !!!!!!!!

NHỚ L.I.K.E cho mk nha

 a) (x+2)(x^2-2x+4)-x(x^2+2)=15 
<=> x^3 + 8 - x^3 - 2x = 15 
<=> -2x = 7 
<=> x = -7/2 

b) (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28 
<=> x^3 + 9x² + 27x + 27 - x(9x² + 6x + 1) + 8x^3 + 1 = 28 
<=> x^3 + 9x² + 27x + 27 - 9x^3 - 6x² - x + 8x^3 + 1 - 28 = 0 
<=> 3x² + 26x = 0 
<=> x(3x + 26) = 0 
Vậy x = 0 và x = -26/3 

c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0 
<=> (x² - 1)[(x² -1)² - x^4 - x² - 1] = 0 
<=> (x-1)(x+1)(x^4 - 2x² + 1 - x^4 - x² - 1 ) = 0 
<=> -(x-1)(x+1)3x² = 0 
Vậy nghiệm là x = 1 ; -1 ; 0

Biết rồi đăng lên làm j

\(A=x^2+2x+1+x^2-2x+1-2x^2+2=4\)

\(2\cdot\left(x-1\right)+3\cdot\left(x-2\right)=0\\ \Rightarrow2x-2+3x-6=0\\ \Rightarrow5x-8=0\\ \Rightarrow5x=0+8\\ \Rightarrow5x=8\\ \Rightarrow x=\dfrac{8}{5}\)

ok, mik sẽ ''cíu'' bn

\(x=\dfrac{8}{5}\)

\(\dfrac{2}{x+1}=\dfrac{x+1}{50}\)

`=> (x+1)^2 = 2.50`

`=> (x+1)^2 = 100`

`=> (x+1)^2 =+ -10`

\(\Rightarrow\left[{}\begin{matrix}x+1=10\\x+1=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-11\end{matrix}\right.\)

=>3x(1-1/6)=3/4

=>3x=3/4:5/6=3/4*6/5=18/20=9/10

=>x=3/10

Đề là \(3x\left(\dfrac{1}{1}\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}\right)=\dfrac{3}{4}?\)

`3x (1/1 \times 1/2 + 1/2 \times 1/3 + 1/3 \times 1/4 + 1/4 \times 1/5 + 1/5 \times 1/6) = 3/4`

\(3x\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{3}{4}\)

\(3x\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{3}{4}\)

\(3x\left(1-\dfrac{1}{6}\right)=\dfrac{3}{4}\)

\(3x\times\dfrac{5}{6}=\dfrac{3}{4}\)

`3x=3/4 \div 5/6`

`3x = 9/10`

`x = 9/10 \div 3`

`x = 3/10.`

\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)