1/9 - 2/3y^4 + y^8
lưa ý / í là ví dụ 1 phần 9
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\(\frac{4}{9}:8-\left(\frac{1}{4}+\frac{9}{20}\right):1\frac{4}{5}\)
\(=\frac{1}{18}-\frac{7}{10}\cdot\frac{5}{9}=\frac{1}{18}-\frac{7}{18}=-\frac{1}{3}\)
\(\frac{4}{9}:8-\left(\frac{1}{4}+\frac{9}{20}\right):1\frac{4}{5}\)
\(=\frac{1}{18}-\frac{5+9}{20}\times\frac{5}{9}\)
\(=\frac{1}{18}-\frac{14\times5}{20\times9}\)
\(\frac{1}{18}-\frac{14}{36}=\frac{2-14}{36}=\frac{-12}{36}=\frac{-1}{3}\)
A=1/2^2+1/3^2+...+1/23^2
=>A<1-1/2+1/2-1/3+...+1/22-1/23
=>A<22/23
a: =35/11-13/11-7/13-6/13
=2-1=1
b: =2+1/7+6/7+7/9-25/9
=2+1-2=1
c: \(=\dfrac{132}{321}\left(-\dfrac{103}{13}-\dfrac{27}{13}\right)=\dfrac{-132}{321}\cdot10=-\dfrac{440}{107}\)
9:
a: -x^3+3x^2-3x+1
=(-x)^3+3*(-x)^2*1+3*(-x)*1^2+1^3
=(-x+1)^3
b: z^3-z^2+1/3z-1/27
=z^3-3*z^2*1/3+3*z*(1/3)^2-(1/3)^3
=(z-1/3)^3
c: x^6-3x^4y+3x^2y^2-y^3
=(x^2)^3-3*(x^2)^2*y+3*x^2*y^2-y^3
=(x^2-y)^3
d: =(x-y)^3+3*(x-y)^2*1/3+3*(x-y)*(1/3)^2+(1/3)^3
=(x-y+1/3)^3
Ví dụ 9:
a) \(-x^3+3x^2-3x+1\)
\(=-\left(x^3-3x^2+3x-1\right)\)
\(=-\left(x-1\right)^3\)
b) \(x^3-x^2+\dfrac{1}{3}x-\dfrac{1}{27}\)
\(=x^3-3\cdot\dfrac{1}{3}\cdot x^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot x-\left(\dfrac{1}{3}\right)^3\)
\(=\left(x-\dfrac{1}{3}\right)^3\)
c) \(x^6-3x^4y+3x^2y^2-y^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot y+3\cdot x^2\cdot y^2-y^3\)
\(=\left(x^2-y\right)^3\)
d) \(\left(x-y\right)^3+\left(x-y\right)^2+\dfrac{1}{3}\left(x-y\right)+\dfrac{1}{27}\)
\(=\left(x-y\right)^3+3\cdot\dfrac{1}{3}\cdot\left(x-y\right)^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot\left(x-y\right)+\left(\dfrac{1}{3}\right)^3\)
\(=\left(x-y+\dfrac{1}{3}\right)^3\)
b= 59/10 : 3/2 - [ 21/3 - 41/2 - 2 x 21/3] : 7/4
=9/15 - [ 21/3 - 21/3 x 41/2 - 2]
=9/15 - [ 0 x 41/2 - 2]
=9/15 - 0
=9/15
(1/3)^2 - 1/3y^4 - 1/3y^4 - (y^4)^2
1/3(1/3 - y^4) - y^4(1/3 - y^4)
(1/3 - y^4)(1/3-y^4)
còn lại bạn làm.