Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
a: \(=x^{n+19-14}=x^{n+5}\)
b: \(=x^{94-17-65}=x^{12}\)
Áp dụng : (A + B)3 = A3 + 3A2B + 3AB2 + B3
11) \(\left(x^2+\frac{3}{xy}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{3}{xy}+3\cdot x^2\cdot\left(\frac{3}{xy}\right)^2+\left(\frac{3}{xy}\right)^3\)
\(=x^6+3\cdot x^4\cdot\frac{3}{xy}+3\cdot x^2\cdot\frac{9}{x^2y^2}+\frac{27}{x^3y^3}\)
\(=x^6+\frac{9x^4}{xy}+\frac{27\cdot x^2}{x^2y^2}+\frac{27}{x^3y^3}\)
\(=x^6+\frac{9x^3}{y}+\frac{27}{y^2}+\frac{27}{x^3y^3}\)
12) \(\left(x^2+\frac{2}{x}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{2}{x}+3\cdot x^2\cdot\left(\frac{2}{x}\right)^2+\left(\frac{2}{x}\right)^3\)
\(=x^6+3\cdot x^4\cdot\frac{2}{x}+3\cdot x^2\cdot\frac{4}{x^2}+\frac{8}{x^3}\)
\(=x^6+\frac{6\cdot x^4}{x}+\frac{12\cdot x^2}{x^2}+\frac{8}{x^3}\)
\(=x^6+6x^3+12+8x^3\)
13) \(\left(3y+\frac{x}{2}\right)^3=\left(3y\right)^3+3\cdot3y^2\cdot\frac{x}{2}+3\cdot3y+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3\)
\(=27y^3+\frac{9y^2\cdot x}{2}+9y+\frac{x^2}{4}+\frac{x^3}{8}\)
14) \(\left(1\frac{1}{2}xy+1\right)^3=\left(\frac{3}{2}xy+1\right)^3=\left(\frac{3}{2}xy\right)^3+3\cdot\left(\frac{3}{2}xy\right)^2\cdot1+3\cdot\frac{3}{2}xy\cdot1^2+1^3\)
\(=\frac{27}{8}x^3y^3+3\cdot\frac{9}{4}x^2y^2+\frac{9}{2}xy+1\)
\(=\frac{27}{8}x^3y^3+\frac{27}{4}x^2y^2+\frac{9}{2}xy+1\)
15) \(\left(\frac{x^2}{2}+\frac{2}{y}\right)^3=\left(\frac{x^2}{2}\right)^3+3\cdot\left(\frac{x^2}{2}\right)^2\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\left(\frac{2}{y}\right)^2+\left(\frac{2}{y}\right)^3\)
\(=\frac{x^6}{8}+3\cdot\frac{x^4}{4}\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\frac{4}{y^2}+\frac{8}{y^3}\)
\(=\frac{x^6}{8}+\frac{3x^4}{2y}+\frac{6x^2}{y^2}+\frac{8}{y^3}\)
Còn 5 bài cuối áp dụng tương tự như thế :)
a: \(M=2x^2-6xy-3xy-6y-2x^2+6y+8xy\)
\(=-xy\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{2}\)
b: x=16 nên x+1=17
\(N=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)
\(=x^4-x^3-x^3+x^3+x^2-x^2-x+20\)
=20-x
=20-16=4
c: Ta có: x=16
nên x+1=17
Ta có: \(C=x^4-17x^3+17x^2-17x+20\)
\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)
=20-x
=4
A = x4 - 17x3 + 17x2 - 17x + 20 tại x = 16
Ta có: x = 16 => x + 1 = 17
=> A = x4 - (x + 1)x3 + (x + 1)x2 - (x + 1)x + 20
= x4 - x4 - x3 + x3 + x2 - x2 - x +20
= 20 - x
Tại x = 16 thì A = 20 - 16 = 4
B = x5 - 15x4 + 16x3 - 29x2 + 13x tại x = 14
Ta có: x = 14 => x + 1 = 15; x + 2 = 16; 2x + 1 = 29; x - 1 = 13
=> B = x5 - (x + 1)x4 + (x + 2)x3 - (2x + 1)x2 + (x - 1)x
= x5 - x5 - x4 + x4 + 2x3 - 2x3 - x2 + x2 - x
= x
Tại x = 14 thì B = 14