Mn help me câu này vs:(x-3y)(x^2+3xy+9y^2)+(x+3y)(x^2-3xy+9y^2)tại x=-1 và y=-2019
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AM
29 tháng 6 2015
\(\left(x+3y\right)^3-\left(x+3y\right)\left(x^2-3xy+9y^2\right)-2x\left(x-2\right)^2=\left(x+3y\right)^3-\left(x^3+27y^3\right)-2x\left(x-2\right)^2\)
Thay x=1 y=2 ta có:
\(\left(1+3.2\right)^3-\left(1^3+27.2^3\right)-2.1.\left(1-2\right)^2=7^3-\left(1+216\right)-2=343-217-2=124\)
30 tháng 11 2019
1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0
Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)
A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)
A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)
b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)
B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
B = \(\frac{x+3y}{x\left(x-3y\right)}\)
30 tháng 11 2019
\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)
\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)
\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)
\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)
T
11 tháng 12 2023
\(A=(x+3y)(x^2-3xy+9y^2)+3y(x+3y)(x-3y)-x(3xy+x^2-5)-5x+1\\A=(x+3y)[x^2-x\cdot3y+(3y)^2]+3y[x^2-(3y)^2]-3x^2y-x^3+5x-5x+1\\A=x^3+(3y)^3+3y(x^2-9y^2)-3x^2y-x^3+1\\A=x^3+27y^3+3x^2y-27y^3-3x^2y-x^3+1\\A=1\)$\Rightarrow$ Giá trị của $A$ không phụ thuộc vào giá trị của biến.
27 tháng 9 2021
a: \(\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
b: \(\left(x-3y\right)^3=x^3-9x^2y+27xy^2-27y^3\)
NK
1
KT
10 tháng 8 2018
đk: \(x\ne0\); \(x\ne\pm3y\)
\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
\(=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
\(=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x+3y}{x\left(x-3y\right)}\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)+\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
\(=x^3-27y^3+x^3+27y^3=2x^3=2.\left(-1\right)^3=-2\)
\(\Leftrightarrow x^3-3y^3+x^3+3y^3\)
\(\Leftrightarrow2x^3\)