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\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\\ =\left[\left(x^n\right)^2+x^ny^n+\left(y^n\right)^2\right]\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\\ =\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)=x^{6n}-y^{6n}\)
a) 2x^2 + 3( x-1)(x+1) - 5x(x+1)
= 2x^2 + 3( x^2 -1 ) - 5x(x+1)
= 2x^2 + 3x^2 - 3 - 5x^2 - 5x
= -5x -3
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
Bài làm :
\(a,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)
\(=8x+16-5x^2-10x+\left(4x-8\right)\left(x+1\right)+2\left(x^2-2^2\right)+10\)
\(=8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8+10\)
\(=\left(8x-10x+4x-8x\right)+\left(-5x^2+4x^2+2x^2\right)+\left(16-8-8+10\right)\)
\(=-6x+x^2+10\)
a)\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)\(=8x+16-5x^2-2+4x-8x-8+2x-4x-4+10\)\(=\left(8x+4x-8x+2x-4x\right)+\left(16-2-8-4+10\right)+5x^2\)
\(=2x+12+5x^2\)
b)\(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=4x-4x-20-\left[x^2+5x+2x+10\right]-3\left[x^2+2x-1x-2\right]\)
\(=4x-4x-20-x^2-5x-2x-10-3x^2-6x+3x+6\)
\(=\left(4x-4x-5x-2x-6x+3x\right)+\left(-20-10+6\right)+\left(-x^2-3x^2\right)\)
\(=-10x-24-4x^2\)
c)\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)
Xét tích \(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\Leftrightarrow\left(x^n\right)^3-\left(y^n\right)^3=x^{3n}-y^{3n}\)
Thay vào bt đã cho ta có \(\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(\Leftrightarrow\left(x^{3n}\right)^2-\left(y^{3n}\right)^2=x^{6n}-y^{6n}\)