Cho x - y = 4 và x.y = 1
Tính x^3 - y^3
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1.(x+3)^2=9(2x-1)^2
=>(x+3)^2-9(2x-1)^2=0
=>(x+3)^2-[3(2x-1)]^2=0
=>(x+3)^2-(6x-3)^2=0
=>(x+3-6x+3).(x+3+6x-3)=0
=>(-5x+6).7x=0
=> 2 TH
*-5x+6=0
=>-5x=0-6
=>-5x=-6
=>x=6/5
*7x=0
=>x=0
vậy x=6/5 hoặc x=0
2) 8^3-50x=0
=>x.(8x^2-50)=0
=>8x^2-50=0
=>8x^2=50
=x^2=50/8
=>x^2=25/4
=>x=5/2
vậy x=5/2
1.
\(\left(x-1\right)^2+2.\left(x-1\right).\left(5-x\right)+\left(5-x\right)^2\)
\(=[\left(x-1\right)+\left(5-x\right)]^2\)
\(=\left(x-1+5-x\right)^2\)
\(=16\)
2.
\(x^2-16+2.\left(x+4\right)\)
\(=\left(x^2-16\right)+2.\left(x+4\right)\)
\(=\left(x-4\right).\left(x+4\right)+2.\left(x+4\right)\)
\(=\left(x+4\right).\left(x-4+2\right)\)
\(=\left(x+4\right).\left(x-2\right)\)
3.
\(\left(x+3\right)^3\)
\(=x^3+3x^2.3+3x.3^2+3^3\)
\(=x^3+9x^2+27x+27\)
Bài 1: Phân tích đa thức thành nhân tử
\(6xy-9x^2y=3xy.\left(2-3x\right)\)
\(x^2+2xy+y^2-9=\left(x+y\right)^2-3^2=\left(x+y-3\right).\left(x+y+3\right)\)
\(3.\left(x-2\right)-2x+4\)
\(=3.\left(x-2\right)-\left(2x-4\right)\)
\(=3.\left(x-2\right)-2.\left(x-2\right)\)
\(=\left(x-2\right).\left(3-2\right)\)
\(=x-2\)
Bài 2: Tìm x biết
\(\left(x-2\right)^2-4x+8=0\)
\(\Rightarrow\left(x-2\right)^2-4.\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right).\left(x-2-4\right)=0\)
\(\Rightarrow\left(x-2\right).\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)
\(3x^2-5x=0\)
\(\Rightarrow x.\left(3x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\3x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)
Ta có \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=4\left(x^2+y^2+1\right)\)(*)
Mà \(\left(x-y\right)^2=x^2+y^2-2xy\Rightarrow x^2+y^2=\left(x-y\right)^2+2xy=4^2+2.1=18\)
Thay \(x^2+y^2=18\)vào (*), ta có: \(x^3-y^3=4\left(18+1\right)=76\)