Bài 1:
a) \(x^2+9y^2-y^4-6xy\)
\(=\left(x^2-6xy+9y^2\right)-y^4\)
\(=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(y^2\right)^2\)
\(=\left(x-3y\right)^2-\left(y^2\right)^2\)
\(=\left(x-3y-y^2\right)\left(x-3y+y^2\right)\)
b) \(2x^2-x-28\)
\(=2x^2-8x+7x-28\)
\(=2x\left(x-4\right)+7\left(x-4\right)\)
\(=\left(x-4\right)\left(2x+7\right)\)
Bài 2:
a) \(2x\left(x^2-2x+3\right)-2x^3\)
\(=2x\left(x^2-2x+3-x^2\right)\)
\(=2x\left(3-2x\right)\)
b) \(2x\left(x-3\right)-\left(x+5\right)\left(2x-1\right)\)
\(=\left(2x^2-6x\right)-\left(2x^2+9x-5\right)\)
\(=2x^2-6x-2x^2-9x+5\)
\(=-15x+5\)
\(=-5\left(3x-1\right)\)
c) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x+5\right)\left(x-5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
Bài 3:
a) \(x-2=\left(x-2\right)^2\)
\(\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\)
\(\Rightarrow\left(x-2\right)\left(1-x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(-3x+9\right)x^2-7x+21=0\)
\(\Rightarrow-3\left(x-3\right)x^2-7\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(-3x^2-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\-3x^2-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-\dfrac{7}{3}\end{matrix}\right.\)
Mà x2 > 0 hoặc x2 = 0 với mọi x
=> x2 = -7/3 không thỏa mãn
=> x= 3
Phân tích đa thức
a, x^2+9y^2-y^4-6xy
=(x^2-6xy+9y^2)-y^4
=(x-3y)^2-y^4
=(x-3y-y^2)(x-3y+y^2)
b, 2x^2-x-28
=(2x^2-8x)+(7x-28)
=2x(x-4)+7(x-4)
=(x-4)(2x+7)
Rút gọn
a,2x(x^2-2x+3)-2x^3
=2x(x^2-2x+3-x^2)
=2x(-2x+3)
b,2x(x-3)-(x+5)(2x-1)
=2x^2-6x-2x^2-9x+5
=-15x+5
=-5(3x-1)
c,(5-x)^2+(x+5)^2-(2x+10)(x-5)
Ta có:(5-x)^2=(x-5)^2
=(x-5)^2-2(x+5)(x-5)+(x+5)^2
=(x-5-x-5)^2
=100
Tìm x
a,x-2=(x-2)^2=0
=>x-2=0=>x=2
b,(-3x+9)x^2-7x+21=0
=>-3(x-3)x^2-7(x-3)=0
=>(x-3)(-3x^2-7)=0
=>\(\left[{}\begin{matrix}x-3=0=>x=3\\-3x^2-7=0=>x=\sqrt{\dfrac{-7}{3}}\end{matrix}\right.\)
