Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B= \(2x^2-4x+3=2x^2-2x.\sqrt{2}.\sqrt{2}+2+3-2\)-2
\(=\left[\sqrt{2}x-\sqrt{2}\right]^2+1>=1\)
Min B=1.Dấu "=" xảy ra khi và chỉ khi \(\sqrt{2}x=\sqrt{2}\Leftrightarrow x=1\)
Phân tích đa thức thành nhân tử
\(x^4+1=x^4+2x^2+1-2x^2=\left[x^2+1\right]^2-2x^2\)
\(=\left[x^2+1+\sqrt{2}x\right]\left[x^2+1-\sqrt{2}x\right]\)
b)\(\left(x^2-8\right)^2+36\)
\(=x^4-16x^2+100\)
\(=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-36x^2\)
\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
c)81x4+4
=81x4+36x2+4-36x2
=(9x2+2)2-(6x)2
=(9x2+6x+2)(9x2-6x+2)
a)x3+3x2+3x+1-27z3
=(x+1)3-(3z)3
=(x+1-3z)[(x+1)2+3z(x+1)+9z2
b)81x4+4
=(9x2)2+22
=(9x2+2)2-(6x)2
=(9x2-6x+2)(9x2+6x+2)
B1:
a, \(4x^2+y\left(y-4x\right)-9\)
\(=4x^2+y^2-4xy-9\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
1.
b) \(a^2-b^2+a-b\)
\(=\left(a^2-b^2\right)+\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+1\right)\)
1. \(x^2-x+\frac{1}{4}-\frac{485}{4}=\left(x-\frac{1}{2}\right)^2-\frac{485}{4}=\left(x-\frac{1}{2}-\frac{\sqrt{485}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{485}}{2}\right)=\left(x-\frac{1+\sqrt{485}}{2}\right)\left(x+\frac{\sqrt{485}-1}{2}\right)\)
2) \(81x^2+4=4\left(\frac{81}{4}x^2+1\right)\)
3) \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\)=> Min A =-3 <=> x=2
. Nhớ L I K E
1.
\(a,x^2-x-121\)\(=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{485}{4}\)\(=\left(x-\frac{1}{2}\right)^2-\frac{485}{4}\)\(=\left(x-\frac{1}{2}-\frac{\sqrt{485}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{485}}{2}\right)\)
\(b,81x^2+4\)\(=\left(9x^2\right)^2+2^2=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2\)
\(=\left(9x^2+2\right)^2-\left(6x\right)^2\)\(=\left(9x^2+2-6x\right)\left(9x^2+2+6x\right)\)
2.
\(A=x^2-4x+1=\left(x^2-2.x.2+4\right)-3\)\(=\left(x-2\right)^2-3\)
Vì \(\left(x-2\right)^2\ge0\)\(\Rightarrow\left(x-2\right)^2-3\ge-3\)
Dấu ''='' xảy ra khi x-2=0 => x=2
Vậy GTNN của A là A=-3 khi x=2