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\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a. \(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
vì \(\left(x-1\right)^2\ge0\) với mọi x
=> (x-1)^2 +4 \(\ge\) vợi mọi x
Pmin=4 <=> x-1=0 <=>x=1
1.
b)\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\) và \(y+3=0\)
\(\Leftrightarrow x=\frac{1}{2}\) và \(y=-3\)
Vậy GTNN của M là \(\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)và \(y=-3\)
1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
Bài 1:
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+1\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3-x+y\right)\)
\(=2\left(x-y\right)\left(2x+3+y\right)\)
Bài 2:
\(P=\left(3x-1\right)^2+2\left(3x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(3x-1-x-1\right)^2\)
\(=\left(2x-2\right)^2\)(1)
b) Thay \(x=\frac{9}{4}\)vào (1) ta được:
\(\left(2.\frac{9}{4}-2\right)^2\)
\(=\frac{25}{4}\)
Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)
Bài 3:
Ta có: \(M=x^2+4x+5\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)
Hay \(M\ge1;\forall x\)
Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(M_{min}=1\Leftrightarrow x=-2\)
Bài 1 : trên là sai nha mình làm lại
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3y-x+y\right)\)
\(=2\left(x-y\right)\left(2x+4y\right)\)
\(=4\left(x-y\right)\left(x+2y\right)\)
a) x^3−3x^2−4x+12
=(x^3-3x^2)-(4x-12)
=x^2(x-3)-4(x-3)
=(x-3)(x^2-4)=(x-3)(x-2)(x+2)
b) x^4-5x^2+4=x^4-x^2-4x^2+4
=(x^4-x^2) - ( 4x^2-4)
=x^2(x^2-1) - 4(x^2-1)
=(x^2-1)(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
c) (x+y+z)^3-x^3-y^3-z^3
=x^3+y^3+z^3+3x^2yz+3xy^2z+3xyz^2-x^3-y^3-z^3
=3x^2yz+3xy^2z+3xyz^2
3xyz(x+y+z)
1)
\(A=x^2-3x+5\)
\(=x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
Có: \(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu = xảy ra khi:
\(x-\frac{3}{2}=0\)
\(\Rightarrow x=\frac{3}{2}\)
Vậy: \(MIN_A=\frac{11}{4}\) tại \(x=\frac{3}{2}\)
b) \(B=6x-4x^2-2\)
\(=-\left(4x^2-6x+2\right)\)
\(=-\left[\left(2x\right)^2-2.2x.2+4-2\right]\)
\(=-\left(2x-2\right)^2+2\)
Có: \(-\left(2x-2\right)^2\le0\)
\(\Rightarrow-\left(2x-2\right)^2+2\le2\)
Dấu = xảy ra khi:
\(2x-2=0\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
Vậy: ..................
1)
a) \(A=x^2-3x+5=x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{11}{4}\)\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy \(Min_A=\frac{11}{4}\) khi \(x=\frac{3}{2}\)
b) \(B=6x-4x^2-2=-4\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)\) \(=-4\left(x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{1}{16}\right)\) \(=\frac{1}{4}-4\left(x-\frac{3}{4}\right)^2\le\frac{1}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_B=\frac{1}{4}\) khi \(x=\frac{3}{4}\)