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\(A_{min}=8-\frac{25}{4}\) khi x=5/2
Bmin=xem lại đề đúng như đề Bmin=5 khi x=0
C=8+25-(2x+5)^2
Cmax=8+25 khi x=-5/2
Dmax=9 khi x=0
Cụ thể mức nào nhỉ tất cả dự trên HĐT \(\left(a+-b\right)^2=a^2+-2ab+b^2\)
cụ thể con A
\(A=x^2-2.\frac{5}{2}x+\left(\frac{5^2}{2^2}\right)+8-\frac{25}{4}\) đã thêm 25/4 =b vào phần đầu => trừ đi
\(A=\left(x-\frac{5}{2}\right)^2+8-\frac{25}{4}=\left(x-\frac{5}{2}\right)^2+\frac{7}{4}\)
\(\left(x-\frac{5}{2}\right)^2\ge0\Rightarrow A\ge\frac{7}{4}\)đẳng thức khi x-5/2=0=> x=5/2
A=(x-5/2)^2+8-25/4=> Amin=7/4 khi x=5/2
B --> xem lại theo đề Bmin =5 khi x=0
C =8+25-(2x+5)^2=> C max=32 khi x=-5/2
D max=9 khi x=0
Câu 1:
\(A=x^2-3x+9\\ =x^2-3x+\dfrac{9}{4}+\dfrac{27}{4}\\ =\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{27}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge0\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\text{ }A_{\left(Min\right)}=\dfrac{27}{4}\text{ }khi\text{ }x=\dfrac{3}{2}\)
\(B=9x^2-6x+2\\ =9x^2-6x+1+1\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\\ Do\text{ }\left(3x-1\right)^2\ge0\forall x\\ \Rightarrow B=\left(3x-1\right)^2+1\ge1\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(3x-1\right)^2=0\\ \Leftrightarrow3x-1=0\\ \Leftrightarrow3x=1\\ \Leftrightarrow x=\dfrac{1}{3}\\ Vậy\text{ }B_{\left(Min\right)}=1\text{ }khi\text{ }x=\dfrac{1}{3}\)
\(C=-x^2+2x+4\\ =-x^2+2x-1+5\\ =-\left(x^2-2x+1\right)+5\\ =-\left(x-1\right)^2+5\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \Rightarrow-\left(x-1\right)^2\le0\forall x\\ \Rightarrow C=-\left(x-1\right)^2+5\le5\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\ \text{Vậy }C_{\left(Max\right)}=5\text{ }khi\text{ }x=1\)
\(D=-x^2+4x\\ =-x^2+4x-4+4\\ =-\left(x^2-4x+4\right)+4\\ =-\left(x-2\right)^2+4\\ \\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(x-2\right)^2\le0\forall x\\ \Rightarrow C=-\left(x-2\right)^2+4\le4\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }C_{\left(Max\right)}=4\text{ }khi\text{ }x=2\)
Câu 2:
\(\text{Ta có : }x+y=2\\ \Rightarrow\left(x+y\right)^2=2^2\\ \Rightarrow x^2+2xy+y^2=4\\ Thay\text{ }x^2+y^2=10\text{ }vào\\ \Rightarrow2xy+10=4\\ \Rightarrow2xy=-6\\ \Rightarrow xy=-3\\ \text{Ta lại có : }x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\\ Thay\text{ }x^2+y^2=10;x+y=2;xy=-3\text{ }ta\text{ }được:\\ x^3+y^3=2\cdot\left(10+3\right)=26\)
Vậy \(x^3+y^3=26\text{ }tại\text{ }x+y=2;x^2+y^2=10\)
\(A=x^2-6x-4=x^2-6x+9-13=\left(x-3\right)^2-13\ge-13\)
Vậy \(A_{min}=-13\Leftrightarrow x=3\)
\(B=x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
a) x4-4x3+4x2
= x2(x2-4x+4)
= x2(x-2)2
b) 2ab2-a2b-b3
= -b(-2ab+a2+b2)
=-b(a-b)2
e) x3+3x2-3x-1
= x3-x2+4x2-4x+x-1
=(x3-x2)+(4x2-4x)+(x-1)
=x2(x-1)+4x(x-1)+(x-1)
(x-1)(x2+4x+1)
f) x3-3x2-3x+1
=x3+x2-4x2-4x+x+1
=(x3+x2)-(4x2+4x)+(x+1)
=x2(x+1)-4x(x+1)+(x+1)
=(x+1)(x2-4x+1)
g)x3-4x2+4x-1
=x3-x2-3x2+3x+x-1
=(x3-x2)-(3x2-3x)+(x-1)
=x2(x-1)-3x(x-1)+(x-1)
=(x-1)(x2-3x+1)
\(A=4x^2-4x+2017=4\left(x^2-x\right)+2017=4\left(x^2-\dfrac{1}{2}.x.2+\dfrac{1}{4}\right)+2016=4\left(x-\dfrac{1}{2}\right)^2+2016\ge2016\)
\(B=3x-x^2-15\\=-\left(x^2-3x+15\right)=-\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{51}{4}\right)\\ =-\left(x-\dfrac{3}{2}\right)^2-\dfrac{51}{4}\le-\dfrac{51}{4}\)
\(C=3a^2-2ab+b^2-4a+4\\ =\left(a^2-2ab+b^2\right)+\left(2a^2-4a+4\right)\\ =\left(a-b\right)^2+2\left(a^2-2.a.1+1+1\right)\\ =\left(a-b\right)^2+2\left(a-1\right)^2+2\ge2\)