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x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x)
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1]
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4
Vay gia tri nho nhat P=4 khi x=1
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4]
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2
Vay gia tri nho nhat Q= -9/2 khi x= 3/2
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4
= ( x-1/2)^2 + (y+3)^2 +3/4
M>= 3/4
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7]
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7
Vay GTLN A=7 khi x=2
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4]
GTLN B= 1/4 khi x=1/2
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4)
= -2[(x-1/2)^2 +9/4]
GTLN N= -9/2 khi x=1/2
Bài 2 :
a) Ta có : \(P=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
Nên : \(P=\left(x-1\right)^2+4\ge4\forall x\)
Vậy GTNN của P là 4 khi x = 1
Bài 1:
Ta có: \(4x-x^2-5\)
\(=-x^2+4x-5=-x^2+4x-4-1\)
\(=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\)
Vì \(-\left(x-2\right)^2< 0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-1< 0\forall x\)
\(\Rightarrow4x-x^2-5< 0\forall x\)
Bài 1:
\(4x-x^2-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-2.x.2+4+1\right)\)
\(=-\left(x-2\right)^2-1\)
Vì \(-\left(x-2\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-2\right)^2-1\le-1\)
\(\Rightarrow4x-x^2-5< 0\) với mọi x
Bài 2:
a) \(M=x^2+y^2-x+6y+10\)
\(M=x^2-2.x\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+y^2+2.y.3+9-9+10\)
\(M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\left(y+3\right)^2\ge0\) với mọi y
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\) với mọi x và y
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow Mmin=\dfrac{3}{4}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)
b) \(Q=2x^2-6x\)
\(Q=2\left(x^2-3x\right)\)
\(Q=2\left(x^2-2.x\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(Q=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Vì \(2\left(x-\dfrac{3}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(\Rightarrow Qmin=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
\(A=x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)
Vậy A > 0 với mọi x.
\(B=x^2-2xy+y^2+1\)
\(=\left(x-y\right)^2+1\)
\(\left(x-y\right)^2\ge0\)
\(\Rightarrow\left(x-y\right)^2+1\ge1>0\)
Vậy B > 0 với mọi x, y.
\(M=x^2-6x+12\)
\(=x^2-6x+9+3\)
\(=\left(x-3\right)^2+3\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+3\ge3\)
\(MinB=3\Leftrightarrow x=3\)
\(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)
\(x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=7\)
\(2x^2+6x+5-2x^2+4x-2=7\)
\(10x=7+3\)
\(10x=10\)
\(x=1\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
\(x^3-\frac{1}{4}x=0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)
\(\left(x+10\right)^2-\left(x^2+2x\right)\)
\(=x^2+20x+100-x^2-2x\)
\(=18x+100\)
\(\left(x+2\right)\left(x-2\right)+\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+x\right)\)
\(=x^2-4+x^3-1-x^3-x^2\)
\(=-5\)
a : x2 + 4x + 7 = (x + 2)2 + 3 > 0
b : 4x2 - 4x + 5 = (2x - 1)2 + 4 > 0
c : x2 + 2y2 + 2xy - 2y + 3 = (x + y)2 + (y - 1)2 + 2 > 0
d : 2x2 - 4x + 10 = 2(x - 1)2 + 8 > 0
e : x2 + x + 1 = (x + 0,5)2 + 0,75 > 0
f : 2x2 - 6x + 5 = 2(x - 1,5)2 + 0,5 > 0
bạn phải tách từng câu ra. chứ kiểu này k ai trả lời cho đâu
2)
a)x2-y2=(x+y).(x-y)=(87+13).(87-13)=100.74=7400
b)x3-3x2+3x-1=(x-1)3=(101-1)3=1003=1000000
c)x3+9x2+27x+27=(x+3)3=(97+3)3=1003=1000000
4)
a)x2-6x+10=x2-6x+9+1=(x-3)2+1>=1>0 voi moi x
b)4x-x2-5= -(x2-4x+5)= -(x2-4x+4+1)= -(x-2)2 - 1<0 voi moi x
Bài 1:
Ta có:
VT=\(\left(a^2+b^2\right)\left(c^2+d^2\right)\)
=\(a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
=\(\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)
=\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\) = VP
Vậy đẳng thức được chứng minh
Bài 2:
a/P=\(x^2-2x+5\)
=\(\left(x^2-2x+1\right)+4\)
=\(\left(x-1\right)^2+4\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow P\ge4\forall x\)
Vậy GTNN của P là 4 khi \(\left(x-1\right)^2=0\) hay x=1
b/Q=\(2x^2-6x\)
=\(2\left(x^2-3x\right)\)
=\(2\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
=\(2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
Vì \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)
\(\Rightarrow Q\ge-\dfrac{9}{2}\forall x\)
Vậy GTNN của Q là \(-\dfrac{9}{2}\) khi \(\left(x-\dfrac{3}{2}\right)^2=0\) hay \(x=\dfrac{3}{2}\)
c/\(M=x^2+y^2-x+6y+10\)
=\(x^2-x+\dfrac{1}{4}+y^2+6y+9+\dfrac{3}{4}\)
=\(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)
\(\Rightarrow M\ge\dfrac{3}{4}\forall x,y\)
Vậy GTNN của M là \(\dfrac{3}{4}\) khi \(\left(x-\dfrac{1}{2}\right)^2=0\) và \(\left(y+3\right)^2=0\) hay \(x=\dfrac{1}{2}\) và y = -3
Bài 3:
a/Đặt A=\(x^2-6x+10\)
A=\(x^2-6x+9+1=\left(x-3\right)^2+1\)
Vì \(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+1\ge1>0\forall x\)
\(\Rightarrow A>0\forall x\)
\(\Rightarrow x^2-6x+10>0\forall x\)
b/Đặt B=\(4x-x^2-5\)
B=\(-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1\)
Vì \(\left(x-2\right)^2\ge0\forall x\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-1\le-1< 0\forall x\)
\(\Rightarrow B< 0\forall x\)
\(\Rightarrow4x-x^2-5< 0\forall x\)
cho tớ hỏi là ở câu b, bài 2 í cậu lấy 9/4 ở đâu vậy ???