tim GTLN cua cac bieu thuc
a)1+6x-x\(^{^2}\)
b)-5x\(^2\)-4x+1
tim gia tri lon nhat ( hoac nho nhat ) cua cac bieu thuc sau:
a) A = x^2 - 6x + 11
b) B = 2x^2 + 10x - 1
c) C = 5x - x^2
giai jup mik voi
tim GTLN hoac GTNN cua bieu thuc C= -x2+6x+1
Tim x , y , z biet : x^2 +y^2 -2z+4y +5=0
Tim GTLN cua bieu thuc P = -x^2 +6x +1
\(P=-x^2+6x+1=-\left(x^2-6x+9\right)+10=-\left(x-3\right)^2+10\le10\)Vậy \(Max_P=10\) khi \(x-3=0\Rightarrow x=3\)
b, \(P=-x^2+6x+1=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-3x-3x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-10\ge-10\)
\(\Rightarrow-\left[\left(x-3\right)^2-10\right]\ge10\)
Hay \(P\ge10\) với mọi giá trị của \(x\in R\).
Để \(P=10\) thì \(-\left[\left(x-3\right)^2-10\right]=10\)
\(\Rightarrow\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy.....
Chúc bạn học tốt!!!
tim gia tri cua x de bieu thuc
A=\(\dfrac{-4}{x^2-4x+10}\) co GTNN
B= -2 + 4x +1 co GTLN
C= \(\dfrac{2}{x^2+4x+5}\) co GTLN
D= \(\dfrac{5}{x^2-6x+12}\) co GTLN
E=\(\dfrac{x^2-2x+2018}{x^2}\) co GTNN
\(A=-\dfrac{4}{x^2-4x+10}\\ =-\dfrac{4}{\left(x^2-2.x.2+4+6\right)}\\ =-\dfrac{4}{\left(x-2\right)^2+6}\)
\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2+6\ge6\\ \Rightarrow\dfrac{4}{\left(x-2\right)^2+6}\le\dfrac{2}{3}\\ \Rightarrow A=-\dfrac{4}{\left(x-2\right)^2+6}\ge-\dfrac{2}{3}\)
Min A=-2/3 khi x=2
\(C=\dfrac{2}{x^2+4x+5}=\dfrac{2}{\left(x+2\right)^2+1}\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1\)
\(\Rightarrow C\le2\)
Dấu ''='' xảy ra \(\Leftrightarrow x=-2\)
Vậy Min C = 2 kjhi x = -2
1) Tim GTNN cua bieu thuc sau
a) M = x^2 + 4x + 9
b) N = x^2 - 20x +101
5) Tim GTLN cua bieu thuc sau
a) C = -y^2 + 6y -15
b) B = -x^2 + 9x - 12
c) D = 3x - x^2
Bài 1:
a: \(M=x^2+4x+4+5=\left(x+2\right)^2+5>=5\)
Dấu '=' xảy ra khi x=-2
b: \(N=x^2-20x+101=x^2-20x+100+1=\left(x-10\right)^2+1>=1\)
Dấu '=' xảy ra khi x=10
Tim x de gia tri cua cac bieu thuc sau =0
a. x^3-x^2-4x-4
b. x^3-x^2+2x-2
c. x^3+6x^2+11x+6
b: \(\Leftrightarrow x^2\left(x-1\right)+2\left(x-1\right)=0\)
=>x-1=0
=>x=1
c: \(\Leftrightarrow x^3+x^2+5x^2+5x+6x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\)
hay \(x\in\left\{-1;-3;-2\right\}\)
Tim gia tri nho nhat cua cac bieu thuc sau:
B=2x2+10x-1
C=5x-x2
\(2x^2+10x-1\)
\(=2\left(x^2+5x-\frac{1}{2}\right)\)
\(=2\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{27}{4}\right)\)
\(=2\left(\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right)\)
\(=\frac{-27}{2}-2\left(x+\frac{5}{2}\right)^2\le\frac{-27}{2}\)
\(MinB=\frac{-27}{2}\Leftrightarrow x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)
+) \(B=2.\left(x^2+5x-\frac{1}{2}\right)\)
\(B=2.\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{27}{4}\right)\)
\(B=2.\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=1.\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)
Vậy Min B=-27/2 khi và chỉ khi x=-5/2
Tim nghien cua da thuc
a)A(x)=-4x-5 h)K(x)=/3x-2/+/4-6x/
b)B(x)=3(2x-1)-2(x + 1) i)M(x)=/x-1/+(x2-1)2
c)C(x)=(2x2-8)(-x2+1) j)N(x)=4x2-3x+7
d)D(x)=3x-x3 k)Pk(x)=7x2-2x-9
l)Q(x)=5x2-11x+6
e)E(x)=2x3+4x
f)G(x)=x3-x2+x-1
a) Đặt A(x)=0
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=-\dfrac{5}{4}\)
b) Đặt B(x)=0
\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow6x-3-2x-2=0\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
TIM GTLN HOAC GTNN CUA CAC BIEU THUC SAU
B=5-2Z^2
C=/X-3/+/5-X/
B = 5 - 2z2
Vì 2z2 ≥ 0 => B = 5 - 2z2 ≤ 5
Dấu "=" xảy ra khi 2z2 = 0 => z = 0
Vậy Bmax là 5 tại z = 0
C = |x - 3| + |5 - x| ≥ |x - 3 + 5 - x| = 2
Dấu "=" xảy ra khi (x - 3)(5 - x) ≥ 0 <=> 5 ≥ x ≥ 3
Vậy Cmin = 2 tại 5 ≥ x ≥ 3