1) tim GTNN cua : a) /2.x-1/ +3 b) -4+ /x-5/ c)(2-5.x)2 +1
2) tim GTLN cua : a)5-/4.x-7/ b) -/x+5/- 6 c) 12- (4.x-5)2
HELP ME !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
1, tim GTLN cua A=13/(x+5)^2+7
2, tim GTNN cua B=|x+2017|+(y+3)^2+2017
3, cho a-1/2=b+3/4=c-5/6 va 5a-3b-4c=46. Tim a,b,c.
a)tim GTNN cua
A=/x-2019/+(y-1)^2020-2
C=/x-3/+/x+4/-5
b)tim GTLN
B=3^2-4/x^2-25/
D=x-4/x-5
a, 1, Vì |x - 2019| ≥ 0 ; (y - 1)2020 ≥ 0 => |x - 2019| + (y - 1)2020 ≥ 0 => |x - 2019| + (y - 1)2020 + (-2) ≥ (-2) => A ≥ -2
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-2019=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2019\\y=1\end{cases}}\)
Vậy GTNN A = -2 khi x = 2019 và y = 1
2, Ta có: |x - 3| = |3 - x|
Vì |x - 3| + |x + 4| ≥ |x - 3 + x + 4| = |1| = 1
=> C ≥ 1 - 5 => C ≥ -4
Dấu " = " xảy ra <=> (3 - x)(x + 4) ≥ 0
+) Th1: \(\hept{\begin{cases}3-x\ge0\\x+4\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\le3\\x\ge-4\end{cases}\Rightarrow}-4\le x\le3\)
+) Th2: \(\hept{\begin{cases}3-x\le0\\x+4\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge3\\x\le-4\end{cases}}\)(Vô lý)
Vậy GTNN của C = -4 khi -4 ≤ x ≤ 3
b,
1, Vì |x2 - 25| ≥ 0 => 4|x2 - 25| ≥ 0 => 32 - 4|x2 - 25| ≤ 32 = 9
Dấu " = " xảy ra <=> x2 - 25 = 0 <=> x2 = 25 <=> x = 5 hoặc x = -5
Vậy GTLN B = 9 khi x = 5 hoặc x = -5
2, Đk: x ≠ 5
\(D=\frac{x-4}{x-5}=\frac{\left(x-5\right)+1}{x-5}=1+\frac{1}{x-5}\)
Để D mang giá trị lớn nhất <=> \(\frac{1}{x-5}\)mang giá trị lớn nhất <=> x - 5 mang giá trị nhỏ nhất <=> x - 5 = 1 <=> x = 6
=> \(D=1+1=2\)
Vậy GTLN của D = 2 khi x = 6
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 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
Bai 1: Tim so nguyen x biet:
x+(x+1)+(x+2)+...+35=0
Bai 2: Tim GTLN:
a) 8-(x+2)^2=E
b) -|x+2|+10=F
Bai 3: Tim x \(\in\)Z:
a)(2x-4)(x+4)<0
b)(x+5)(3x-12)>0
Bai 11: Cho:
S=1-2+3-4+5-6+...+19-20
a) S co\(⋮\)2; 3; 5 khong?
b) Tim tat ca cac uoc cua S
chi tiet gi minh nha
a,tìm GTNN
d=5|x+3y+1|^11+7(x^2-4)^12+20
b, tim GTLN
E=27-3(x^2+1)^2012-3|2x-y+4|
bai 1:tim GTNN cua bieu thuc
A=x2+3x+7
B=(x-2)(x-5)(x2-7x-10)
bai 2:tim GTLN cua bieu thuc
A=11-10x-x2
B=[x-4](2-[x-4])
bai 3:tim x,y sao cho
A=2x2+9y2-6xy-6x-12y+2016 co GTNN
B=-x2+2xy-4y2+2x+10y-8 co GTLN
bai 4 :
a)cho x+y=3;x2+y2=5.tinh x3+y3
b)cho x-y=5;x2+y2=15.tinh x3-y3
1)tim x biet rang:
a)3^x-1=1/243
b)2^x+2^x+3=144
c)81^-2x.27x=9^5
2)tim tiep so ghang thu 5 cua day so sau:-1/a^2;2/a^3;-6/a^4;24/a^5;...
3)tim so tu nhien x biet :
a)4^x+4^x+3=4160
b)2^x-1+5.2^x-2=7/32
a) Tim xThuoc Z
5 . (x/3-4) =15
2x+3 chia het cho x+1
b) Tim GTLN cua 7 phan (x+1)^2+1
c)Chung to neu a,b nguyen to thi a^2 -b^2 chia het cho 24
a) \(5\cdot\left(\frac{x}{3}-4\right)=15\)
\(\Leftrightarrow\)\(\frac{x-12}{3}=3\)
\(\Leftrightarrow x-12=9\)
\(\Leftrightarrow x=21\)
Vạy x=21
+) 2x+3 chia hét cho x+1
Bạn chia cột dọc 2x+3 : x+1 =2 dư 1
Vậy để 2x+3 \(⋮\) x+1 thì x+1 \(\in\) Ư(1)
Mà Ư(1)={1;-1}
=> x+1={1;-1}
*)TH1: x+1=1<=>x=0
*)TH2: x+1=-1<=>x=-2
Vậy x={-2;0} thì 2x+3\(⋮\) x+1
b)Tìm GTLN của \(\frac{7}{\left(x+1\right)^2+1}\)
Vì \(\left(x+1\right)^2\ge0\) với mọi x
=>\(\left(x+1\right)^2+1\ge1\)
=> \(\frac{7}{\left(x+1\right)^2+1}\le\frac{7}{1}=7\)