P=-|5/3-X|
Q=9-|x-1/10|
tim GTLN
Tìm GTLN của biểu thức : a) P = | 5/3 - x | ; b) Q = 9 - | x - 1/10 |
a) vì | \(\frac{5}{3}-x\)| \(\ge\)0 \(\forall\)x
x không tính được thì phải. sai đề rồi
b) | x - \(\frac{1}{10}\)| \(\ge\)0 \(\forall\)x
\(\Rightarrow\)9 - | x - \(\frac{1}{10}\)| \(\le\)9
\(\Rightarrow\)Qmax \(\Leftrightarrow\)Q = 9 \(\Rightarrow\)x = \(\frac{1}{10}\)
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) tim GTNN, GTLN cua A = \(\sqrt{\left(x-1\right)}\)+\(\sqrt{\left(5-x\right)}\)
b) cho cac so duong x,y thoa man x+y>=3
CM: x+y+1/2x+2/y>=9/2
a ) Tìm GTLN : Áp dụng BĐT bunhiacopski, ta có :
Dầu bằng xảy ra khi \(x-1=5-x\Leftrightarrow x=3\).
Sao ko hiện làm lại :
\(\left(\sqrt{x-1}.1+\sqrt{5-x}.1\right)^2\le\) bé hơn hoặc bằng ( 1 + 1 ) ( x - 1 + 5 -x ) = 8
Tìm x biết |x+1|+|x+2|+.....+|x+9|=x+10
Câu 2 nha
tim gtln cua A và B biết
A=3x+1/x-1 với x€z
B= 5- 3x-5/x-1
Cho bieu thuc \(Q=\frac{x}{x^2-3x+9}-\frac{11}{x^3+27}+\frac{1}{x+3}:\frac{x^2-1}{x+3}\)
a)Tim DKXD cua bt Q
b)Rut gon
c)Tim GTLN
d)Co gtri nguyen nao cua x de Q có gtri nguyên hay k
a: ĐKXĐ: x<>-3
b: \(Q=\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{\left(x+3\right)\left(x^2-3x+9\right)}+\dfrac{1}{x+3}\right)\cdot\dfrac{x+3}{x^2-1}\)
\(=\dfrac{x^2+3x-11+x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\cdot\dfrac{x+3}{x^2-1}\)
\(=\dfrac{2x^2-2}{x^2-1}\cdot\dfrac{1}{x^2-3x+9}=\dfrac{2}{x^2-3x+9}\)
Tim GTNN hoac GTLN cua cac bieu thuc sau :
a) A = 3|2x - 1| - 5 b) B = 10 - 5 |x - 2| c) C =\(\frac{1}{\left|x-2\right|+3}\)
a,Tim GTNN cua bieu thuc \(C=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
b,Tim GTLN cua bieu thuc \(D=\frac{4}{\left(2x-3\right)^2+5}\)
\(\text{a)Để C đạt GTNN}\)
\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)
\(\Rightarrow C\ge-10\)
\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)
b)\(\text{Để D đạt GTLN}\)
=>(2x-3)2+5 đạt GTNN
Mà (2x-3)2\(\ge\)5
\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)
1. Tim \(x\in Z\)de \(A\in Z\)
A= \(\frac{x+5}{x+3}\)
2. Tim GTLN hoac GTNN cua bieu thuc:
A= -7+/2-x/
B= 5- /2x+3/
3. Tim \(x\in Z\)
a) \(7⋮x+3\)
b) \(x+4⋮x+1\)
c) \(4x+3⋮x+2\)
d) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+9\right)+\left(x+10\right)=5\)
e) \(\left(x-10\right)+\left(x-9\right)+...+\left(x-1\right)=-2015\)
i) (-4)3 + /x/=2010
k) /x+5/ \(\le2\)
l) /x+3/ = x-1
4. Chung minh rang:
\(A=1-3+3^2-3^2+...+3^{98}-3^{99}⋮4\)
(giai ho mk nhe! thu 2 phai nop roi! ban nao giai duoc nho giai nhanh gium mk nhe, mk mang on suot doi luon!)
1,
để A thuộc Z thì
x+5 chia het cho x+3
co x+3 chia het cho x+3
=>(x+5)-(x+3)chia het cho x+3
hay2 chia het cho x+3
=>x+3 thuộc ước của 2
=>x+3 thuoc {1,-1,2,-2}
ta co bang
x+3 | 1 | -1 | 2 | -2 |
x | -2 | -4 | -1 | -5 |
vay de A thuoc Z thi x thuoc {-2,-4,-1,-2}