Để  \(\frac{1}{x^2+2010}\)đạt GTLN thì \(x^2+2010\)đạt GTNN mà \(x^2\)\(\ge\)0

\(\Leftrightarrow\)\(x^2+2010\ge\)2010

\(\Rightarrow\)\(\frac{1}{x^2+2010}\le\frac{1}{2010}\)khi x = 0

Vậy \(\frac{1}{x^2+2010}\)đạt GTLN bằng \(\frac{1}{2010}\)khi x = 0

cam on ban nha

Ta có : |x-2013| ≥ 0 với mọi x

=> |x-2013|+2≥ 2

=>\(\frac{2016}{\left|x-2013\right|+2}\)≤ \(\frac{2016}{2}\)

=> Max A =1008

<=> x-2013=0 

<=> x=2013

=1008

nha anh của cậu rất đẹp tớ rất thích susuca

không biet luon

Min A = -1 <-> x=2/3

Min B =2 <-> x=0 ; y=1

Max C = 5 <-> x=1/2

Max D = 1/3 <-> x=2

a)\(\left|3x-2\right|\ge0\Rightarrow2\left|3x-2\right|\ge0\Rightarrow A=2\left|3x-2\right|-1\ge-1\)

=>A_min=-1 <=>|3x-2|=0 <=>3x-2=0 <=>3x=2<=>x=2/3

b)\(x^2\ge0;\left|2y-2\right|\ge0\Leftrightarrow3\left|3y-2\right|\ge0\)

=>\(x^2+3\left|2y-2\right|\ge0\Rightarrow B=x^2+3\left|2y-2\right|-1\ge-1\)

=>B_min=-1 <=>x²=0 và |2y-2|=0 <=> x=0 và y=1

tach 14-x = 10-4-x roi sau do chac ban cung phai tu biet lam

a) Ta có : \(1-4x-2x^2=-\left(2x^2+4x-1\right)=-[2(x^2+2x+1)-3]=-[2(x+1)^2-3]\)

Lại có \(2\left(x+1\right)^2\ge0=>-[2(x+1)^2-3]\le-3\)

Dấu"=" xảy ra khi và chỉ khi \(x+1=0=>x=-1\)

Vậy giá trị lớn nhất của biểu thức đã cho bằng -3 khi x=-1

b)\(x^2-4x+y^2+2y-5=\left(x-2\right)^2+\left(y+1\right)^2-10\)

Lại có : \(\left(x-2\right)^2\ge0;\left(y+1\right)^2\ge0=>\left(x-2\right)^2+\left(y+1\right)^2-10\ge-10\)

Dấu "=" xảy ra khi và chỉ khi \(x-2=y+1=0=>x=2;y=-1\)

\(\text{a) }1-4x-2x^2\)

\(=\left(-2x^2-4x-2\right)+3\)

\(=-2\left(x^2+2x+1\right)+3\)

\(=-2\left(x+1\right)^2+3\)

\(\text{Vì }-2\left(x+1\right)^2\le0\)

\(\text{nên }-2\left(x+1\right)^2+3\le3\)

\(\text{Do đó: }GTLN=3\), dấu bằng  xảy ra khi \(x=-1\)

\(\text{b) }x^2-4x+y^2+2y-5\)

\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)-10\)

\(=\left(x-2\right)^2+\left(y+1\right)^2-10\)

\(\text{Vì }\left(x-2\right)^2\ge0;\left(y+1\right)^2\ge0\)

\(\text{nên }\left(x-2\right)^2+\left(y+1\right)^2\ge0\)

\(\text{hay }\left(x-2\right)^2+\left(y+1\right)^2-10\ge-10\)

\(\text{Do đó: }GTNN=-10\), dấu bằng xảy ra tai \(x=2\)và  \(y=-1\)

a) \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\left(ĐK:x\ne-3;x\ne2\right)\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

Để \(A=-\frac{3}{4}\)

\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\Leftrightarrow x=\frac{22}{7}\left(tm\right)\)

Vậy \(x=\frac{22}{7}\) thì \(A=-\frac{3}{4}\)

b) \(A=\frac{x-4}{x-2}=\frac{\left(x-2\right)-2}{x-2}=1-\frac{2}{x-2}\)

Để \(A\in Z\Rightarrow\frac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)\)

Mà: \(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)

=> \(x-2\in\left\{1;-1;2;-2\right\}\)

+) \(x-2=1\Rightarrow x=3\left(tm\right)\)

+) \(x-2=-1\Rightarrow x=1\left(tm\right)\)

+) \(x-2=2\Rightarrow x=4\left(tm\right)\)

+) \(x-2=-2\Rightarrow x=0\left(tm\right)\)

Vậy \(x\in\left\{0;1;3;4\right\}\) thì \(A\in Z\)

A=x+2/x+3-5/(x-2)(x+3)-1/x-2

A=(x+2)(x-2)-5-x-3/(x-2)(x+3)

A=x^2-4-5-x-3/(x-2)(x+3)

A=x^2-x-12/(x-2)(x+3)

A=(x+3)(x-4)/(x-2)(x+3)

A=x-4/x-2

Để A=-3/4 thì x-4/x-2=-3/4

Từ đó suy ra (x-4)4=-3(x-2)

4x-16=-3x+6

7x=22

x=22/7

b,Do A nguyên nên x-4/x-2 nguyên(x#2)

suy ra x-4-x+2 chia hết cho x-2

nên 2 chia hết cho x-2

mà ước 2=-2;-1;1;2

nên x=0;1;3;4

hình như bạn cho đề sai

đúng đè mà!