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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(Q=\frac{12}{x^2+2x+15}=\frac{12}{x^2+2x+1+14}=\frac{12}{\left(x+1\right)^2+14}\)
Nhận thấy: (x+1)2 \(\ge\)0 với mọi x
=> (x+1)2+14 \(\ge\)14 với mọi x => \(Q\le\frac{12}{14}\) với mọi x
=> GTNN của Q là: \(\frac{12}{14}=\frac{6}{7}\)
ĐS: Qmin=6/7
![](https://rs.olm.vn/images/avt/0.png?1311)
X2-\(\frac{7}{9}\)X=0 <=> X(X-\(\frac{7}{9}\))=0
=> x=0 hoặc x-\(\frac{7}{9}\)=0
x-\(\frac{7}{9}\)=0 <=>X=0+\(\frac{7}{9}\)=\(\frac{7}{9}\)
=> X=0 hoặc \(\frac{7}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{-157}{623}=\frac{-331441}{132699}\)
\(\frac{-47}{213}=\frac{-29281}{132699}\)
Vì \(-331441< -29281\)nên\(\frac{-331441}{132699}< \frac{-29281}{132699}\)
hay\(\frac{-157}{623}< \frac{-47}{213}\)
Vậy \(\frac{-157}{623}< \frac{-47}{213}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{101}}\)
\(\Rightarrow25A=5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\)
\(\Rightarrow25A-A=\left(5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\)
hay \(24A=5-\frac{1}{5^{101}}\)
\(\Rightarrow A=\frac{5-\frac{1}{5^{101}}}{24}\)
\(\Rightarrow A:\left(1-\frac{1}{5^{102}}\right)=\frac{5-\frac{1}{5^{101}}}{24}.\frac{1}{1-\frac{1}{5^{102}}}\)
\(=\frac{5\left(1-\frac{1}{5^{102}}\right)}{24}.\frac{1}{1-\frac{1}{5^{102}}}=\frac{5}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(A=12-\left|x-3\right|-\left|y+7\right|\)
\(-\left|x-3\right|\le0;-\left|y+7\right|\le0\)
\(\Rightarrow A\le12-0-0=12\)
Vậy Max A = 12 <=> x = 3 ; y = -7
b)\(B=-\left(x-2018\right)^6-1\)
\(-\left(x-2018\right)^6\le0\)
\(B\le0-1=-1\)
Vậy Max B = -1 <=> x = 2018
a) \(A=12-\left|x-3\right|-\left|y+7\right|\)
Nhận thấy: \(\left|x-3\right|\ge0;\)\(\left|y+7\right|\ge0\)
suy ra: \(A=12-\left|x-3\right|-\left|y+7\right|\le12\)
Vậy MIN A = 12
Dấu "=" xảy ra <=> \(x=3;y=-7\)
b) \(B=-\left(x-2018\right)^6-1\)
Nhận thấy: \(\left(x-2018\right)^6\ge0\)
suy ra: \(B=-\left(x-2018\right)^2-1\le-1\)
Vậy MIN B = -1
Dấu "=" xảy ra <=> \(x=2018\)
c) \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\)
Nhận thấy: \(\left|x+8\right|\ge0\) \(\left(3y+7\right)^{2016}\ge0\)
suy ra: \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\le\frac{20}{7}\)
Vậy MIN C = 20/7
Dấu "=" xảy ra <=> \(x=-8;y=-\frac{7}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề ra ,ta có :
- 1 / 12 < x < 1 / 8 mà x có giá trị nguyên
=> x = 0
\(x:\frac{1}{3}=\frac{12}{99}:\frac{15}{90}\)
\(x:\frac{1}{3}=\frac{12}{99}:\frac{1}{6}\)
\(x:\frac{1}{3}=\frac{8}{11}\)
\(x=\frac{8}{11}X\frac{1}{3}\)
\(x=\frac{8}{33}\)
tk mk nhe
Ta có: x:1/3=12/99:15/90
x:1/3=8/11
x=8/33
Vậy x=8/33