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a)\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\)
Vì \(-\left|x+\frac{3}{2}\right|\)\(\le\)0
Suy ra:\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)
Dấu = xảy ra khi \(x+\frac{3}{2}=0\)
\(x=-\frac{3}{2}\)
Vậy Max A=\(\frac{1}{4}\) khi \(x=-\frac{3}{2}\)
b)\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)
Vì \(-\left|x-\frac{4}{3}\right|\le0;-\left|y+\frac{1}{2}\right|\le0\)
Suy ra:\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)
Dấu = xảy ra khi \(x-\frac{4}{3}=0;x=\frac{4}{3}\)
\(y+\frac{1}{2}=0;y=-\frac{1}{2}\)
Vậy Max B=\(\frac{5}{3}\) khi \(x=\frac{4}{3};y=-\frac{1}{2}\)
a/ Ta có ; \(\left|x+\frac{3}{2}\right|\ge0\Rightarrow-\left|x+\frac{3}{2}\right|\le0\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)
Vậy BT đạt giá trị lớn nhất bằng 1/4 khi x = -3/2
b/ \(\begin{cases}\left|x-\frac{4}{3}\right|\ge0\\\left|y+\frac{1}{2}\right|\ge0\end{cases}\) \(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\le0\\-\left|y+\frac{1}{2}\right|\le0\end{cases}\)
\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)
\(\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)
Vậy BT đạt giá trị lớn nhất bằng 5/3 khi x = 4/3 , y = -1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
6 là số chẵn nên \(-\left[\frac{4}{9}x-\frac{2}{15}\right]^6\le0\)
=> B ≥ 3
=> GTLN của B = 3 khi x = 3/10
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\)
=>\(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\)
=>\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\le3\)
=>GTLN của B=3 <=>\(\left(\frac{4}{9}x-\frac{2}{15}\right)^6=0\Leftrightarrow\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow\frac{4}{9}x=\frac{2}{15}\Leftrightarrow x=\frac{2}{15}:\frac{4}{9}=\frac{2}{15}\cdot\frac{9}{4}=\frac{3}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét mẫu (x-2)2+(x-y)4+3
R đạt GTLN khi (x-2)2+(x-y)4+3 nhỏ nhất
Ta có \(\left(x-2\right)^2\ge0\)
\(\left(x-y\right)^4\ge0\)
=>(x-2)2+(x-y)4+3\(\ge3\)
Vậy mẫu số đạt GTNN là 3 khi x=y=2
Khi đó GTLN của R là 2013/3
Vì \(\left(x-2\right)^2\ge0\forall x\in R\)
\(\left(x-y\right)^4\ge0\forall x;y\in R\)
\(\Rightarrow\left(x-2\right)^2+\left(x-y\right)^2+3\ge3\forall x;y\in R\)
Để biểu thức\(R_{max}\Leftrightarrow\)\(\left(x-2\right)^2+\left(x-y\right)^4+3=3\Rightarrow\left(x-2\right)^2=\left(x-y\right)^4=0\)
Ta có \(:\)\(\left(x-2\right)^2=0\Rightarrow x=0+2=2\)
Thay \(x=2\)vào \(\left(x-y\right)^4=0\)ta có \(:\)
\(\left(x-y\right)^4=\left(2-y\right)^4=0\Rightarrow y=2-0=2\)
\(\Rightarrow R_{max}=\frac{2013}{\left(2-2\right)^2+\left(2-2\right)^2+3}=\frac{2013}{3}\)
Vậy GTLN của \(R=\frac{2013}{3}\)tại \(x=2;y=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{4}{\left(x^2-1\right)^2+9}\)
vì \(\left(x^2-1\right)^2\ge0\)
\(\Rightarrow\left(x^2-1\right)^2+9\ge9\)
để A lớn nhất => \(\left(x^2-1\right)^2+9\)nhỏ nhất
dấu "=" xảy ra khi \(x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\)
vậy GTLN của A=\(\frac{4}{9}\)khi và chỉ khi x=+-1
Gọi B= \(\left(x^2-1\right)^2+9\)
Ta có\(\left(x^2-1\right)^2\ge0\)\(\forall x\)
\(\Rightarrow\left(x^2-1\right)^2+9\ge9\)\(\forall x\)
\(\Rightarrow B\ge9\)\(\forall x\)
\(\Rightarrow A=\frac{4}{\left(x^2-1\right)^2+9}\)\(\le\frac{4}{9}\)
Dấu = xảy ra khi \(\left(x^2-1\right)^2+9=9\)
\(\Rightarrow\left(x^2-1\right)^2=0\)
\(\Rightarrow x^2-1=0\)
\(\Rightarrow x^2=1\Rightarrow x=1\)
Vậy GTLN của A là \(\frac{4}{9}\)khi x =1
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x^2-1\right)^2+9\ge0\forall x\)
Để A có GTLN thì (x2-1)2+9 phải nhỏ nhất
=>(x2-1)2+9=9
=>x=0
\(\Rightarrow A=\frac{4}{\left(0^2-1\right)^2+}=\frac{4}{10}=0,4\)
Ta có: \(\left(x-\frac{2}{3}\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^2+9\ge9\)
\(\Rightarrow\frac{1}{\left(x-\frac{2}{3}\right)^2+9}\le\frac{1}{9}\)
\(\Rightarrow P=\frac{4}{\left(x-\frac{2}{3}\right)^2+9}\le\frac{4}{9}\)
Dấu "=" xảy ra khi x-2/3=0 => x=2/3
Vậy GTLN của P = 4/9 khi x=2/3
\(\left(x-\frac{2}{3}\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^2+9\ge9\)
\(\Rightarrow\frac{4}{\left(x-\frac{2}{3}\right)^2+9}\le\frac{4}{9}\)
Dấu "=" xảy ra khi \(x-\frac{2}{3}=0\Rightarrow x=\frac{2}{3}\)
Vậy GTLN của P = \(\frac{4}{9}\)khi x = \(\frac{2}{3}\)