bài1:tìm giá trị lớn nhất
a)C=\(\dfrac{3}{4}\)-/x+\(\dfrac{5}{2}\)/
b)D=-/x+\(\dfrac{3}{2}\)/-\(\dfrac{5}{7}\)
c)E=\(\dfrac{-5}{3}\)-/x=\(\dfrac{3}{5}\)/
Tìm giá trị lớn nhất , giá trị nhỏ nhất của biểu thức :
a)\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\)
b)B=\(\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\)
c)C=\(-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ai lm đc câu nào thì giúp mk với , cảm ơn !!
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)
Bài 2:Tìm x biết:
a)\(\dfrac{1}{7}+x=\dfrac{-2}{3}\)
b)\(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
c)\(\left\{\dfrac{3}{5}-2x\right\}.\dfrac{5}{8}=1\)
d)\(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
e)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
f)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
g)\(\left|X+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
h)\(\left(\dfrac{1}{32}\right)^x.8^{2x}=512\)
i)\(5,3x+\left(-3,3\right)x+1,7=-4,9\)
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
Tìm x, biết:
a) \(\dfrac{-2}{5}\) + \(\dfrac{4}{5}\) . x = \(\dfrac{3}{5}\)
b) \(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\) : x = \(\dfrac{2}{5}\)
c) \(\dfrac{4}{7}\) . x + \(\dfrac{2}{3}\) = \(\dfrac{-1}{5}\)
d) \(\dfrac{5}{7}\) : x -1 = \(\dfrac{2}{3}\)
a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)
\(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)
\(\dfrac{4}{5}\).\(x\) = 1
\(x\) = \(\dfrac{5}{4}\)
b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)
\(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )
\(x\) = - \(\dfrac{20}{29}\)
c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)
\(\dfrac{4}{7}\).\(x\) = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{4}{7}\).\(x\) = - \(\dfrac{13}{15}\)
\(x\) = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)
\(x\) = - \(\dfrac{91}{60}\)
d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)
\(\dfrac{5}{7}\): \(x\) = \(\dfrac{2}{3}\)+ 1
\(\dfrac{5}{7}\): \(x\) = \(\dfrac{5}{3}\)
\(x\) = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)
\(x\) = \(\dfrac{3}{7}\)
Tìm giá trị lớn nhất (GTNN) của các biểu thức sau:
A= \(\dfrac{4+5\left|1-2x\right|}{7}\)
B= \(\dfrac{x^2+4x-6}{3}\)
C= \(\dfrac{5}{x^2-2x+3}\)
Tìm các giá trị lượng giác, biết:
a) \(cos\alpha=\dfrac{2}{\sqrt{5}}\); \(-\dfrac{\pi}{2}< \alpha< 0\)
b) \(sinx=\dfrac{3}{5};\dfrac{\pi}{2}< x< \pi\)
c) \(tanx=\dfrac{4}{5};-\pi< x< -\dfrac{\pi}{2}\)
d) \(cotx=-\dfrac{3}{4};\dfrac{3\pi}{2}< x< \pi\)
e) \(tanx=\dfrac{4}{5};\pi< x< \dfrac{3\pi}{2}\)
f) \(cosx=\dfrac{4}{5};270^o< x< 360^o\)
g) \(sinx=-\dfrac{3}{5};180^o< x< 270^o\)
a: -pi/2<a<0
=>sin a<0
=>sin a=-1/căn 5
tan a=-1/2
cot a=-2
b: pi/2<x<pi
=>cosx<0
=>cosx=-4/5
=>tan x=-3/4
cot x=-4/3
c: -pi<x<-pi/2
=>cosx<0 và sin x<0
1+tan^2x=1/cos^2x
=>1/cos^2x=1+16/25=41/25
=>cosx=-5/căn 41
sin x=-6/căn 41
cot x=5/4
g: 180 độ<x<270 độ
=>cosx <0
=>cosx=-4/5
tan x=3/4
cot x=4/3
Tìm giá trị lớn nhất hoặc giá trị nhỏ nhất của các biểu thức sau:
a) S= \(\dfrac{3}{2x^2+2x+3}\)
b) T= \(\dfrac{5}{3x^2+4x+15}\)
c) V= \(\dfrac{1}{-x^2+2x-2}\)
d) X= \(\dfrac{2}{-4x^2+8x-5}\)
Tìm giá trị lớn nhất hoặc nhỏ nhất của các đa thức sau:
A =\(\dfrac{37}{x^2-2x+3}\) B \(=\dfrac{-26}{x^2-5x+10}\) C \(=\dfrac{-2023}{x^2-x+6}\) D \(=\dfrac{0,75}{x^2+x+5}\) E \(=\dfrac{13}{2x^2-x+37}\) F \(=\dfrac{-61}{3x^2-x+19}\)
Tìm x, biết:
a) \(\dfrac{2}{5}\) + \(\dfrac{3}{4}\): x = \(\dfrac{-1}{2}\)
b) \(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\) x + \(\dfrac{2}{3}\) x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\)x - x= \(\dfrac{-9}{14}\)
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
c và d mình làm dược nhưng ko ghi được cái suy ra
a,\(\dfrac{4}{9}\) : \(\dfrac{3}{10}\)
b,\(\dfrac{3}{7}\) x \(\dfrac{5}{7}\)
c,\(\dfrac{4}{9}\) : \(\dfrac{3}{2}\)
D \(\dfrac{3}{9}\) + \(\dfrac{5}{6}\) x \(\dfrac{2}{3}\)
E \(\dfrac{2}{7}\) x \(\dfrac{4}{3}\) + \(\dfrac{2}{6}\)
G \(\dfrac{4}{7}\) : 2 + \(\dfrac{4}{7}\)
a)\(=\dfrac{4}{9}\times\dfrac{10}{3}=\dfrac{40}{27}\)
b)\(\dfrac{3}{7}\times\dfrac{5}{7}=\dfrac{15}{49}\)
c)\(\dfrac{4}{9}:\dfrac{3}{2}=\dfrac{4}{9}\times\dfrac{2}{3}=\dfrac{8}{27}\)
a, =40/27
b, =15/49
c, =8/27
d, =3/9 +5/9
e, = 8/21 +2/6=5/7
g, =2/7+4/7=6/7
tính giá trị lớn nhất của
A = -( x + 1 ) \(^2\)+ 5
tìm x
2. x - 0, 7 = 1, 3
x - √25 = \(\left(\dfrac{2}{5}-\dfrac{6}{5}\right)\)
\(\dfrac{3}{4}+\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
\(\text{#ID07 - DNfil}\)
`A = -(x + 1)^2 + 5`
Ta có: `(x + 1)^2 \ge 0` `AA` `x`
`=> -(x + 1)^2 \le 0` `AA` `x`
`=> -(x + 1)^2 + 5 \le 5` `AA` `x`
Vậy, GTLN của A là `5` khi `(x + 1)^2 = 0 => x + 1 = 0 => x = -1`
________
2.
`2x - 0,7 = 1,3`
`=> 2x = 1,3 + 0,7`
`=> 2x = 2`
`=> x = 1`
Vậy, `x = 1`
__
`x - \sqrt{25} = (2/5 - 6/5)`
`=> x - \sqrt{25} = -3/5`
`=> x = -3/5 + \sqrt{25}`
`=> x = -3/5 + 5`
`=> x = 22/5`
Vậy, `x = 22/5`
__
`3/4 + 1/4 \div x = 2/5`
`=> 1/4 \div x = 2/5 - 3/4`
`=> 1/4 \div x = -7/20`
`=> x = 1/4 \div (-7/20)`
`=> x = -5/7`
Vậy, `x = -5/7.`