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a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
a)
\(4x-10< 0\\ 4x< 10\\ x< \dfrac{10}{4}=\dfrac{5}{2}\)
b)
\(2x+x+12\ge0\\ 3x\ge-12\\ x\ge-\dfrac{12}{3}=-4\)
c)
\(x-5\ge3-x\\ 2x\ge8\\ x\ge4\)
d)
\(7-3x>9-x\\ -2>2x\\ x< -1\)
đ)
\(2x-\left(3-5x\right)\le4\left(x+3\right)\\ 2x-3+5x\le4x+12\\ 3x\le15\\ x\le5\)
e)
\(3x-6+x< 9-x\\ 5x< 15\\ x< 3\)
f)
\(2t-3+5t\ge4t+12\\ 3t\ge15\\ t\ge5\)
g)
\(3y-2\le2y-3\\ y\le-1\)
h)
\(3-4x+24+6x\ge x+27+3x\\ 0\ge2x\\ 0\ge x\)
i)
\(5-\left(6-x\right)\le4\left(3-2x\right)\\ 5-6+x\le12-8x\\ \\ 9x\le13\\ x\le\dfrac{13}{9}\)
k)
\(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\ 10x-15-20x+28\ge19-2x-22\\ 13-10x\ge-2x-3\\ -8x\ge-16\\ x\le\dfrac{-16}{-8}=2\)
l)
\(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\ \dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\ \Rightarrow40x-100-90x+30< 36-12x-30x+15\\ 130-50x< 51-42x\\ 92x< -79\\ x< -\dfrac{79}{92}\)
m)
\(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\ \dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\ \Rightarrow10x-3+2x>7x-5+2x\\ 12x-3>9x-5\\ 3x>-2\\ x>-\dfrac{2}{3}\)
n)
\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\ \dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\ \Rightarrow28x-8-24x< 60-3x+6\\ 4x-8< -3x+66\\ 7x< 74\\ x< \dfrac{74}{7}\)
a) \(4x-10< 0\)
\(\Leftrightarrow4x< 10\)
\(\Leftrightarrow x< \dfrac{5}{2}\)
b) ???
c) \(x-5\ge3-x\)
\(\Leftrightarrow2x-5\ge3\)
\(\Leftrightarrow2x\ge8\)
\(\Leftrightarrow x\ge4\)
d) \(7-3x>9-x\)
\(\Leftrightarrow7-2x>9\)
\(\Leftrightarrow-2x>2\)
\(\Leftrightarrow x< -1\)
đ) ???
e) \(3x-6+x< 9-x\)
\(\Leftrightarrow4x-6< 9-x\)
\(\Leftrightarrow5x-6< 9\)
\(\Leftrightarrow5x< 15\)
\(\Leftrightarrow x< 3\)
f) ???
g) ???
h) \(3-4x+24+6x\ge x+27+3x\)
\(\Leftrightarrow2x+27\ge4x+27\)
\(\Leftrightarrow-2x\ge0\)
\(\Leftrightarrow x\le0\)
i) \(5-\left(6-x\right)\le4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x\le12-8x\)
\(\Leftrightarrow x-1\le12-8x\)
\(\Leftrightarrow9x-1\le12\)
\(\Leftrightarrow9x\le13\)
\(\Leftrightarrow x\le\dfrac{13}{9}\)
k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\)
\(\Leftrightarrow10x-15-20x+28\ge19-2x-22\)
\(\Leftrightarrow-10x+23\ge-3-2x\)
\(\Leftrightarrow-8x+13\ge-3\)
\(\Leftrightarrow-8x\ge-16\)
\(\Leftrightarrow x\ge2\)
l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)
\(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\)
\(\Leftrightarrow x>-\dfrac{121}{8}\)
m, n) làm tương tự:
đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
Mấy này bạn quy đồng lên cùng mẫu xong khử mẫu rồi giải. Dễ mà.
a)Ta thấy: \(x^2\ge0\forall x\)\(\Rightarrow-x^2\le0\forall x\)\(\Rightarrow5-x^2\le5\forall x\)
Đẳng thức xảy ra khi \(-x^2=0\Rightarrow x=0\)
b)Ta thấy:\(x^2\ge0\forall x\)\(\Rightarrow5+x^2\ge5\forall x\)\(\Rightarrow\dfrac{1}{5+x^2}\le\dfrac{1}{5}\forall x\)
Đẳng thức xảy ra khi \(x^2=0\Rightarrow x=0\)
c)Ta có: \(x^2-4x+7=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\ge3\forall x\)\(\Rightarrow\dfrac{1}{\left(x-2\right)^2+3}\le\dfrac{1}{3}\forall x\)
\(\Rightarrow\dfrac{3}{\left(x-2\right)^2+3}\le\dfrac{3}{3}=1\forall x\)
Đẳng thức xảy ra khi \(\left(x-2\right)^2=0\Rightarrow x=2\)
d)\(-2x^2+3x+2017\)
\(=\dfrac{16145}{8}-2x^2+3x-\dfrac{9}{8}\)
\(=\dfrac{16145}{8}-2\left(x^2-\dfrac{3x}{2}+\dfrac{9}{16}\right)\)
\(=\dfrac{16145}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{16145}{8}\forall x\)
Đẳng thức xảy ra khi \(-2\left(x-\dfrac{3}{4}\right)^2=0\)\(\Rightarrow x=\dfrac{3}{4}\)
a) ta có: \(-x^2\le0\) với mọi x
=> \(5-x^2\le5\) với mọi x
dấu "=" xảy ra khi x= 0
vậy max = 5 khi x = 0
b) để \(\dfrac{1}{5+x^2}\) nhận max
<=> 5+x2 nhận min
mà x2 \(\ge\) 0 với mọi x
=> 5+x2\(\ge\) 5 với mọi x
dấu "=" xảy ra khi x = 0
vậy Min của 5 +x2 =5 khi x =0
=> max của \(\dfrac{1}{5+x^2}\) = \(\dfrac{1}{5}\) khi x =0
c) để \(\dfrac{3}{x^2-4x+7}\) nhận max
<=> x2-4x+7 nhận min
ta có: x2-4x+7 = (x-2)2+3
mà (x-2)2 \(\ge\) 0 với mọi x
=> (x-2)2+3 \(\ge\) 3 với mọi x
<=> x2-4x+7 \(\ge\) 3 với mọi x
dấu "=" xảy ra khi x=2
=> min của x2 -4x+7 = 3 khi x=2
=> max của \(\dfrac{1}{x^2-4x+7}=\dfrac{1}{3}\) khi x=2
d) Ta có:-2x2+3x+2017
= \(-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)+2018,125\)
= \(-2\left(x-\dfrac{3}{4}\right)^2+2018,125\)
mà \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x
=> \(-2\left(x-\dfrac{3}{4}\right)^2+2018,125\)\(\le\) 2018,125 với mọi x
=> -2x2+3x+2017 \(\le\) 2018,125 với mọi x
dấu "=" xảy ra khi x =\(\dfrac{3}{4}\)
=> max của -2x2+3x+2017 = 2018,125 khi \(x=\dfrac{3}{4}\)