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a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)
b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)
\(-3x=\frac{47}{10}\)
\(x=\frac{-47}{30}\)
c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)
\(-7x=\frac{-1}{6}\)
\(x=\frac{1}{42}\)
d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(3x=\frac{1}{6}\)
\(x=\frac{1}{18}\)
Học tốt nhé bn!
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Bài 1:
A = 3(x + 1)2 + 5
Ta có: (x + 1)2 \(\ge\) 0 Với mọi x
\(\Rightarrow\) 3(x + 1)2 \(\ge\) 0 với mọi x
\(\Rightarrow\) 3(x + 1)2 + 5 \(\ge\) 5 với mọi x
Hay A \(\ge\) 5
Dấu "=" xảy ra khi và chỉ khi x + 1 = 5 hay x = -1
Vậy...
B = 2|x + y| + 3x2 - 10
Ta có: 2|x + y| \(\ge\) 0 với mọi x, y
3x2 \(\ge\) 0 với mọi x
\(\Rightarrow\) 2|x + y| + 3x2 - 10 \(\ge\) -10 với mọi x,y
Dấu "=" xảy ra khi và chỉ khi x + y = 0; x = 0
\(\Rightarrow\) x = y = 0
Vậy ...
C = 12(x - y)2 + x2 - 6
Ta có: 12(x - y)2 \(\ge\) 0 với mọi x; y
x2 \(\ge\) 0 với mọi x
\(\Rightarrow\) 12(x - y)2 + x2 - 6 \(\ge\) -6 với mọi x, y
Dấu "=" xảy ra khi và chỉ khi x = y = 0
Phần D ko rõ đầu bài nha vì D luôn có một giá trị duy nhất
Bài 2:
Phần A ko rõ đầu bài!
B = 3 - (x + 1)2 - 3(x + 2y)2
Ta có: -(x + 1)2 \(\le\) 0 với mọi x
-3(x + 2y)2 \(\le\) 0 với mọi x, y
\(\Rightarrow\) 3 - (x + 1)2 - 3(x + 2y)2 \(\le\) 3 với mọi x, y
Dấu "=" xảy ra khi và chỉ khi x = 2y; x + 1 = 0
\(\Rightarrow\) x = -1; y = \(\dfrac{-1}{2}\)
Vậy ...
C = -12 - 3|x + 1| - 2(y - 1)2
Ta có: -3|x + 1| \(\le\) 0 với mọi x
-2(y - 1)2 \(\le\) 0 với mọi y
\(\Rightarrow\) -12 - 3|x + 1| - 2(y - 1)2 \(\le\) -12 với mọi x, y
Dấu "=" xảy ra khi và chỉ khi x + 1 = 0; y - 1 = 0
\(\Rightarrow\) x = -1; y = 1
Vậy ...
Phần D đề ko rõ là \(\dfrac{5}{2x^2}-3\) hay \(\dfrac{5}{2}\)x2 - 3 nữa
F = \(\dfrac{-5}{3}\) - 2x2
Ta có: -2x2 \(\le\) 0 với mọi x
\(\Rightarrow\) \(\dfrac{-5}{3}-2x^2\) \(\le\) \(\dfrac{-5}{3}\) với mọi x
Dấu "=" xảy ra khi và chỉ khi x = 0
Vậy ...
Chúc bn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, 1/2 +x=3/4`
`=> x= 3/4 -1/2`
`=> x= 3/4-2/4`
`=>x= 1/4`
`b, 5/2 -x=1/3`
`=> x= 5/2 -1/3`
`=> x= 15/6 - 2/6`
`=>x= 13/6`
`c, 2 . (1/3 +x)=1/5`
`=> 1/3 +x=1/5:2`
`=> 1/3 +x= 1/10`
`=>x= 1/10-1/3`
`=>x= 3/30 - 10/30`
`=>x=-7/30`
`d, 2/3 - (1/2 -x)=1/5`
`=> 1/2-x= 2/3 -1/5`
`=>1/2-x= 10/15 - 3/15`
`=>1/2-x=7/15`
`=>x= 1/2-7/15`
`=>x=1/30`
`1/2 + x = 3/4`
`=> x = 3/4 - 1/2`
`=> x = 1/4`
`5/2 - x = 1/3`
`=> x = 5/2 - 1/3`
`=> x = 13/6`
`2.(1/3 + x) = 1/5`
`=>1/3 + x = 1/10 `
`=> x = 1/10 - 1/3`
`=> x = -7/30`
`2/3 - (1/2 -x)= 1/5`
`=> 1/2 - x = 7/15`
`=> x = 1/2 - 7/15`
`=> x = 1/30`
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x/5=6/5
=>x=6
2: =>x^2=36
=>x=6 hoặc x=-6
3: =>x-5/2=3+1/6=19/6
=>x=19/6+5/2=19/6+15/6=34/6=17/3
4: =>x=1/2:2/5=1/2*5/2=5/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow-5x-\frac{1}{5}-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\left(-5x-\frac{1}{2}x\right)+\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\left(\frac{-10}{2}x-\frac{1}{2}x\right)+\left(\frac{5}{15}-\frac{3}{15}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\frac{-11}{2}x+\frac{2}{15}=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\frac{-11}{2}x-\frac{3}{2}x=-\frac{5}{6}-\frac{2}{15}\)
\(\Leftrightarrow\frac{-14}{2}x=-\frac{25}{30}-\frac{4}{30}\)
\(\Leftrightarrow-7x=-\frac{29}{30}\)
\(\Leftrightarrow x=-\frac{29}{30}\times\frac{-1}{7}\)
\(\Leftrightarrow x=\frac{29}{210}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Leftrightarrow3x-\frac{3}{2}-5x-3=\frac{1}{5}-x\)
\(\Leftrightarrow\left(3x-5x\right)-\left(\frac{3}{2}+3\right)=\frac{1}{5}-x\)
\(\Leftrightarrow-2x-\left(\frac{3}{2}+\frac{6}{2}\right)=\frac{1}{5}-x\)
\(\Leftrightarrow-2x-\frac{9}{2}=\frac{1}{5}-x\)
\(\Leftrightarrow-2x+x=\frac{1}{5}+\frac{9}{2}\)
\(\Leftrightarrow-x=\frac{2}{10}+\frac{45}{10}\)
\(\Leftrightarrow-x=\frac{47}{10}\)
\(\Leftrightarrow x=\frac{-47}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. x:(1/2+2/3)=6/5
=>x:7/6=6/5
=>x=6/5*7/6=>x=7/5
b.(x-1/2)-5(x-2/3)=3/2x
=>x-1/2-5x+10/3=3/2x
=>-4x+17/6=3/2x
=>17/6x=3/2x--4x
=>17/6=x(3/2+4)=>17/6=11/2x=>x=17/33
c.-5(x+1/5)-1/2(x-2/3)=x
=>-5x-1-1/2x+1/3=x=>-11/2x-2/3=x
=>-2/3=x+11/2x=>-2/3=x(1+11/2)=>-2/3=13/2x
=>x=-4/39
a) x : (1/2 + 2/3) = 6/5
=> x : 7/6 = 6/5
=> x = 6/5 x 7/6
x = 7/5
b) , c) ko bít hihi
\(2\times\left(\frac{1}{2}\times x-\frac{1}{3}\right)-\frac{3}{2}=\frac{1}{5}\)
\(2\times\left(\frac{1}{2}\times x-\frac{1}{3}\right)=\frac{1}{5}+\frac{3}{2}\)
\(2\times\left(\frac{1}{2}\times x-\frac{1}{3}\right)=\frac{17}{10}\)
\(\frac{1}{2}\times x-\frac{1}{3}=\frac{17}{10}:2\)
\(\frac{1}{2}\times x-\frac{1}{3}=\frac{17}{20}\)
\(\frac{1}{2}\times x=\frac{17}{20}+\frac{1}{3}\)
\(\frac{1}{2}\times x=\frac{71}{60}\)
\(x=\frac{71}{60}:\frac{1}{2}\)
\(x=\frac{71}{30}\)