\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{12}< 2x< \dfrac{-12}{31}+\dfrac{-136}{-31}\)
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`a)1/2+[-1]/[-3]-5/12 < 2x < 12/[-31]+136/31`
`186/372+124/372-155/372 < [744x]/372 < [-144]/372+1632/372`
`186+124-155 < 744x < -144+1632`
`155 < 744x < 1488`
`155:744 < 744x:744 < 1488:744`
`5/24 < x < 2`
Vậy `5/24 < x < 2`
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`b)[-2]/5 < x/15 < 1/6`
`[-12]/30 < [2x]/30 < 5/30`
`-12 < 2x < 5`
`-12:2 < 2x:2 < 5:2`
`-6 < x < 5/2`
Vậy `-6 < x < 5/2`
Giải:
a) x - \(\dfrac{9}{25}\)= \(\dfrac{16}{25}\)
x = \(\dfrac{16}{25}\)+\(\dfrac{9}{25}\)
x = \(\dfrac{25}{25}\)
x = 1
b) \(\dfrac{-12}{30}\)<\(\dfrac{x}{30}\)<\(\dfrac{5}{30}\)
=> x có thể bằng \(\dfrac{-11}{30}\) đến \(\dfrac{4}{30}\)
=> x bằng -5; -4; -3; -2; -1;0;1;2
a: -5/5=-1=-7/7
8/7=8/7
b: -3/15=-1/5=-6/30
5/6=25/30
c: -34/136=-1/4=-9/36
-12/108=-1/9=-4/36
26/-156=-1/6=-6/36
a) Do \(\left|1+2x\right|\ge0\Rightarrow\dfrac{-1}{4}\left|1+2x\right|\le0\)
\(\Rightarrow A=2,25-\dfrac{1}{4}\left|1+2x\right|\le2,25\)
\(maxA=2,25\Leftrightarrow x=-\dfrac{1}{2}\)
b) Do \(\left|2x-3\right|\ge0\Rightarrow3+\dfrac{1}{2}\left|2x-3\right|\ge3\)
\(\Rightarrow B=\dfrac{1}{3+\dfrac{1}{2}\left|2x-3\right|}\le\dfrac{1}{3}\)
\(maxB=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}\)
1) = (8/31 + 23/31)+ ( -12/25+-13/25) = 1 + (-1) = 0
2) = 1/2 +3/4 -3/4 + 4/5 = 1/2 +4/5 = 13/10
3)= ( 7/3 x 15/21) x ( -5/2 x 4/-5 ) = ( 7/3 x 5/7 ) x 2 = 5/3 x 2 = 10/3
4) = 1/4 + 3/4 x -7/6 = 1/4 + -7/8 = -5/8
5)= 3/29 x 29/3 - 1/5 x 29/3 = 1 x 29/15 = 29/15
6)= 5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5 /11
7) = 11/12 x -1/8 + 11/12 x -3/16 - 11/12 = 11/12 x ( -1/8 + -3/16 - 1) = 11/12 x -21/16 = -77/64 ( mk ko chắc , bạn ấn máy tính để thử lại )
8) = 203/20 - 7/4 + 41/13 = 198/20 + 41/13 = 99/10 + 41/13 = 1697/130 ( câu này cứ lỏ lỏ kiểu gì ý :v )
1) = (8/31 + 23/31)+ ( -12/25+-13/25) = 1 + (-1) = 0
2) = 1/2 +3/4 -3/4 + 4/5 = 1/2 +4/5 = 13/10
3)= ( 7/3 x 15/21) x ( -5/2 x 4/-5 ) = ( 7/3 x 5/7 ) x 2 = 5/3 x 2 = 10/3
4) = 1/4 + 3/4 x -7/6 = 1/4 + -7/8 = -5/8
5)= 3/29 x 29/3 - 1/5 x 29/3 = 1 x 29/15 = 29/15
6)= 5/7 x ( 5/11 + 2/11 - 14/11) = 5/7 x -7/11 = -5 /11
7) = 11/12 x -1/8 + 11/12 x -3/16 - 11/12 = 11/12 x ( -1/8 + -3/16 - 1) = 11/12 x -21/16 = -77/64 ( mk ko chắc , bạn ấn máy tính để thử lại )
8) = 203/20 - 7/4 + 41/13 = 198/20 + 41/13 = 99/10 + 41/13 = 1697/130 ( câu này cứ lỏ lỏ kiểu gì ý :v )
\(\dfrac{2}{5}\times15\dfrac{1}{3}-\dfrac{2}{5}\times10\dfrac{1}{3}\)
\(=\dfrac{2}{5}\times\left(15\dfrac{1}{3}-10\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\times5\)
\(=2\)
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\(12\dfrac{5}{11}-\left(3\dfrac{1}{4}+2\dfrac{5}{11}\right)\)
\(=12\dfrac{5}{11}-3\dfrac{1}{4}-2\dfrac{5}{11}\)
\(=\left(12\dfrac{5}{11}-2\dfrac{5}{11}\right)-3\dfrac{1}{4}\)
\(=10-3\dfrac{1}{4}\)
\(=\dfrac{27}{4}\)
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\(\dfrac{34}{31}-\dfrac{19}{28}-\dfrac{3}{31}\)
\(=\left(\dfrac{34}{31}-\dfrac{3}{31}\right)-\dfrac{19}{28}\)
\(=\dfrac{31}{31}-\dfrac{19}{28}\)
\(=1-\dfrac{19}{28}\)
\(=\dfrac{9}{28}\)
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
\(\Leftrightarrow6x-3-4x+20< =4x-1+24\)
=>2x+17-4x-23<=0
=>-2x-6<=0
=>-2x<=6
hay x>=-3
\(\Leftrightarrow\dfrac{6x-3-4x+20-4x+1-24}{12}\le0\)
\(\Rightarrow\left\{{}\begin{matrix}-2x-6< 0\\-2x-6\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\ne-3\end{matrix}\right.\)
Nếu bạn bổ sung thêm điều kiện x ∈ N thì làm như này , còn ko có điều kiện của x thì kết luận luôn chứ không suy ra x=1
\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{12}< 2x< \dfrac{-12}{31}+\dfrac{-136}{-31}\)
\(\Rightarrow\dfrac{5}{12}< 2x< 4 \)
\(\Rightarrow\dfrac{5}{24}< x< 2\)
Mà x ∈ N => x = 1
21+31−125<2x<31−12+−31−136
\Rightarrow\dfrac{5}{12}< 2x< 4⇒125<2x<4
\Rightarrow\dfrac{5}{24}< x< 2⇒245<x<2
Mà x ∈ N => x = 1