Tìm xϵZ biết:
\(\dfrac{1-x}4\)=\(dfrac-3\8\)
K
Khách
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17 tháng 7 2021
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
17 tháng 7 2021
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
AT
16 tháng 7 2021
a) \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,9\right)\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) \(\sqrt{x}=\sqrt{6+4\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
\(\Rightarrow Q=\dfrac{2+\sqrt{2}+1}{2+\sqrt{2}-3}=\dfrac{3+\sqrt{2}}{\sqrt{2}-1}=\dfrac{\left(3+\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=4\sqrt{2}+5\)
c) \(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Để \(Q\in Z\Rightarrow4⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{4;5;7;2;1\right\}\Rightarrow x\in\left\{16;25;49;4;1\right\}\)
16 tháng 7 2021
a) Ta có: \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
AH
Akai Haruma
Giáo viên
16 tháng 7 2021
Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn!
NT
1
TC
12 tháng 3 2023
`C = (x+4)/(x+1) = (x+1+3)/(x+1) = 1+3/(x+1)`
Để `C in ZZ`
`=> x+1 in Ư(3)=(+-1,+-3)`
`@ x+1 =1 => x =0`
`@ x+1=-1 => x = -2`
`@x+1 =3 => x = 2`
`@x+1 =-3 =>x=-4`
`B = (x-4)/(x+2) = (x+2-6)/(x+2) = 1-6/(x+2)`
Để `B in ZZ`
`=> x+2 in Ư(6) = {+-1,+-2,+-3,+-6)`
`@ x+2 =1 => x = -1`
`@x+2 =-1 => x=-3`
`@ x+2 =2 => x=0`
`@ x+2 =-2 => x=-4`
`@x+2 =3 => x = 1`
`@ x +2 =-3 => x = -5`
`@ x+2 =6 => x=4`
`@x+2 =-6 => x= -8`
2
25 tháng 9 2023
`a)` Với `x >= 0,x ne 4` có:
`Q=[2(2-\sqrt{x})+2+\sqrt{x}-2\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`
`Q=[4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`
`Q=[6-3\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`
`Q=3/[2+\sqrt{x}]`
`b)` Với `x >= 0,x ne 4` có:
`Q=6/5<=>3/[2+\sqrt{x}]=6/5`
`=>12+6\sqrt{x}=15`
`<=>x=1/4` (t/m)
`c)` Với `x >= 0,x ne 4` có:
`Q in Z<=>3/[2+\sqrt{x}] in ZZ`
`=>2+\sqrt{x} in Ư_{3}`
Mà `Ư_{3}={+-1;+-3}`
`@2+\sqrt{x}=1=>\sqrt{x}=-1` (Vô lý)
`@2+\sqrt{x}=-1=>\sqrt{x}=-3` (Vô lý)
`@2+\sqrt{x}=-2=>\sqrt{x}=-4` (Vô lý)
`@2+\sqrt{x}=2=>\sqrt{x}=0<=>x=0` (t/m)
Vậy `x=0`
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 7 2021
ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\dfrac{3x-2\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(2\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-2\sqrt{x}-4-x+1-2x-6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-8\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
Đề bài có vẻ không hợp lý
AT
13 tháng 7 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-1}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
b) \(P=\sqrt{x}-1\Rightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\Rightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(\Rightarrow4\sqrt{x}=x-1\Rightarrow x-4\sqrt{x}-1=0\)
\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\\\sqrt{x}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+2\sqrt{5}}{2}=2+\sqrt{5}\end{matrix}\right.\)
mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=2+\sqrt{5}\Rightarrow x=9+4\sqrt{5}\)
c) \(P=\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=4-\dfrac{4}{\sqrt{x}+1}\)
Để \(P\in Z\Rightarrow4⋮\sqrt{x}+1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\left(\sqrt{x}+1\ge1\right)\)
\(\Rightarrow x\in\left\{0;1;9\right\}\) mà \(x\ne1\Rightarrow x\in\left\{0;9\right\}\)
AT
13 tháng 7 2021
Từ khúc có \(x-4\sqrt{x}-1=0\)
Ta có: \(\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1\)
Thế vào \(\Rightarrow x-4\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow x-\sqrt{x}\left(2-\sqrt{5}+2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow x-\left(2-\sqrt{5}\right)\sqrt{x}-\left(2+\sqrt{5}\right)\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)-\left(2+\sqrt{5}\right)\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)=0\)
\(\Rightarrow\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)\left(\sqrt{x}-\left(2+\sqrt{5}\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\\\sqrt{x}=2+\sqrt{5}\end{matrix}\right.\) rồi khúc sau như trên
AT
12 tháng 7 2021
a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)
\(\Rightarrow x=0\)
c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)
\(\Rightarrow P_{max}=4\) khi \(x=0\)
LH
0
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