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8 tháng 10 2017
1/ Ta có \(\frac{1}{3}< \frac{9}{x}< \frac{1}{2}\)
\(\Rightarrow\frac{9}{27}< \frac{9}{x}< \frac{9}{18}\)
\(\Rightarrow27>x>18\)
Vì \(x\in Z\Rightarrow x\in\left\{19,20,...,26\right\}\)
Vậy....
DK
1
2 tháng 7 2023
a: Để E nguyên thì -x+3 chia hết cho x-1
=>-x+1+2 chia hết cho x-1
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
b: \(E=\dfrac{-\left(x-3\right)}{x-1}=\dfrac{-\left(x-1-2\right)}{x-1}=-1+\dfrac{2}{x-1}\)
Để E min thì x-1=-1
=>x=0
1 tháng 2 2022
a) Ta có: \(M=\dfrac{8-x}{x+3}=\dfrac{-\left(x+3\right)+11}{x+3}=-1+\dfrac{11}{x+3}\) (ĐK: \(x\ne-3\))
Để \(M\in Z\) thì \(\left(x+3\right)\inƯ\left(11\right)=\left\{1;-1;11;-;11\right\}\)
\(\Rightarrow x\in\left\{-2;-4;8;-14\right\}\) (TMĐK)
Vậy \(x\in\left\{-2;-4;8;-14\right\}\) thì \(M\in Z\)
6 tháng 2 2018
\(A=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
a) \(A=\frac{\sqrt{\frac{1}{4}}-5}{\sqrt{\frac{1}{4}}+3}\)
\(A=\frac{\frac{1}{2}-5}{\frac{1}{2}+3}\)
\(A=\frac{\frac{-9}{2}}{\frac{7}{2}}\)
\(A=\frac{-9}{2}.\frac{2}{7}\)
\(A=\frac{-9}{7}\)
b) \(A=-1\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+3}=-1\)
\(\Leftrightarrow-\sqrt{x}-3=\sqrt{x}-5\)
\(\Leftrightarrow-\sqrt{x}-\sqrt{x}=-5+3\)
\(\Leftrightarrow-2\sqrt{x}=-2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)
vậy \(x=1\)
c) \(A=\frac{\sqrt{x}+3-8}{\sqrt{x}+3}\)
\(A=1-\frac{8}{\sqrt{x}+3}\)
\(\Leftrightarrow\sqrt{x}+3\inƯ\left(8\right)\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
lập bảng tự làm
DL
6 tháng 2 2018
\(A=\frac{\sqrt{\frac{1}{4}}-5}{\sqrt{\frac{1}{4}}+3}\)
\(A=\frac{\frac{1}{2}-5}{\frac{1}{2}+3}\)
\(A=\frac{-\frac{9}{2}}{\frac{7}{2}}=-\frac{9}{2}\cdot\frac{2}{7}=-\frac{9}{7}\)
SG
2
7 tháng 7 2021
ĐK: x \(\ne\)-3
Ta có: \(\frac{x+5}{x+3}< 1\) <=> \(\frac{x+5}{x+3}-1< 0\)
<=> \(\frac{x+5-x-3}{x+3}< 0\) <=> \(\frac{2}{x+3}< 0\) <=> \(x+3< 0\)(vì 2 > 0)
<=> \(x< -3\)
b)Đk: x \(\ne\)-4
\(\frac{x+3}{x+4}>1\)<=> \(\frac{x+3}{x+4}-1>0\) <=> \(\frac{x+3-x-4}{x+4}>0\)
<=> \(-\frac{1}{x+4}>0\) <=> \(x+4< 0\)(vì -1 < 0)
<=> \(x< -4\)
21 tháng 2 2022
Câu 1:
Để A>1 thì \(\dfrac{x+5}{x+8}-1>0\)
=>-3/x+8>0
=>x+8<0
hay x<-8
8 tháng 3 2016
câu 1 : 0 số cặp x y
câu 2 : ko có giá trị x thỏa mãn
câu 3 : GTLN A=2013
câu 4 : AB=2cm
câu 5: x+y=16
k cho mik nha bạn
18 tháng 6 2019
\(a,\frac{-24}{x}+\frac{18}{x}=\frac{-24+18}{x}=\frac{-6}{x}\)
\(\Leftrightarrow x\inƯ(-6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(b,\frac{2x-5}{x+1}=\frac{2x+2-7}{x+1}=\frac{2(x+1)-7}{x+1}=2-\frac{7}{x+1}\)
\(\Leftrightarrow7⋮x+1\Leftrightarrow x+1\inƯ(7)=\left\{\pm1;\pm7\right\}\)
Xét các trường hợp rồi tìm được x thôi :>
\(c,\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-x-5}{x-1}=\frac{2x+7}{x-1}=\frac{2x-2+9}{x-1}=\frac{2(x-1)+9}{x-1}=2+\frac{9}{x-1}\)
\(\Leftrightarrow9⋮x-1\Leftrightarrow x-1\inƯ(9)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2;10;-8\right\}\)
d, TT
14 tháng 11 2023
a: \(\left|7-2x\right|+7=2x\)
=>\(\left|2x-7\right|+7=2x\)
=>\(\left|2x-7\right|=2x-7\)
=>2x-7>=0
=>\(x>=\dfrac{7}{2}\)
b: \(\left|1-x\right|=4x+1\)
=>\(\left|x-1\right|=4x+1\)
=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)
=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)
d: \(\left|x-7\right|+2x+5=6\)
=>\(\left|x-7\right|=6-2x-5=-2x+1\)
=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
e: 3x-|2x-1|=2
=>|2x-1|=3x-2
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)