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a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}-\dfrac{1213}{100}=2\cdot\left[\left(x-\dfrac{10}{7}\right)\cdot\dfrac{49}{50}+\dfrac{2}{5}\right]\)
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}=\dfrac{49}{25}\left(x-\dfrac{10}{7}\right)+\dfrac{4}{5}+\dfrac{1213}{100}\)
\(\Leftrightarrow\dfrac{1313}{100}-\dfrac{101}{25}x=\dfrac{49}{25}x-\dfrac{490}{175}+\dfrac{1293}{100}\)
=>-6x=13/5
hay x=-13/30
b:
ĐKXĐ: x<>0
\(\dfrac{2}{x}+\dfrac{y}{3}=\dfrac{1}{6}\)
=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)
=>\(6\left(6+xy\right)=3x\)
=>\(x=2\left(6+xy\right)=12+2xy\)
=>\(x\left(1-2y\right)=12\)
mà x,y là các số nguyên
nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)
c: ĐKXĐ: y<>-1
\(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)
=>\(\dfrac{xy+x+3}{3\left(y+1\right)}=\dfrac{1}{6}\)
=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)
=>\(2xy+2x+6=y+1\)
=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)
=>\(\left(2x-1\right)\left(y+1\right)=-6\)
mà x,y là các số nguyên
nên \(\left(2x-1;y+1\right)\in\left\{\left(1;-6\right);\left(-1;6\right);\left(3;-2\right);\left(-3;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(0;5\right);\left(2;-3\right);\left(-1;1\right)\right\}\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)
=>-x-6-3=4
=>-x-9=4
=>-x=5
hay x=-5
b: =>(x+1)2=16
=>x+1=4 hoặc x+1=-4
=>x=3 hoặc x=-5
c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)
=>x-29=0
hay x=29
a)\(3\dfrac{1}{3}:2\dfrac{1}{2}-1< x< 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\)
\(\dfrac{4}{3}-1< x< \dfrac{23}{7}+\dfrac{5}{2}\)
\(\dfrac{1}{3}< x< \dfrac{81}{14}\)
Vì\(\dfrac{1}{3}=0,333333333333333333333333...\)
\(\dfrac{81}{14}=5,785714286\)
=>\(x=\left\{1;2;3;4;5\right\}\)
b)\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}\)
\(-\dfrac{1}{12}< x< \dfrac{1}{8}\)
Vì\(-\dfrac{1}{12}=-0.08333333333333333\)
\(\dfrac{1}{8}=0.125\)
=> \(x=\left\{0\right\}\)
a.\(3\dfrac{1}{3}:2\dfrac{1}{2}-1< x< 7\dfrac{2}{3}.\dfrac{3}{7}+\dfrac{5}{2}\)
\(\dfrac{4}{3}-1< x< \dfrac{23}{7}+\dfrac{5}{2}\)
\(\dfrac{1}{3}< x< \dfrac{81}{14}\)
\(0,3333...< x< 5,7857...\)
Vì \(x\in Z\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
Vậy........
b. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\dfrac{-1}{12}< x< \dfrac{1}{8}\)
\(-0,0833...< x< 0,125\)
Vì \(x\in Z\Rightarrow x\in\left\{0\right\}\)
Vậy............
\(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right):\left(x-\dfrac{13}{7}\right)< 0\)
\(TH1:\left\{{}\begin{matrix}x-\dfrac{1}{5}>0\\x-\dfrac{13}{7}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{13}{7}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{5}< x< \dfrac{13}{7}\)
\(TH2:\left\{{}\begin{matrix}x-\dfrac{1}{5}< 0\\x-\dfrac{13}{7}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\) (vô lý nên loại)
Vậy \(\dfrac{1}{5}< x< \dfrac{13}{7}\) thỏa mãn đề bài
\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
Mà \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ..
\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
=> \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}\)= 0
(x + 1).(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\)) = 0
Ta thấy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) > 0
=> x + 1 = 0
x = 0 - 1
x = -1