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\(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\Leftrightarrow\dfrac{9}{3x}+\dfrac{xy}{3x}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{xy+9}{3x}=\dfrac{5}{6}\Leftrightarrow6\left(xy+9\right)=5\cdot3x\)
\(\Leftrightarrow6xy+54=15x\)\(\Leftrightarrow6xy-15x=-54\)
\(\Leftrightarrow3x\left(2y-5\right)=-54\)
\(\Leftrightarrow x\left(2y-5\right)=-18\)
Ta có:\(y\left(x-1\right)=x^2+2\)
\(\Rightarrow y\left(x-1\right)-x^2=2\)
\(\Rightarrow y\left(x-1\right)-x^2+1=3\)
\(\Rightarrow y\left(x-1\right)-\left(x^2-1\right)=3\)
\(\Rightarrow y\left(x-1\right)-\left(x+1\right)\left(x-1\right)=3\)
\(\Rightarrow\left(y-x-1\right)\left(x-1\right)=3\)
Vì x,y nguyên nên ta có bảng
| x-1 | 3 | 1 | -1 | -3 |
| y-x-1 | 1 | 3 | -3 | -1 |
| x | 4 | 2 | 0 | -2 |
| y | 6 | 8 | 2 | 4 |
Vậy \(\left(x,y\right)\in\left\{\left(4,6\right);\left(2,8\right);\left(0,2\right);\left(-2,4\right)\right\}\) thỏa mãn
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
Giải :
\(\dfrac{x-3}{y-2}=\dfrac{3}{2}\) nên 2(x-3) = 3(y-2)
Do đó : 2x - 6 = 3y - 6 nên 2x = 3y
\(\Rightarrow\) 2x - 2y = y hay 2(x-y) = y
Nên 2.4 = y
Vậy : \(y=8;x=\dfrac{3y}{2}=\dfrac{3.8}{2}=12\)
\(\dfrac{x-3}{y-2}=\dfrac{3}{2}\)
\(\Rightarrow\left(x-3\right)\cdot2=3\cdot\left(y-2\right)\)
\(\Rightarrow2x-6=3y-6\)
\(\Rightarrow2x=3y\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{3}{2}\)
mà x - y = 4
\(\Rightarrow\left\{{}\begin{matrix}x=4:\left(3-2\right)\cdot3=12\\y=4:\left(3-2\right)\cdot2=8\end{matrix}\right.\)
Bài 1
a) \(\frac{5}{6}=\frac{x-1}{x}\)
<=> 5x=6x-6
<=> 5x-6x=-6
<=> -11x=-6
<=> \(x=\frac{6}{11}\)
b)c)d) nhân chéo làm tương tự
Mk giúp pn bài 1 thui nha...
a) A=3+32+33+...+3100
<=>A=(3+32) +(33+34) +...+(399+3100)
<=>A=12+32.(3+32)+...+398.(3+32)
<=>A=12+32.12+...+398.12
<=>A=12.(32+33+...+398)
Ta có 12 chia hết cho 4 => 12.(32+33+...+398) chia hết cho 4 => A chia hết cho 4
Vậy A chia hết cho 4
b) A=3+32+33+...+3100
<=> 3A=32+33+...+3101
<=>3A-A=32+33+...+3101-3-32-33-...-3100
<=>2A=3101-3
<=>A=(3101-3)/2
Thay A=(3101-3)/2 vào 2A+3=3x-1 ta có:
2.[(3101-3)/2]+3=3x-1
<=>3101-3+3=3x-1
<=>3101=3x-1
<=>x-1=101
<=>x=102
vậy x=102
Ai thấy đúng tích nha , mấy pn kb +theo dõi mk vs ạ....
a,\(\dfrac{x}{3}-\dfrac{1}{y}=\dfrac{1}{2}\)
=> \(\dfrac{1}{y}=\dfrac{x}{3}-\dfrac{1}{2}=>\dfrac{1}{y}=\dfrac{2x-3}{6}\)
=> y(2x-3)=6.1=6
=> y và 2x-3 là Ư (6)= {+-1,+-2,+-3,+-6}
| 2x-3 | -1 | 1 | 2 | -2 | 3 | -3 | 6 | -6 |
| x | 1 | 2 | 2,5 | 1/2 | 3 | 0 | 9/2 | -3/2 |
| y | -6 | 6 | 3 | -3 | 2 | -2 | 1 |
-1 |
vậy (x;y)= .......................
b,c làm tương tự
chúc bn học tốt 
Giải:
x/2-2/y=1/2
-2/y=1/2-x/2
-2/y=1-x/2
=>y.(1-x)=-2.2
y.(1-x)= -4
=>y và 1-x thuộc Ư(-4)=(1;-1;2;-2;4;-4)
Ta có bảng tương ứng:
1-x =1 thì x=0;y=-4
1-x=-1 (loại)
1-x=2 thì x=-1;y=2
1-x=-2 thì x=3;y=-2
1-x=4 thì x=-3;y=1
1-x=-4 thì x=5;y=-1
Vậy (x;y)=(0;4);(-1;2);(3;-2);(-3;1);(5;-1)
Chúc bạn học tốt!
ảm ơn bạn nha