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a) Ta có: (x-3)(y+2)=5
nên (x-3) và (y+2) là ước của 5
\(\Leftrightarrow x-3;y+2\in\left\{1;-5;-1;5\right\}\)
Trường hợp 1:
\(\left\{{}\begin{matrix}x-3=1\\y+2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x-3=5\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=-1\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}x-3=-1\\y+2=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-7\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}x-3=-5\\y+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\left\{\left(4;3\right);\left(8;-1\right);\left(2;-7\right);\left(-2;-3\right)\right\}\)
b) Ta có: (x-2)(y+1)=5
nên x-2 và y+1 là các ước của 5
\(\Leftrightarrow x-2;y+1\in\left\{1;-1;5;-5\right\}\)
Trường hợp 1:
\(\left\{{}\begin{matrix}x-2=1\\y+1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x-2=5\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=0\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}x-2=-1\\y+1=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-6\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}x-2=-5\\y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\left\{\left(3;4\right);\left(7;0\right);\left(1;-6\right);\left(-3;-2\right)\right\}\)
a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
a) 2/3x +1/2 = 1/10
=> 2/3x = 1/10 - 1/2
2/3x = -2/5
=> x = -2/5 : 2/3
x = -3/5
b) (4,5 - 2x) : 3/4 = \(1\frac{1}{3}=\frac{4}{3}\)
=> 4,5 - 2x = 4/3 x 3/4
4,5 - 2x = 1
-2x = 4,5 - 1
-2x = 3,5
=> x = 3,5 : (-2)
x = - 1,75
c) x/4 = 5/20
=> x = 5/20 x 4
x = 1
d) ?????
2/3x+1/2=1/10
<=>2/3x=1/10-1/2
<=>2/3x=-2/10
=>x=-2/10 chia 2/3=-3/10
ý kia tương tự
1)\(\left(2^5:2^3\right).2^x=64\)
\(\Rightarrow2^{5-3+x}=2^6\)
\(\Rightarrow2^{2+x}=2^6\)
\(\Rightarrow.2^22^x=2^6\)
\(\Rightarrow2^x=2^6:2^2\)
\(\Rightarrow2^x=2^4\Rightarrow x=4\)
2)Tính:
\(F=3^0+3^1+...+3^9\)
\(\Rightarrow3F=3\left(3^0+3^1+...+3^9\right)=3+3^2+3^3+...+3^{10}\)
\(3F-F=3+3^2+...+3^{10}-3^0-3^1-...-3^9\)
\(2F=3^{10}-3^0=3^{10}-1\)
\(F=\frac{3^{10}-1}{2}\)
2
ta có : F = 1 + 3 + 32 + ..... + 39
=> 3F = 3 + 32 + 33 +..... + 310
=> 3F - F = 310 - 1
=> 2F = 310 - 1
=> F = \(\frac{3^{10}-1}{2}\)
câu 1:
3.(x+2) + 5x = 22
=> 3x + 6 + 5x = 22
=> 8x = 22 - 6 = 16
=> x = 16/8 = 2
câu 2:
2(x + 1) + 5(x + 2) = 61
=> 2x + 2 + 5x + 10 = 61
=> 7x + 12 = 61
=>7x = 61 - 12 = 49
=> x = 49/7 = 7
hok tốt
# kiseki no enzeru #
C1:
3( x + 2 ) + 5x = 22
3x + 6 + 5x = 22
3x + 5x = 22 - 6
8x = 16
x = 16 : 8
x = 2
C2:
2( x + 1 ) + 5( x +2 ) = 61
2x + 2 + 5x + 10 = 61
2x + 5x = 61 - 2 - 10
7x = 49
x = 49 : 7
x = 7
~ Hok tốt ~
\(\frac{1}{2}.x+\frac{3}{5}.\left(x-2\right)=3\)
\(\frac{1}{2}.x+\frac{3}{5}.x-\frac{6}{5}=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{21}{5}\)
\(\frac{11}{10}.x=\frac{21}{5}\)
\(\Rightarrow x=\frac{42}{11}\)
x = 42/11 đó bạn.
Nhớ k cho mình nha!!!