Tìm x ; y thuộc z .Biết :
a.(x+2) * (y-5) = -7
b.(x-1) * (xy-3) = -5
a. (x + 2) * (y - 5) = -7
<=> (y - 5) = -\(\dfrac{7}{x+2}\)
x ∈ Z => 7 chia hết cho (x + 2)
=> x = 5
<=> y -5 = -1
y = -1 + 5
y = 4
Vậy x = 5 và y = 4
b. (x-1) * (xy-3) = -5
<=> (xy-3) = -\(\dfrac{5}{x-1}\)
x ∈ Z => 5 chia hết cho x-1
=> x =6 ; -4; 2
TH1 : x = 6 => 6y-3
<=> 6y - 3 = -\(\dfrac{5}{6-1}\)
=> 6y - 3 = -1
6y = -1+3
6y = 2
y = 6:2
y = 3
TH2 : x = -4
<=> -4y - 3 = - \(\dfrac{5}{-4-1}\)
<=> -4y - 3 = 1
-4y = 1 + 3
-4y = 4
y = 4 : -4
y = -1
TH3 : x = 2
<=> 2y - 3 = -\(\dfrac{5}{2-1}\)
<=> 2y - 3 = -5
2y = -5 + 3
2y = -2
y = -2 : 2
y = -1
Vậy x =2 và y = -1 hoặc x = -4 và y = -1
a) x+7=-12
x=(-12)-7
x=-19
b)x-15=-21
x=(-21)+15
x=-6
c)13-x=20
x=13-20
x=-7
d)17-(2+x)=3
x=17-3
x=14
x=14-2
x=12
a,x+7=-12
=>x= -12-7
=>x= -19
b,x-15= -21
=>x= -21+15
=>x= -6
c,13-x=20
=>x=13-20
=>x= -7
d, 17-(2+x)=3
=>2+x=17-3
=>2+x=14
=>x=14-2
=>x=12
a) \(x+7=-12\)
\(x=-12-7\)
\(x=-19\)
b) \(x-15=-21\)
\(x=-21+15\)
\(x=-6\)
c) \(13-x=20\)
\(x=13-20\)
\(x=-7\)
d) \(17-\left(2+x\right)=3\)
\(2+x=17-3\)
\(2+x=14\)
\(x=14-2\)
\(x=12\)
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
Lời giải:
Đặt $\frac{x}{2}=\frac{y}{5}=\frac{z}{3}=k$
$\Rightarrow x=2k; y=5k; z=3k$
Thay vào điều kiện $2x-y-3z=10$ có:
$2.2k-5k-3.3k=10$
$\Leftrightarrow -10k=10$
$\Leftrightarrow k=-1$
$\Rightarrow x=-2; y=-5; z=-3$
Vậy.........
=> \(\left[\frac{x-1}{2}+1\right]^2=1600\)
=> \(\left[\frac{x-1}{2}+1\right]^2=40^2\)
=> \(\left[\frac{x-1}{2}+1\right]=40^{ }\)
=> \(\frac{x-1}{2}=39\)
=> x - 1 = 39.2
=> x - 1 = 78
=> x=79
TB mỗi bn đc số tiền là:
[(102000 + 231000 + 177000) : 2] : 3 =85000 đ
Đ/S: 85000đ
Tổng số tiền 3 bạn là: (102000 + 231000 + 177000) :2 = 255000 đồng
Trung bình mỗi người có: 255000 : 3 = 85000 đồng
Đặt tử số là A
Ta có: 2A = 2 + 22 + 23 + 24+...+22009
=> A = 2A - A= 22009 - 1
=> B = \(\frac{2^{2009}-1}{1-2^{2009}}\) = -1
Ta có: \(\frac{1}{2}\left(3x-\frac{1}{2}\right)-\frac{1}{4}x=7\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{4}-\frac{1}{4}x=7\)
\(\Leftrightarrow\frac{5}{4}x-\frac{1}{4}=7\)
\(\Leftrightarrow\frac{5}{4}x=7+\frac{1}{4}=\frac{29}{4}\)
hay \(x=\frac{29}{4}:\frac{5}{4}=\frac{29}{4}\cdot\frac{4}{5}=\frac{29}{5}\)
Vậy: \(x=\frac{29}{5}\)