Tìm x ∈ Z, biết :
a) x + 9 = -7
b) x – 4 = -8
c) | x | - 2 = 5
d) |x – 2| = 5
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a) 9 x 1=9 9 x 4=36 9 x 7=63
b) 9 x 2=18 9 x 5=45 9 x 8=72
c) 9 x 3=27 9 x 6=54 9 x 981
d) 9 x 10=90 9 x 0=0 0 x 9=0
a) Ta có: \(x:2=y:\left(-5\right)\)
nên \(\dfrac{x}{2}=\dfrac{y}{-5}\)
mà x-y=-7
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-1\\\dfrac{y}{-5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)
Vậy: (x,y)=(-2;5)
b) Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)
nên \(\dfrac{x}{8}=\dfrac{y}{12}\)(1)
Ta có: \(\dfrac{y}{4}=\dfrac{z}{5}\)
nên \(\dfrac{y}{12}=\dfrac{z}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
mà x+y-z=10
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{8}=2\\\dfrac{y}{12}=2\\\dfrac{z}{15}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)
Vậy: (x,y,z)=(16;24;30)
b)
Do đó ta có
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(a,x\in\left\{-5;-4;-3;-2;-1\right\}\\ b,x\in\left\{-3;-2;-1;...;5;6\right\}\\ c,x\in\left\{-4;-3;...;3;4\right\}\\ d,x\in\left\{-3;-2;-1;0;1;2\right\}\)
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) \(x+\dfrac{4}{9}=\dfrac{5}{27}\)
\(x=\dfrac{5}{27}-\dfrac{4}{9}\)
\(x=-\dfrac{7}{27}\)
b) \(x-\dfrac{4}{11}=\dfrac{7}{33}\)
\(x=\dfrac{7}{33}+\dfrac{4}{11}\)
\(x=\dfrac{19}{33}\)
c) \(\dfrac{8}{5}-x=\dfrac{1}{3}\times\dfrac{2}{5}\)
\(\dfrac{8}{5}-x=\dfrac{2}{15}\)
\(x=\dfrac{8}{5}-\dfrac{2}{15}\)
\(x=\dfrac{22}{15}\)
d) \(x-\dfrac{3}{4}=\dfrac{1}{2}+\dfrac{2}{6}\)
\(x-\dfrac{3}{4}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{3}{4}\)
\(z=\dfrac{19}{12}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)
\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{25}{28}\)
\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)
\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)
\(\Rightarrow x=-\dfrac{37}{30}\)
\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)
\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)
\(\Rightarrow35x=-70\)
\(\Rightarrow x=-2\)
\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)
\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)
\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)
\(\Rightarrow x=-\dfrac{8}{9}\)
B1:Tìm x,biết
a)\(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-3\right)7=\left(x+5\right)5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-21-5x-25=0\)
\(\Leftrightarrow2x-46=0\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=23\)
b)\(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow7.9=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow63=x^2-1\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x=\sqrt{64}=8\)
c)\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=20.5\)
\(\Leftrightarrow\left(x+4\right)^2-100=0\)
\(\Leftrightarrow\left(x+4\right)^2-10^2=0\)
\(\Leftrightarrow\left(x+4-10\right)\left(x+4+10\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+14=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
a) x + 9 = -7
x = -7 – 9
x = -16
b) x – 4 = -8
x = -8 + 4
x = -4
c) | x | - 2 = 5
| x | = 5 + 2
| x | = 7
x = ±7
d) | x – 2 | = 5
x – 2 = 5 hoặc x – 2 = -5
x = 5 + 2 hoặc x = -5 + 2
x = 7 hoặc x = -3