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Bài 1:
x/-3=9/4
nên x=-9/4*3=-27/4
2x+y=-4
=>y=-4-2x=-4-2*(-27/4)=-4+27/2=27/2-8/2=19/2
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a) 6 ⋮ x
⇒ x ∈ Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}
b) 8 ⋮ (x + 1)
⇒ x + 1 ∈ Ư(8) ={-8; -4; -2; -1; 1; 2; 4; 8}
⇒ x ∈ {-9; -5; -3; -2; 0; 1; 3; 7}
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B = {a \(\in\) Z| (a2 + 3a + 6) ⋮ (a + 3)}
a2 + 3a + 6 ⋮ a + 3
a.(a + 3) + 6 ⋮ a + 3
6 ⋮ a + 3
a + 3 \(\in\) Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}
Lập bảng ta có:
a + 3 | - 6 | - 3 | -2 | -1 | 1 | 2 | 3 | 6 |
a | - 9 | - 6 | -5 | -4 | -2 | -1 | 0 | 3 |
Theo bảng trên ta có: a \(\in\) {-9; -6; -5; -4; -2; -1; 0; 3}
B = {-9; -6; -5; -4; -2; -1; 0; 3}
Vậy số phần tử tập B là 8 phần tử.
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b) 42x=39.42-37.42
42x=42(39-37)
42x=42.2
Vì 42x=42.2
nên x=2
Vậy x=2
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\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)
a) \(\left(x-6\right)^2=9=3^2\)
\(\Rightarrow x-6=3\) hay \(x-6=-3\)
\(\Rightarrow x=9\) hay \(x=3\)
b) \(4^{2x-6}=1=4^0\)
\(\Rightarrow2x-6=0\Rightarrow x=3\)
a) \(\left(x-6\right)^2=9\)
\(\Rightarrow\left(x-6\right)^2=3^2\)
\(\Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=9\end{matrix}\right.\)
b) \(4^{2x-6}=1\)
\(\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)