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a) 2ˣ + 2ˣ⁺³ = 72
2ˣ.(1 + 2³) = 72
2ˣ.9 = 72
2ˣ = 72 : 9
2ˣ = 8
2ˣ = 2³
x = 3
b) Để số đã cho là số nguyên thì (x - 2) ⋮ (x + 1)
Ta có:
x - 2 = x + 1 - 3
Để (x - 2) ⋮ (x + 1) thì 3 ⋮ (x + 1)
⇒ x + 1 ∈ Ư(3) = {-3; -1; 1; 3}
⇒ x ∈ {-4; -2; 0; 2}
Vậy x ∈ {-4; -2; 0; 2} thì số đã cho là số nguyên
c) P = |2x + 7| + 2/5
Ta có:
|2x + 7| ≥ 0 với mọi x ∈ R
|2x + 7| + 2/5 ≥ 2/5 với mọi x ∈ R
Vậy GTNN của P là 2/5 khi x = -7/2
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
a) \(A=\frac{5}{\sqrt{x}+1}\)
A nguyên\(\Leftrightarrow\frac{5}{\sqrt{x}+1}\)nguyên\(\Leftrightarrow5⋮\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Mà \(\sqrt{x}+1\ge1\)nên \(\sqrt{x}+1\in\left\{1;5\right\}\)
\(TH1:\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
\(TH2:\sqrt{x}+1=5\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)
b) \(B=\frac{7}{\sqrt{x}-3}\)
A nguyên \(\Leftrightarrow\frac{7}{\sqrt{x}-3}\)nguyên\(\Leftrightarrow7⋮\left(\sqrt{x}-3\right)\)
\(\Leftrightarrow\sqrt{x}-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
c) \(C=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\)
\(=1+\frac{4}{\sqrt{x}-3}\)
C nguyên\(\Leftrightarrow\frac{4}{\sqrt{x}-3}\in Z\Leftrightarrow4⋮\sqrt{x}-3\)
Tương tự hai câu a,b
d) \(D=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\sqrt{x}-1+3}{\sqrt{x}-1}\)
\(=1+\frac{3}{\sqrt{x}-1}\)
D nguyên\(\Leftrightarrow\frac{3}{\sqrt{x}-1}\)nguyên
Tương tự
\(a,\frac{-24}{x}+\frac{18}{x}=\frac{-24+18}{x}=\frac{-6}{x}\)
\(\Leftrightarrow x\inƯ(-6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(b,\frac{2x-5}{x+1}=\frac{2x+2-7}{x+1}=\frac{2(x+1)-7}{x+1}=2-\frac{7}{x+1}\)
\(\Leftrightarrow7⋮x+1\Leftrightarrow x+1\inƯ(7)=\left\{\pm1;\pm7\right\}\)
Xét các trường hợp rồi tìm được x thôi :>
\(c,\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-x-5}{x-1}=\frac{2x+7}{x-1}=\frac{2x-2+9}{x-1}=\frac{2(x-1)+9}{x-1}=2+\frac{9}{x-1}\)
\(\Leftrightarrow9⋮x-1\Leftrightarrow x-1\inƯ(9)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2;10;-8\right\}\)
d, TT
\(a,\Rightarrow2x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Rightarrow x\in\left\{-2;1;2;5\right\}\\ b,=\dfrac{2\left(x-1\right)+1}{x-1}=2+\dfrac{1}{x-1}\in Z\\ \Rightarrow x-1\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Rightarrow x\in\left\{0;2\right\}\\ c,\Rightarrow x^2-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow x^2\in\left\{2;4;8\right\}\\ \Rightarrow x^2=4\left(x\in Z\right)\\ \Rightarrow x=\pm2\)
`a)A` nguyên `<=>x+2 in Ư_5`
Mà `Ư_5 ={+-1;+-5}`
`@x+2=1=>x=-1`
`@x+2=-1=>x=-3`
`@x+2=5=>x=3`
`@x+2=-5=>x=-7`
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`b)B=[x-5]/x=1-5/x`
`B` nguyên `<=>x in Ư_{5}`
Mà `Ư_{5}={+-1;+-5}`
`=>x in {+-1;+-5}`
______________________________________________
`c)C=[x-2]/[x+1]=[x+1-3]/[x+1]=1-3/[x+1]`
`C` nguyên `<=>x+1 in Ư_3`
Mà `Ư_3={+-1;+-3}`
`@x+1=1=>x=0`
`@x+1=-1=>x=-2`
`@x+1=3=>x=2`
`@x+1=-3=>x=-4`
______________________________________________
`d)D=[2x-7]/[x+1]=[2x+2-9]/[x+1]=2-9/[x+1]`
`D` nguyên `<=>x+1 in Ư_9`
Mà `Ư_9 ={+-1;+-3;+-9}`
`@x+1=1=>x=0`
`@x+1=-1=>x=-2`
`@x+1=3=>x=2`
`@x+1=-3=>x=-4`
`@x+1=9=>x=8`
`@x+1=-9=>x=-10`