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a) \(9< 3^x< 243\)
\(\Leftrightarrow3^2< 3^x< 3^5\)
\(\Rightarrow x\in\left\{3;4\right\}\)
b) Sửa đề: \(3^4.3^x\div9=27\)
\(\Leftrightarrow3^{x+4}=3\)
\(\Rightarrow x+4=1\)
\(\Rightarrow x=-3\)
c) \(3^x\div3^2=243\)
\(\Leftrightarrow3^{x-2}=3^5\)
\(\Rightarrow x-2=5\)
\(\Rightarrow x=7\)
d) \(25< 5^x< 3125\)
\(\Leftrightarrow5^2< 5^x< 5^5\)
\(\Rightarrow x\in\left\{3;4\right\}\)
e) \(2^x-64=2^6\)
\(\Leftrightarrow2^x=64+64=128\)
\(\Leftrightarrow2^x=2^7\)
\(\Rightarrow x=7\)
f) \(2^x\div16=128\)
\(\Leftrightarrow2^x=2^7.2^4\)
\(\Leftrightarrow2^x=2^{11}\)
\(\Rightarrow x=11\)
a) \(5.\left(x-3\right)=15\)
\(x-3=15:5\)
\(x-3=3\)
\(x=6\)
b)\(10+2.x=4^5:4^3\)
\(10+2.x=16\)
\(2x=16-10\)
\(2x=6\)
\(x=3\)
c) \(5^{x+1}=125\)
\(5^{x+1}=5^3\)
\(x+1=3\)
\(x=2\)
d) \(5^{2x-3}-2.5^2=5^2.3\)
\(5^{2x-3}=2.5^2+5^2.3\)
\(5^{2x-3}=125\)
\(5^{2x-3}=5^2\)
\(2x-3=2\)
\(2x=6\)
\(x=3\)
Mk nhanh nek bn
bài 2: (x-3).(y+2) = -5
Vì x, y \(\in\)Z => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}
Ta có bảng:
| x-3 | 5 | -5 | -1 | 1 |
| y+2 | 1 | -1 | -5 | 5 |
| x | 8 | -2 | 2 | 4 |
| y | -1 | -3 | -7 | 3 |
bài 3: a(a+2)<0
TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)
TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
Vậy -2<a<0
Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)
TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2
TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại
Vậy 1<a<2
\(a,\left(x-2\right)^2=4^2\)
\(\Rightarrow\hept{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=6\\x=-2\end{cases}}\)
Bài 1:
a) \(\left(x-2\right)^2=16\)
\(\Rightarrow\left(x-2\right)^2=4^2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=4+2=6\)
b) \(\left(2x-3\right)^2=9\)
\(\Rightarrow\left(2x-3\right)^2=3^2\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=3+3=6\)
\(\Rightarrow x=6:2=3\)
Bài 2 tương tự nhé em
P/s: Chỉ cần phân tích vế phải sao cho cùng số mũ với vế trái là được nhé!
Chúc em học tốt!
a) \(\text{2(x-51)=2.2^2+20}\)
\(2\left(x-51\right)=2.4+20\)
\(2\left(x-51\right)=28\)
\(x-51=28\div2\)
\(x-51=14\)
\(x=14+51\)
\(\text{b)3.(x+1)-26=541}\)
\(3.\left(x+1\right)=541+26\)
\(3\cdot\left(x+1\right)=567\)
\(x+1=567\div3\)
\(x+1=189\)
\(x=189-1\)
\(x=188\)
\(x=65\)
\(\text{c)4(x-3)=7^2-1^10}\)
\(4\left(x-3\right)=49-1\)
\(4\left(x-3\right)=48\)
\(x-3=48\div4\)
\(x-3=12\)
\(x=12+3\)
\(x=15\)
\(\text{e)2x-138=2^3.3^2}\)
\(2x-138=8\cdot9\)
\(2x-138=72\)
\(2x=72+138\)
\(2x=210\)
\(x=210\div2\)
\(x=105\)
\(\text{f)(x-1)^4=16}\)
\(\left(x-1\right)^4=2^4\)
\(x-1=2\)
\(x=2+1\)
\(x=3\)
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
Bài 1:
\(y^{10}=y\Rightarrow y^{10}-y=0\)
\(\Rightarrow y\left(y^9-1\right)=0\Rightarrow\orbr{\begin{cases}y=0\\y^9-1=0\Rightarrow y^9=1\Rightarrow y=1\end{cases}}\)
Bài 2:
\(a)16^x< 32^4\)
Ta có:\(16^x=\left(2^4\right)^x=2^{4x};32^4=\left(2^5\right)^4=2^{20}\)
\(\Rightarrow2^{4x}< 2^{20}\Rightarrow4x< 20=4.5\)mà \(x\inℕ\Rightarrow x\in\left\{0;1;2;3;4\right\}\)
\(b)9< 3^x< 81\)
\(\Rightarrow3^2< 3^x< 3^4\)
\(\Rightarrow2< x< 4\)mà \(x\inℕ\Rightarrow x=3\)
\(c)25< 5^x< 125\)
\(\Rightarrow5^2< 5^x< 5^3\)
\(\Rightarrow2< x< 3\)mà\(x\inℕ\Rightarrow\)không có giá trị x thõa mãn
y10 = y
<=> y10 - y = 0
<=> y(y - 1)(y2 + y + 1)(y6 + y3 + 1) = 0
=> y = 0, y = 1