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a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
1,
Ta có: \(x^2\ge0;\left|y-13\right|\ge0\)
\(\Rightarrow x^2+\left|y-13\right|\ge0\)
\(\Rightarrow x^2+\left|y-13\right|+14\ge14\)
\(\Rightarrow\frac{1}{x^2+\left|y-13\right|+14}\le\frac{1}{14}\)
\(\Rightarrow P=\frac{12}{x^2+\left|y-13\right|+14}\le\frac{12}{14}=\frac{6}{7}\)
Dấu "=" xảy ra khi x = 0, y = 13
Vậy Pmin = 6/7 khi x = 0, y = 13
2, \(P=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=1+\frac{7}{n-5}\)
Để P có GTLN thì\(\frac{7}{n-5}\) có GTLN => n - 5 có GTNN và n - 5 > 0 => n = 6
3,
Ta có: \(10\le n\le99\)
\(\Rightarrow20\le2n\le198\)
\(\Rightarrow2n\in\left\{36;64;100;144;196\right\}\)
\(\Rightarrow n\in\left\{18;32;50;72;98\right\}\)
\(\Rightarrow n+4\in\left\{22;36;50;72;98\right\}\)
Ta thấy chỉ có 36 là số chính phương
Vậy n = 32
4,
ÁP dụng TCDTSBN ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\) (vì a+b+c khác 0)
\(\Rightarrow\hept{\begin{cases}\frac{a+b-c}{c}=1\\\frac{b+c-a}{a}=1\\\frac{a+c-b}{b}=1\end{cases}\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\\a+c-b=b\end{cases}\Rightarrow}\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}}\)
\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}\cdot\frac{a+c}{c}\cdot\frac{b+c}{b}=\frac{2c}{a}\cdot\frac{2b}{c}\cdot\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy B = 8
Câu 1:
Ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2=\left(x-1\right)^x\cdot\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[1-\left(x-1\right)\right]\cdot\left[1+\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(1-x+1\right)\cdot\left(1+x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(2-x\right)\cdot x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\2-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=2\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy: x\(\in\){0;1;2}
Câu 2:
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\left(y-3\right)^2\ge0\forall y\)
Do đó: \(\left(x+2\right)^2+2\left(y-3\right)^2\ge0\forall x,y\)
mà \(\left(x+2\right)^2+2\left(y-3\right)^2< 4\)
và các số chính phương nhỏ hơn 4 là 0 và 1
nên \(\left(x+2\right)^2+2\left(y-3\right)^2\in\left\{0;1;2\right\}\)
*Trường hợp 1: (x+2)2=2(y-3)2=0
\(\Leftrightarrow\left(x+2\right)^2+2\left(y-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)
*Trường hợp 2: \(\left(x+2\right)^2=0\) và \(\left(y-3\right)^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\\left[{}\begin{matrix}y-3=1\\y-3=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\\left[{}\begin{matrix}y=4\\y=2\end{matrix}\right.\end{matrix}\right.\)
*Trường hợp 3: \(\left(x+2\right)^2=1\) và \(\left(y-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\\y=3\end{matrix}\right.\)
Vậy: (x,y)\(\in\){(-2;3);(-2;4);(-2;2);(-1;3);(-3;3)}
Câu 1 bạn làm nhầm rồi.
$(x-1)^x(x-1)^2=(x-1)^x(x-1)^4$ không tương đương với $(x-1)^2=(x-1)^4$
Mà từ đây suy ra \(\left[\begin{matrix} (x-1)^x=0\\ (x-1)^2=(x-1)^4\end{matrix}\right.\)
Đối với TH $(x-1)^x=0$ thì có thể xảy ra 2TH: $x-1=0$ hoặc $x=0$
a) \(\frac{-32}{\left(-2\right)^n}=4\)
\(\frac{\left(-2\right)^5}{\left(-2\right)^n}=4\)
\(\left(-2\right)^{5-n}=\left(-2\right)^2\)
=> 5-n = 2
n = 3
b) \(\frac{8}{2^n}=2\)
\(\frac{2^3}{2^n}=2\)
\(2^{3-n}=2^1\)
=> 3 -n = 1
n = 2
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
=> 2n -1 = 3
2n = 4
n = 2
a) \(\frac{-32}{\left(-2\right)^n}=4\Leftrightarrow\left(-2\right)^n=\frac{-32}{4}\)
\(\left(-2\right)^n=-8\)Mà \(-8=2^{-3}\)
\(\Rightarrow x=-3\)
b) \(\frac{8}{2^n}=2\Leftrightarrow2^n=\frac{8}{2}\)
\(2^n=4\) Mà \(4=2^2\Rightarrow x=2\)
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\Rightarrow\left(\frac{1}{2}\right)^{2n}:\frac{1}{2}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{8}\cdot\frac{1}{2}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{16}\Leftrightarrow\frac{1}{2^{2n}}=\frac{1}{16}\) mà\(16=2^4\)
\(2n=4\Rightarrow n=2\)
Vậy .........................
(1/32)n.16n=1024-1
=> (1/32.16)n=1/1024
=> (1/2)n=1/1024
=> (1/2)n=(1/2)10
=> n=10
\(\left(\frac{1}{32}\right)^n.16^n=1024^{-1}\)
\(\left(\frac{1}{32}.16\right)^n=\frac{1}{1024}\)
\(\left(\frac{1}{2}\right)^n=\frac{1}{1024}\)
\(\frac{1^n}{2^n}=\frac{1}{1024}\)
<=> 1n = 1 => n thuộc N
<=> 2n = 1024
=> 2n = 1024 = 210 ( 2n = 210 )
<=> n = 10