\(2008^n=1\Rightarrow n=0\)

\(2^n=1024\Rightarrow2^n=2^{10}\Rightarrow n=10\)

tíc mình nha

\(a,2008^n=1=>n=0.\)

\(b,5^n+5^{n+2}=650\)

\(=>5^n\left(1+5^2\right)=650\)

\(=>5^n.26=650\)

\(=>5^n=650:26\)

tự tính tiếp nhé !!

\(c,2^n=1024=2^{10}=>n=10\)

\(d,3^{n-1}5.3^{n-1}=162\)mình nghĩ đề thiếu !

Ừ ! Mình xem lại đề bài rồi. Câu d, phải là 3^n-1 + 5.3^n-1 = 162

\(f\)) \(32^{-x}.16^x=1024\)

\(\left(2\right)^{-5x}.2^{4x}=2^{10}\)

\(\Leftrightarrow2^{4x-5x}=2^{10}\)

\(\Leftrightarrow2^{-x}=2^{10}\)

\(\Leftrightarrow-x=10\)

\(\Leftrightarrow x=-10\)

\(g\)) \(3^{x-1}.5+3^{x-1}=162\)

\(3^{x-1}.\left(5+1\right)=162\)

\(3^{x-1}.6=162\)

\(3^{x-1}=162:6\)

\(3^{x-1}=27\)

\(\Leftrightarrow3^{x-1}=3^3\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

\(h\)) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)

\(\Leftrightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=\left(1,-1\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=-1\\2x-1=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=0\\2x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)

\(i\)) \(5^x+5^{x+2}=650\)

\(5^x.\left(1+5^2\right)=650\)

\(5^x.26=650\)

\(5^x=650:26\)

\(5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

 Mình làm đc mỗi 1 câu, Thông cảm

7^6+7^5+7^4 chia hết cho 11

= 7^4.2^2+7^4.7+7^4

= 7^4.(2^2+7+1)

= 7^4. 11

Vì tích này có số 11 nên => chia hết cho 7

tìm hai số x và y biết x:2=y:(-5) và x-y=-7

tìm hai số x;y.Biết 7x=3y và x-y=16

tìm ba số x,y,z.Biết 2a=4b và 3b=5c và a+2b-3c=99

a) \(15+2^n=31\)

   \(2^n=16\Rightarrow n=4\)

b) \(2.2^n+4.2^n=6.2^5\)

    \(2^n\left(2+4\right)=6.2^5\)

    \(2^n.6=6.2^5\Rightarrow n=5\)

c) \(32^n:16^n=1024\)

    \(\left(2^5\right)^n:\left(2^4\right)^n=2^{10}\)

     \(2^{5n}:2^{4n}=2^{10}\)

     \(2^n=2^{10}\Rightarrow n=10\)

d) \(5^n+5^{n+2}=650\)

   \(5^n+5^n.25=650\)

   \(5^n\left(1+25\right)=650\)

   \(5^n.26=650\)

   \(5^n=25\Rightarrow n=2\)

e) \(3^n+5.3^{n+1}=432\)

    \(3^n+5.3^n.3=432\)

    \(3^n\left(1+15\right)=432\)

   \(3^n.16=432\)

   \(3^n=27\Rightarrow n=3\)

\(a,15+2^n=31\)

\(2^n=16\)

\(x=4\)

a)

\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)

\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)

\(\Rightarrow9^n=3^n\)

\(\Rightarrow\left(3^2\right)^n=3^n\)

\(\Rightarrow3^{2n}=3^n\)

\(\Rightarrow2n=n\)

\(\Leftrightarrow n=0\)

Vậy \(n=0\)

d) Ta có:

\(6^{3-n}=216\)

\(\Rightarrow6^{3-n}=6^3\)

\(\Rightarrow3-n=3\)

\(\Rightarrow n=3-3\)

\(\Rightarrow n=0\)

Vậy \(n=0\)\(\text{ }\)

e) Ta có:

\(5^n+5^{n+2}=650\)

\(\Rightarrow5^n+5^n\cdot5^2=650\)

\(\Rightarrow5^n\cdot\left(1+5^2\right)=650\)

\(\Rightarrow5^n\cdot26=25\cdot26\)

\(\Rightarrow5^n=25\)

\(\Rightarrow5^n=5^2\)

\(\Rightarrow n=2\)

Vậy \(n=2\)

1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)

=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

Bài 2. Ta có: (3x - 5)¹⁰⁰ \(\ge\)0 \(\forall\)x

       (2y + 1)¹⁰⁰ \(\ge\)0 \(\forall\)y

=> (3x - 5)¹⁰⁰ + (2y + 1)¹⁰⁰ \(\ge\)0 \(\forall\)x;y

Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)

Vậy ...

1. x( 6 - x )²⁰⁰³ = ( 6 - x )²⁰⁰³

<=> x( 6 - x )²⁰⁰³ - ( 6 - x )²⁰⁰³ = 0

<=> ( x - 1 )( 6 - x )²⁰⁰³ = 0

<=> \(\orbr{\begin{cases}x-1=0\\6-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=6\end{cases}}\)

2. \(\left(3x-5\right)^{100}+\left(2y+1\right)^{100}\le0\)

\(\hept{\begin{cases}\left(3x-5\right)^{100}\ge0\forall x\\\left(2y+1\right)^{100}\ge0\forall y\end{cases}\Rightarrow}\left(3x-5\right)^{100}+\left(2y+1\right)^{100}\ge0\forall x,y\)

Dấu " = " xảy ra <=>  \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)

\(A=1+3+3^2+3^3+...+3^{101}\)

\(3A=3+3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\left(3^{101}-1\right):2\)

Thu gọn tổng sau:

A=1+3+3²+3³+...+3¹⁰⁰ 

B= 2¹⁰⁰-2⁹⁹-2⁹⁸-2⁹⁷-...-2²-2

C= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷-...+3²-3+1