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Put \(2^6+2^9+2^n=576+2^n=m^2\)
\(\Rightarrow2^n=m^2-576=m^2-24^2=\left(m-24\right)\left(m+24\right)\)
Put \(\left\{{}\begin{matrix}m-24=2^a\\m+24=2^b\end{matrix}\right.\) ( a < b ; a + b = n)
Then \(\Rightarrow2^b-2^a=\left(m+24\right)-\left(m-24\right)\)
\(\Rightarrow2^a\left(2^{b-a}-1\right)=48=2^4.3\)
\(2^{b-a}-1⋮̸2\Rightarrow2^{b-a}-1=3\)
\(\Rightarrow\left\{{}\begin{matrix}2^a=2^4\\2^{b-a}=3+1=4=2^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\b-a=2\Rightarrow b=6\end{matrix}\right.\)
\(\Rightarrow n=a+b=4+6=10\)
Check the result: \(2^6+2^9+2^{10}=40^2\).
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Put \(\left(2k-1\right)\left(2k+1\right)=123476543\)
\(\Leftrightarrow4k^2-1-123476543=0\)
\(\Leftrightarrow4k^2-123476544=0\)
\(\Leftrightarrow4\left(k^2-30869136\right)=0\)
\(\Leftrightarrow k^2=30869136=5556^2\)
\(\Leftrightarrow k=5556\)
\(\Rightarrow\left\{{}\begin{matrix}2k-1=11111\\2k+1=11113\end{matrix}\right.\)
Ans : 11111 & 11113.
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\(\overline{ab}+\overline{ba}=10a+b+10b+a=11\left(a+b\right)=n^2\)
\(\Rightarrow\)Put \(a+b=11.k^2\)
We have \(0< a+b\le18\)
\(\Rightarrow a+b=11=2+9=3+8=4+7=5+6\)
So, \(\left(a;b\right)\in\left\{\left(2;9\right);\left(9;2\right);\left(3;8\right);\left(8;3\right);\left(7;4\right);\left(4;7\right);\left(5;6\right);\left(6;5\right)\right\}\)
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Let A be the least number we have to find.
So A - 5 is the LCM(15; 18; 9; 12) = - 180
Then A = - 175.
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\(a^2+a=a\left(a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^{2001}+a^{2000}+7=0+0+7=7\\a^{2001}+a^{2000}+7=\left(-1\right)+1+7=7\end{matrix}\right.\)
Ans : 7.
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Ans : \(x^{n+1}-1\)
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Let m2 and n2 be a + 22 and a - 23 respectively. ( m > n > 0)
So, we have \(m^2-n^2=\left(a+22\right)-\left(a-23\right)\)
\(\Rightarrow\left(m-n\right)\left(m+n\right)=45=1.45=3.15=5.9\)
\(\cdot\) If m - n = 1 and m + n = 45; then m = 23 ; n = 22; so \(a=m^2-22=n^2+23=507\)
\(\cdot\) If m - n = 3 and m + n = 15; then m = 9 and n = 6, so a = 59.
\(\cdot\) If m - n = 5 and m + n = 9; then m = 7 and n = 2; so a = 27.
Ans : \(a\in\left\{27;59;507\right\}.\)
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The area of the square is S2.
The area of the rectangle is \((S+3)(S-3)=S^2 - 9\)
So, the sum of the areas of 2 fingures is \(S^2+\left(S^2-9\right)=2S^2-9\).
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\(\left(m+n\right)^2=5^2\)
\(\Rightarrow m^2+n^2+2mn=25\)
\(\Rightarrow m^2+n^2=25-2mn=25-2.6=25-12=13\)
\(\Rightarrow m^3+n^3=\left(m+n\right)\left(m^2+n^2-mn\right)=5.\left(13-6\right)=35\)
Ans : 35.
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We have :
\(x^2-5x+1=0\)
\(\Rightarrow x^2+1=5x\) ( 1 )
\(x^2+\dfrac{1}{x^2}=\left(x^2+\dfrac{1}{x^2}+2.\dfrac{1}{x}.x\right)-2.\dfrac{1}{x}.x=\left(x+\dfrac{1}{x}\right)^2-2\)
\(=\left(\dfrac{x^2+1}{x}\right)^2-2\) ( 2 )
( 1 )( 2 )\(\Rightarrow x^2+\dfrac{1}{x^2}=\left(\dfrac{5x}{x}\right)^2-2=25-2=23\)
Ans : 23.
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\(\left\{{}\begin{matrix}r-s=4\\t-s=9\end{matrix}\right.\Rightarrow\left(t-s\right)-\left(r-s\right)=t-r=9-4=5\)
We have :
\(r^2+s^2+t^2-rs-st-rt\)
\(=\dfrac{2r^2+2s^2+2t^2-2rs-2st-2rt}{2}\)
\(=\dfrac{\left(r^2+s^2-2rs\right)+\left(r^2+t^2-2rt\right)+\left(t^2+s^2-2ts\right)}{2}\)
\(=\dfrac{\left(r-s\right)^2+\left(t-r\right)^2+\left(t-s\right)^2}{2}\)
\(=\dfrac{4^2+5^2+9^2}{2}=\dfrac{122}{2}=61\)
Ans : 61
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\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right)+\left(1999^2-2000^2\right)\)
\(=\left(-3\right)+\left(-7\right)+\left(-11\right)+...+\left(-3999\right)\)
The number of terms is \(\dfrac{\left(-3\right)-\left(-3999\right)}{4}+1=1000\) ( terms )
So the value of this expression is \(\dfrac{1000.\left[\left(-3\right)+\left(-3999\right)\right]}{2}=-2001000\)
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Let \(A=\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)
\(\Rightarrow A=\dfrac{1}{5}.5.\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)
\(\Rightarrow A=\dfrac{1}{5}.\left(6-1\right).\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)
\(\Rightarrow A=\dfrac{1}{5}.\left[\left(6-1\right)\left(6+1\right)\right]\left(6^2+1\right)...\left(6^{16}+1\right)\)
\(\Rightarrow A=\dfrac{1}{5}.\left(6^2-1\right)\left(6^2+1\right)\left(6^4+1\right)...\left(6^{16}+1\right)\)
\(...\)
\(\Rightarrow A=\dfrac{1}{5}.\left(6^{32}-1\right)=\dfrac{6^{32}-1}{5}\)
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You must check ur problem
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m2 - n2 = ( m + n )( m - n ) = 64
So m + n and m - n \(\in D\left(64\right)=\left\{-64;-32;-16;-8;-4;-2;-1;1;2;4;8;16;32\right\}\) ( D : divisor )
Try all of them, we have pair ( 32 ; 2 ) satisfy the equation.
So we have 2 pairs of m , n ( 17 ; 15 ) and ( - 17; -15).
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Let x, y, z be the weights of three pumpkins :
\(x\le y\le z\)
So we have x + y = 8 ; x + z = 9 ; y + z = 11
=> ( x + z ) - ( x + y ) = 9 - 8
=> z - y = 1
=> z = ( 11 + 1 ) : 2 = 6
y = z - 1 = 5
x = 8 - y = 8 - 5 = 3
So the answer is 3 kg.
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Your answer b isn't complete !
a. \(\left(2x+3\right)^2=\left(2x\right)^2+2.\left(2x\right).3+3^2=4x^2+12x+9\)
b. \(\left(x-2\right)^3=x^3-3.\left(x^2\right).2+3.x.2^2-2^3\)
\(=x^3-6x^2+12x-8\)
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Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;
333 . 335 = 111555, ... It's the first step to solve ur problem.
Now, we have to prove that expression equal 333...3 . 333....5.
We have :
\(111...11111111111555...555555555\)
( 2002 1s) (2002 5s)
=111.....11000....0 + 555.......5
( 2002 1s) (2002 0s) (2002 5s)
= 1111.....111 . ( 10000...000 + 5 )
( 2002 1s) ( 2002 0s)
= 111....111 . 10000...00005
( 2002 1s) (2001 0s )
= 1111...1111 . ( 3 . 333...33335 )
( 2002 1s) (2001 3s )
= 333......3333 . 333333...3335
( 2002 3s) ( 2001 3s )
The sum of these 2 numbers is 6666......68
(2002 6s)
Sorry if my English is bad :>
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We have a + b = 3, so \(\left(a+b\right)^2=a^2+b^2+2ab=9\)
\(\Leftrightarrow a^2+b^2=9-2ab=9-2.4=1\)
Answer : 1.
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It's incomplete :)
We easily have 2 equations :
\(\left\{{}\begin{matrix}5k-4m=7\Rightarrow4\left(5k-4m\right)=20k-16m=28\\5m+4k=22\Rightarrow5\left(5m+4k\right)=25m+20k=110\end{matrix}\right.\)
\(\Leftrightarrow25m+20k-\left(20k-16m\right)=110-28\)
\(\Leftrightarrow41m=82\)
\(\Leftrightarrow m=2\)
\(\Leftrightarrow k=3\)
Therefore, ...