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Answers ( 46 )
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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 \(A=\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)

    ​ ​\(\Rightarrow A=\dfrac{1}{5}.5.\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)

    ​ ​\(\Rightarrow A=\dfrac{1}{5}.\left(6-1\right).\left(6+1\right)\left(6^2+1\right)...\left(6^{16}+1\right)\)

    \(\Rightarrow A=\dfrac{1}{5}.\left[\left(6-1\right)\left(6+1\right)\right]\left(6^2+1\right)...\left(6^{16}+1\right)\)

    \(\Rightarrow A=\dfrac{1}{5}.\left(6^2-1\right)\left(6^2+1\right)\left(6^4+1\right)...\left(6^{16}+1\right)\)

    \(...\)

    \(\Rightarrow A=\dfrac{1}{5}.\left(6^{32}-1\right)=\dfrac{6^{32}-1}{5}\)

    ok

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    You must check ur problem ok

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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 :>

  • See question detail

    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.

  • See question detail

    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, ...

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