tìm x biết ( 9+x)× 1000=678000
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=> x +2 chia hết cho cả 5;7 và9
=> x+2 là bội chung của 5;7;9
=> x+2 thuộc 315;630 và 945 (vì x<1000)
=> x thuộc 313;628;943
Vì x : 7 dư 5 => x = 7k + 5 ( k thuộc N ) => x + 2 = 7k + 7 = 7 . ( k + 1 ) chia hết cho 7
Vì x : 9 dư 7 => x = 9q + 7 ( q thuộc N ) => x + 2 = 9q + 9 = 9 . ( q + 1 ) chia hết cho 9
Vì x : 11 dư 9 => x = 11n + 9 ( n thuộc N ) => x + 2 = 11n + 11 = 11 . ( n + 1 ) chia hết cho 11
Mà x chia hết cho 7,9,11 => x thuộc B C ( 7,9,11 )
TA có : 7 = 7
9 = 9
11 = 11
=> BCNN ( 7,9,11 ) = 7 . 9 . 11 = 693
=> B C ( 7,9,11 ) = B(693) = { 0 , 693 , 1386 , ... }
MÀ x chia hết cho 7,9,11 và x < 1000 => x = 693
Vậy x = 693
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a) \(x=1000002;1000003;...;9999999\)
b) \(x=9999999\)
c) \(x=1000001;1000002\)
@#$%^&* !
a) x = 1 000 002
b) x = 9 999 999
c) x = 1 000 001 ; 1 000 002
k mik nha
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a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)
\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)
\(\Rightarrow2y-1=3+y\)
\(2y-y=3+1\)
\(y=4\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{2}{3}\)
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\left(\left(2x-1\right)^3\right)^2=\left(\left(2x-1\right)^4\right)^2\)
\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)
\(8x^3-1=16x^4-1\)
\(16x^4-8x^3=0\)
\(8x^3\left(2x-1\right)=0\)
Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)
Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)
Vậy x=0 và x=1/2
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a) \(x=6\times1000+3\times100+2\times10+8=6000+300+20+8=6328\)
b) \(x=5\times1000+6\times10+7=5000+60+7=5067\)
c) \(x=9\times1000+9\times100+9=9000+900+9=9909\)
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f: Ta có: \(x\left(2x-9\right)-4x+18=0\)
\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)
g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)
\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)
f. x(2x - 9) - 4x + 18 = 0
<=> x(2x - 9) - 2(2x - 9) = 0
<=> (x - 2)(2x - 9) = 0
<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)
g. 4x(x - 1000) - x + 1000 = 0
<=> 4x(x - 1000) - (x - 1000) = 0
<=> (4x - 1)(x - 1000) = 0
<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)
h. 2x(x - 4) - 6x2(-x + 4) = 0
<=> 2x(x - 4) + 6x2(x - 4) = 0
<=> (2x + 6x2)(x - 4) = 0
<=> 2x(1 + 3x)(x - 4) = 0
<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)
i. 2x(x - 3) + x2 - 9 = 0
<=> 2x(x - 3) + (x - 3)(x + 3) = 0
<=> (2x + x + 3)(x - 3) = 0
<=> (3x + 3)(x + 3) = 0
<=> \(\left[{}\begin{matrix}3x+3=0\\x+3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
j. 9x - 6x2 + x3 = 0
<=> x(9 - 6x + x2) = 0
<=> x(3 - x)2 = 0
<=> \(\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
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x-992 chia 9=1000(dư 8)
x-992=1000 nhân 9 (dư 8)
x-992=9000(dư 8)
x=9000 nhân 992 +8
x=8928000+8
x=8928008
x-992 chia 9=1000(dư 8)
x-992=1000 nhân 9 (dư 8)
x-992=9000(dư 8)
x=9000 nhân 992 +8
x=8928000+8
x=8928008
( 9 + x ) x 1000 = 6780000
9 + x = 6780000 : 1000
9 + x = 6780
x = 6780 - 9
x = 6771
(9 + x) \(\times\) 1000 = 678000
(9 + x) = 678000 : 1000
(9 + x) = 678
x = 678 - 9
x = 669
Vậy x=669.