tìm x biết:
a,5x=125
b,(x+1)+(x+2)+(x+3)+..........+(x+100)=5750
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\(5x\left(4+2\right)+3x=240\) \(\left(8x+x+63\right):11=9\)
\(20x+10x+3x=240\) \(8x+x+63=99\)
\(33x=240\) \(9x=36\)
\(x=\frac{240}{33}\) \(x=4\)
5.(4x + 2) + 3x = 240
20x + 10 + 3x = 240
23x + 10 = 240
23x = 240 - 10
23x = 230
x = 230 : 23 = 10
31.57.x - 57.14.x = 85.57
(1767 - 798 ).x = 4845
969 . x = 4845
x = 4845 : 969
x = 5
x+x+x+...x+ 1+2+3+...100 = 5750
100.x + 101.50 = 5750
100.x = 5757-5050
100.x = 700
x = 700: 100
x = 7
a: Ta có: \(100-7\left(x-5\right)=58\)
\(\Leftrightarrow7\left(x-5\right)=42\)
\(\Leftrightarrow x-5=6\)
hay x=11
b: Ta có: \(12\left(x-1\right):3=4^3+2^3\)
\(\Leftrightarrow12\left(x-1\right)=216\)
\(\Leftrightarrow x-1=18\)
hay x=19
a) Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)
\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b,\(< =>25x^2+10x+1-25x^2+9-30=0\)
\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)
c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)
\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)
\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)
\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)
a: Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
(x+1)+(x+2)+(x+3)+...+(x+100)=5750
x + 1 + x + 2 + x + 3 + .... + x + 100 = 5750
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750
100x + 5050 = 5750
100x = 5750 - 5050 = 700
x = 700 : 100 = 7
(x+1)+(x+2)+(x+3)+...+(x+100)=5750
x + 1 + x + 2 + x + 3 + .... + x + 100 = 5750
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750
100x + 5050 = 5750
100x = 5750 - 5050 = 700
x = 700 : 100 = 7
Ta có: (x + 1) + (x + 2) +...+ (x + 100) = 5750
=> (x + x + x +...... +x) + (1 + 2 + 3 + 4 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
Ta có: ( x+1) + (x+2) + (x+3) +...+ ( x+100) = 5750
<=> ( x + x + x + ...... + x ) + (1 + 2 + 3 + ..... + 100) = 5750
<=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
[ x + 1 ] + [ x + 2 ] + [ x + 3 ] + ... + [ x + 100 ] = 5750
[ x + x + x + .... + x ] + [ 1 + 2 + 3 + ... + 100 ] = 5750
Đặt A = 1 + 2 + 3 + ... + 100 và B = x + x + x + .... + x
Dãy A có số số hạng là :
( 100 - 1 ) : 1 + 1 = 100 ( số hạng )
=> A = ( 1 + 100 ) x 100 : 2 = 5050
=> B = 100x
Ta có :
[ x + x + x + .... + x ] + [ 1 + 2 + 3 + ... + 100 ] = 5750
100 x + 5050 = 5750
100 x = 700
x = 7
a)5^x=125
5^x=5^3
=> x=3
b) (x+1)+(x+2)+(x+3)+...+(x+100)=5750
(x+x+x+...+x)+(1+2+3+...+100)=5750
100x+5050=5750
100x=5750-5050
100x=700
x=700:100
x=7