Tìm x, Biết::
a)100 - 7 x ( x - 5 ) = 58
b)5x - 206 = 2(mũ 4) x 4
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a)100-7.[x-5]=58
7.[x-5]=100-58=42
x-5=42:7=6
x=6+5
x=11
b)12.[x-1]:3=43+23
12 [x-1]:3=64+8
12.[x-1]:3=72
[x-1]:3=72:12=6
[x-1]=6.3=18
x=18-1
x=17
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a, 100 - 7 ( x - 5 ) = 31 + 33
100 - 7 ( x - 5 ) = 31 + 27
100 - 7 ( x - 5 ) = 58
7 ( x - 5 ) = 100 - 58
7 ( x - 5 ) = 42
x - 5 = 42 : 7
x - 5 = 6
=> x = 6 +5
=> x = 11
Vậy x = 11
b, 12 ( x - 1 ) : 3 = 43 + 23
12 ( x - 1 ) : 3 = 64 + 8
12 ( x - 1 ) : 3 = 72
12 ( x - 1 ) = 72 . 3
12 ( x - 1 ) = 216
x - 1 = 216 : 12
x - 1 = 18
=> x = 18 + 1
=> x = 19
Vậy x = 19
c, 24 + 5x = 75 : 73
24 + 5x = 72
24 + 5x = 49
5x = 49 - 24
5x = 25
=> x = 25 : 5
=> x = 5
Vậy x = 5
d, 5x - 206 = 24 . 4
5x - 206 = 16 . 4
5x - 206 = 64
5x = 64 + 206
5x = 270
=> x = 270 : 5
=> x = 54
Vậy x = 54
e, 125 = x3
53 = x3
=> x = 5
Vậy x = 5
g, 64 = x2
82 = x2
=> x = 8
Vậy x = 8
b ) 20 + 5x = 55 : 53
20 + 5x = 52
20 + 5x = 25
5x = 25 - 20
5x = 5
x = 5 : 5
x = 1
c ) 5x - 201 = 24.4
5x - 201 = 16.4
5x - 201 = 64
5x = 64 + 201
5x = 265
x = 265 : 5
x = 53
a, ( 2x - 3 )2- (2x + 1)2 = -3
4x2-12x+9-4x2+4x-1=-3
-8x-1=-3
-8x=-2
x=\(\frac{1}{4}\)
b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7
25x2-10x+1-25x2+16=7
-10x+17=7
-10x=-10
x=1
c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0
x2-10x+25+x2-9-2x2-4x-2=0
-14x+14=0
-14(x-1)=0
=>x-1=0
x=1
a) \(\left(2x-3\right)^2-\left(2x+1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)
\(\Leftrightarrow-16x+8=-3\)
\(\Leftrightarrow-16x=-11\)
\(\Leftrightarrow x=\frac{11}{16}\)
b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x+17=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)
\(\Leftrightarrow-14x-18=0\)
\(\Leftrightarrow-14x=18\)
\(\Leftrightarrow x=-\frac{9}{7}\)
#H
1.
a) \(5x.5x.5x=\left(5x\right)^3.\)
b) \(x^1.x^2.....x^{2006}=x^{\frac{\left(2006+1\right).2006}{2}=}x^{2013021}.\)
c) \(x^1.x^4.x^7.....x^{100}=x^{\frac{\left(100+1\right).\left(\frac{100-1}{3}+1\right)}{2}}=x^{1717}.\)
d) \(x^2.x^5.x^8.....x^{2003}=x^{\frac{\left(2003+2\right).\left(\frac{2003-2}{3}+1\right)}{2}}=x^{669670}.\)
2.
\(2^x+80=3^y\)
Với \(x>0\Rightarrow2^x\) chẵn
Và 80 chẵn
\(\Rightarrow2^x+80\) chẵn.
Mà \(3^y\) lẻ
\(\Rightarrow x< 0.\)
Mà \(x\in N\)
\(\Rightarrow x=0.\)
\(\Rightarrow2^0+80=3^y\)
\(\Rightarrow1+80=3^y\)
\(\Rightarrow3^y=81\)
\(\Rightarrow3^y=3^4\)
\(\Rightarrow y=4.\)
Vậy \(\left(x;y\right)=\left(0;4\right).\)
Chúc bạn học tốt!