Tìm x:
A. ( x - 5 )² = 16
B. 4^x + 33 = 7²
C. 2^x . 4 = 128
D. 5^x . 3 - 75 = 0
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b. 1500(x-7)=0
x-7=0
x=7
c. (2x-4)(48-12x)=0
2x-4=0 hoặc 48-12x=0
x=2 hoặc x=4
d. (x+12)(x-1)=0
x+12=0 hoặc x-1=0
x=-12 hoặc x=1
bài 2 :
a . 128-3(x+4)=23
3(x+4)=105
x+4=35
x=31
b. [(14X+26).3+55]:5=35
(14x+26).3+55=175
(14x+26).3=120
14x+26=40
14x=14
x=1
d. 720:[41-(2X-5)]=23.5
41-(2x-5)=720:(23.5)
41-(2x-5)=144/23
2x-5=799/23
2x=914/23
x=457/23
b, 1500.(x – 7) = 0
<=>1500x-10500=0
<=>1500x=10500
<=>x=7
Vậy x=7
c,(2.x – 4).(48 – 12.x) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-4=0\\48-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\12x=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy x=2 hoặc x=4
d, (x + 12).(x – 1) =0
\(\Leftrightarrow\left\{{}\begin{matrix}x+12=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-12\\x=1\end{matrix}\right.\)
Vậy x=-12 hoặc x=1
Bài 2:
a) 128- 3(x+ 4) = 23
\(\Leftrightarrow\)128-(3x+12)=23
\(\Leftrightarrow\)128-3x-12=23
\(\Leftrightarrow\)116-3x=23
\(\Leftrightarrow\)3x=116-23
\(\Leftrightarrow\)3x=93
\(\Leftrightarrow\)x=31
Vậy x=31
b) [(14x+ 26). 3+ 55]: 5= 35
\(\Leftrightarrow\)(14x+ 26). 3+ 55=175
\(\Leftrightarrow\)42x+78+55=175
\(\Leftrightarrow\)42x+133=175
\(\Leftrightarrow\)42x=175-133
\(\Leftrightarrow\)42x=42
\(\Leftrightarrow\)x=1
Vậy x=1
d, 720: [41- (2x- 5)]= 23. 5
\(\Leftrightarrow\)720: 41- (2x- 5)=115
\(\Leftrightarrow\)41-(2x- 5)=720:115
\(\Leftrightarrow\)41-(2x- 5)=\(\dfrac{144}{23}\)
\(\Leftrightarrow\)2x-5=\(\dfrac{799}{23}\)
\(\Leftrightarrow\)2x=\(\dfrac{914}{23}\)
\(\Leftrightarrow\)x=\(\dfrac{457}{23}\)
Vậy x=\(\dfrac{457}{23}\)
a,5mũ 36=(5mũ3)mũ12=125 mũ12
11^24=(11^2)12=121^12
vì 121<125 nên 5^36>11^24
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
a) 2x (x - 5) - x (3 + 2x) = 26
=> 2x2 - 10x - (3x - 2x2) = 26
=> 2x2 - 10x - 3x - 2x2 = 26
=> -13x = 26 => x = 26 : (-13) = -2
xin loi nhung hoi nhiu mik viet cau tra loi dc ko - Nguyễn Diệu Thảo
a, 128 - 3.( x + 4 ) = 23 b, [( 6x - 39 ) : 7 ] . 4 = 12 c, ( x : 3 - 4 ) . 5 = 15
3. ( x + 4 ) = 128 - 23 [(6x - 39 ) : 7 ] = 3 ( x : 3 - 4 ) = 3
3. ( x + 4 ) = 105 ( 6x - 39 ) = 21 x : 3 = 7
x + 4 = 35 6x = 21 + 39 x = 7 . 3
x = 35 - 4 6x = 60 x = 21
x = 31 x = 10 Vậy x = 21
Vậy x = 31 Vậy x = 10
Trl:
a. 128 - 3 . ( x + 4 ) = 23
=> 3 . ( x + 4 ) = 128 - 23
=> 3 . ( x + 4 ) = 105
=> x + 4 = 105 : 3
=> x + 4 = 35
=> x = 35 - 4
=> x = 31
Vậy x = 31
b. [( 6x - 39 ) : 7 ] . 4 = 12
=> [( 6x - 39 ) : 7 ] = 12 : 4
=> [( 6x - 39 ) : 7 ] = 3
=> ( 6x - 39 ) = 3 . 7
=> ( 6x - 39 ) = 21
=> 6x = 21 + 39
=> 6x = 60
=> x = 60 : 6
=> x = 10
Vậy x = 10
c. ( x : 3 - 4 ) . 5 = 15
=> ( x : 3 - 4 ) = 15 : 5
=> ( x :3 - 4 ) = 3
=> x : 3 = 3 + 4
=> x : 3 = 7
=> x = 7 . 3
=> x = 21
Vậy x = 21
d. | x + 2 | = 0
=> x + 2 = 0
=> x = 0 - 2
=> x = -2
Vậy x = -2
e. | x - 5 | = |-7|
=> | x - 5 | = 7
=> x - 5 = 7 hoặc x - 5 = -7
=> x = 7 + 5 hoặc x = -7 + 5
=> x = 12 hoặc x = -2
Vậy \(x\in\left\{12;-2\right\}\)
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 0
<=> x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 0
<=> x2 + 3x + 2x + 6 - x2 - 5x + 2x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> (2x + 3)(x - 4) + (x - 5)(x - 2) - (3x - 5)(x - 4) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - (3x2 - 12x - 5x + 20) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - 3x2 + 12x + 5x - 20 = 0
<=> 5x = 12 - 10 + 20
<=> 5x = 22
<=> x = 22/5
c) (8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
<=> 8x + 16 - 5x2 - 10x + (4x - 8)(x + 1) + 2(x2 - 4) = 0
<=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0
<=> x2 - 6x = 0
<=> x(x - 6) = 0
<=> x = 0 hay x - 6 = 0
I<=> x = 6
d) (8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
<=> 24x2 + 16x - 9x - 6 - (4x2 + 16x + 7x + 28) = 10x2 - 2x + 5x - 1 - 33
<=> 24x2 + 16x - 9x - 6 - 4x2 - 16x - 7x - 28 - 10x2 + 2x - 5x + 1 + 33 = 0
<=> 10x2 - 19x = 0
<=> x(10x - 19) = 0
<=> x = 0 hay 10x - 19 = 0
I <=> 10x = 19
I <=> x = 19/10
(\(x\) - 5)2 = 16
(\(x-5\))2 = 42
\(\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 1; 9 }
B. 4\(x\) + 33 = 72
4\(^x\) + 33 = 49
4\(^x\) = 49 - 33
4\(^x\) = 16
4\(^x\) = 16
4\(^x\) = 42
\(x\) = 2
c. 2\(^x\).4 = 128
2\(^x\) = 128 : 4
2\(^x\) = 32
2\(^x\) = 25
\(x\) = 5
D, 5\(^x\) . 3 - 75 = 0
5\(^x\).3 = 75
5\(^x\) = 75: 3
5\(^x\) = 25
5\(^x\) = 52
\(x\) = 2
A) \(\left(x-5\right)^2=16=4^2\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
B) \(4^x+33=7^2\Rightarrow4^x=49-33\Rightarrow4^x=16=4^2\Rightarrow x=2\)
C) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)
D) \(5^x.3-75=0\Rightarrow5^x.3=75\Rightarrow5^x=25=5^2\Rightarrow x=2\)