Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1
a) 155 - 10.(x+1) = 55
=>10 .(x+1) = 100
=>x + 1 = 10
=>x = 9
còn lại tương tự
2 . tìm x :
a ) 52x = 625
52x = 54
=> 2 . x = 4
x = 4 : 2
x = 2
b ) 9x-1 = 9
9x-1 = 91
=> x - 1 = 1
x = 1 + 1
x = 2
c ) 2x : 25 = 1
2x-5 = 1
=> x = 5 , vì 25-5 = 20 = 1
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
A, (X+1)+(x+4)+(x+7)+...+(x+25)+(x+28)=155
b,(x+9)+(x-2)+(x+7)
+(x-4)+(x+5)+(x-6)+(x+3)+(x-8)+(x+1)=95
A,(x+1)+(x+4)+(x+7)+...+(x+25)+(x+28)=155
x+1+x+4+x+7+...+x+25+x+28=155
x.[(28-1):3+1]+(1+4+7+...+25+28)=155
10.x+[(28+1).10:2]=155
10.x+145=155
A
<=>x+1+x+4+x+7+...+x+25+x+28=155
<=>10x.(1+4+7+10+13+16+19+22+25+28)=155
<=>10x.145 =155
<=>10x=155:145=\(\frac{31}{29}\)
x=\(\frac{31}{29}:10=\frac{31}{290}\)
Vậy x=\(\frac{31}{290}\)
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
<=> (x + x + ... + x) + (1 + 4 + ... + 28) = 155
<=> 10x + 145 = 155
<=> 10x = 10
<=> x = 1
b, (x + 9) + (x - 2) + (x + 7) + (x - 4) + ... + (x - 8) + (x + 1) = 95
<=> (x + 9 + x + 7 + ... + x + 1) + (x - 2 + x - 4 + .. + x - 8) = 95
<=> (5x + 25) + (4x - 20) = 95
<=> (5x + 4x) + (25 - 20) = 95
<=> 9x + 5 = 95
<=> 9x = 90
<=> x = 10
@láo như cáo
Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
Bài 3:
a) 11 - (15 + 11) = x - (25 - 9)
=> 11 - 15 - 11 = x - 25 + 9
=> x = 9 - 25 + 15
=> x = -1
b) 2 - x = 17 - (-5)
=> 2 - x = 17 + 5
=> -x = 17 + 5 - 2
=> -x = 20
=> x = -20
c) x - 12 = (-9) - 15
=> x - 12 = -9 - 15
=> x = -9 - 15 + 12
=> x = -12
d) IxI - 7 = 9
=> IxI = 9 + 7
=> IxI = 16
=> x = 16 hay x = -16
e) 9 - 25 = (7 - x) - (25 + 7)
=> 9 - 25 = 7 - x - 25 - 7
=> x = 7 - 7 - 25 + 25 - 9
=> x = -9
Bài 4:
a) 25 . 46 + 54 . 25
= 25(46 + 54)
= 25.100
= 2500
b) 1200 : 25
= (25.48) : 25
= 25:25 . 48
= 1.48
= 48
c) 1356 - 998
= 1356 + 2 - 998 - 2
= 1358 - 1000
= 358
d) 117 + 57 - 17
= 117 - 17 + 57
= 100 + 57
= 157
Bài 5:
a) 34 . 315
= 34 + 15
= 319
b) 88 : 88
= 88 - 8
= 80
= 1
c) 100 - [120 - (15 - 5)2 ]
= 100 - (120 - 102)
= 100 - (120 - 100)
= 100 - 120 + 100
= 80
a) 5.5x = 625
=> 5x+1 = 54
=> x + 1 = 4
=> x = 4 - 1
=> x = 3
Vậy x = 3
b) 8(x + 25) - 155 = 181
=> 8(x + 25) = 181 + 155
=> 8(x + 25) = 336
=> x + 25 = 336 : 8
=> x + 25 = 42
=> x = 42 - 25
=> x = 17
Vậy x = 17
c) (x - 9)4 = (x - 9)2
=> (x - 9)4 - (x - 9)2 = 0
=> (x - 9)2.[(x - 9)2 - 1] = 0
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\\left(x-9\right)^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=1\\x-9=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
a)5*5x=625
\(\Rightarrow5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b)8(x+25)-155 =181
\(\Rightarrow8\left(x+25\right)=336\)
\(\Rightarrow x+25=42\)
\(\Rightarrow x=17\)
c) ( x-9)4 = ( x-9)2
\(\Rightarrow\left(x-9\right)^4-\left(x-9\right)^2=0\)
\(\Rightarrow\left(x-9\right)^2\left[\left(x-9\right)^2-1\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=\pm1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)