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a/ 2x . 7 = 224
2x = 224 : 7
2x = 32
2x = 25
x = 5.
b/ 32x + 1 . 11 = 2673
32x + 1 = 2673 : 11
32x + 1 = 243
32x + 1 = 35
32x = 35 - 1
32x = 34
2x = 4
x = 4 : 2
x = 2.
c/ (3x + 5)2 = 289
(3x + 5)2 = 172
3x + 5 = 17
3x = 17 - 5
3x = 12
x = 12 : 3
x = 4.
d/ x . (x2)3 = x5
x1 . x5 = x5
x6 = x5
=> x = 0 hoặc x = 1
a) 2x x 7=224
2x=224:7
2x=32
2x=25
=> x=5
Vậy x=5
b) 32x+1 x 11=2673
32x+1=2673:11
32x+1=243
32x+1=35
=> 2x+1=5
x=(5-1):2
x=2
Vậy x=2
c) (3*x+5)2=289
(3*x+5)2=172
=> 3*x+5=17
x=(17-5):3
x=4
Vậy x=4
d) x.(x2)3=x5
x.x6=x5
x=x5:x6
x=x-1
2
a) 5 X - 5 mu 3=5
5 X - 125=5
5 X=5+125
5 X=130
X=130:5
X=26
mk chi biet moi bai do thoi sorry ban
- Bài 1:
a)1117-1116:{1240-(2^4-5)^2+[39-9(3^2-7)]:7}
=1117-1116:{1240-121+[39-9.2]:7}
=1117-1116:{1240-121+21:7}
=1117-1116:1122
=\(\frac{208693}{187}\)
b)7+10+13+16+...+2014+2017
Số số hạng của tổng là: (2017-7):3+1=671
Tổng: (2017+7).671=1358104
- Bài 2:
a)5x- 5^3=5 b)3(x-7)-128=157 c)611-11(5x+37)=39 d)3x.3x+1.3x+2=31.32.33.34.35
5x=5+5^3 3(x-7)=157+128 11(5x+37)=611-39 33x+3=315
5x=130 3(x-7)=285 11(5x+37)=572 => 3x+3=15
x=130:5 x-7=285:3 5x+37=572:11 3x=15-3
x=26 x-7=95 5x+37=52 3x=12
x=95+7 5x=52-37=15 x=12:3
x=102 x=15:3=5 x=4
Thấy đúng thì k cho mình nha ^^
a) 2x . 7 = 224
2x . 7 = 234256
2x =234256:7
2x =đề sai
b) ( 3x + 5 )2 = 289
( 3x + 5 )2 = 172
=>3x + 5=17
3x=17-5
3x=12
x=12:3
x=4
d) 32x +1 .11 = 2673
32x +1 .11 = 19034163
32x +1 =19034163:11
32x +1 =đề sai
a) \(\left(x-1\right)^3=125\)
\(\Leftrightarrow\left(x-1\right)^3=5^3\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
b) \(2^{x+2}-2^x=96\)
\(\Leftrightarrow\left(2^2-1\right)\cdot2^x=96\)
\(\Leftrightarrow\left(4-1\right)\cdot2^x=96\)
\(\Leftrightarrow3\cdot2^x=96\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
c) \(\left(2x+1\right)^3=343\)
\(\Leftrightarrow\left(2x+1\right)^3=7^3\)
\(\Leftrightarrow2x+1=7\)
\(\Leftrightarrow2x=7-1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
d) \(720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\left(đk:x\ne23\right)\)
\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5\)
\(\Leftrightarrow720:\left(46-2x\right)=40\)
\(\Leftrightarrow\dfrac{720}{46-2x}=40\)
\(\Leftrightarrow\dfrac{720}{2\left(23-x\right)}=40\)
\(\Leftrightarrow\dfrac{360}{23-x}=40\)
\(\Leftrightarrow360=40\left(23-x\right)\)
\(\Leftrightarrow9=23-x\)
\(\Leftrightarrow x=23-9\)
\(\Leftrightarrow x=14\left(đk:x\ne23\right)\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
e) \(2^x\cdot7=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
f) \(\left(3x+5\right)^2=289\)
\(\Leftrightarrow3x+5=\pm17\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=17\\3x+5=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{22}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{22}{3};x_2=4\)
a)\(\left(x-1\right)^3=125\Leftrightarrow\left(x-1\right)^3=5^3\Leftrightarrow x-1=5\Leftrightarrow x=6\)b)\(2^{x+2}-2^x=96\Leftrightarrow2^x.2^2-2^x=96\Leftrightarrow2^x\left(2^2-1\right)=96\Leftrightarrow2^x.3=96\Leftrightarrow2^x=32\Leftrightarrow x=5\)c)\(\left(2x-1\right)^3=343\Leftrightarrow\left(2x-1\right)^3=7^3\Leftrightarrow2x-1=7\Rightarrow2x=8\Rightarrow x=4\)d)\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-\left(2x-5\right)\right]=40\Leftrightarrow\left[41-\left(2x-5\right)\right]=720:40=18\)
\(\Leftrightarrow41-2x+5=18\Leftrightarrow36-2x=18\Leftrightarrow2x=18\Leftrightarrow x=9\)
e)\(2^x.7=224\Leftrightarrow2^x=224:7=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
f) \(\left(3x+5\right)^2=289\Leftrightarrow\left(3x+5\right)=17^2\Leftrightarrow3x+5=17\Leftrightarrow3x=12\Leftrightarrow x=4\)
a) Ta có: \(\left(x+1\right)^2=1+3+5+...+99\)
\(\Leftrightarrow\left(x+1\right)^2=\frac{\left(1+99\right).50}{2}=2500\)
\(\Leftrightarrow x+1=\pm50\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=-50\\x+1=50\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-51\\x=49\end{cases}}\)
Mình ko biết cách tính tổng phần này nên mình gj=hi luông kết quả bn nha
b) Ta có: \(\left(x+1\right)^2=1^3+2^3+3^3+...+10^3\)
\(\Leftrightarrow\left(x+1\right)^2=3025\)
\(\Leftrightarrow x+1=\pm55\)
\(\Leftrightarrow\orbr{\begin{cases}x=-56\\x=54\end{cases}}\)
a) Bạn xemm lại đề nhé.
b) \(5\left(x+35\right)=515\)
\(x+35=515\div5\)
\(x+35=103\)
\(x=103-35\)
\(x=68\)
Vậy \(x=68\).
c) \(90-3\left(x+1\right)=42\)
\(3\left(x+1\right)=90-42\)
\(3\left(x+1\right)=48\)
\(x+1=48\div3\)
\(x+1=16\)
\(x=16-1\)
\(x=15\)
Vậy \(x=15\).
d) \(12x-33=3^2.3^3\)
\(12x-33=3^5\)
\(12x-33=243\)
\(12x=243+33\)
\(12x=276\)
\(x=276\div12\)
\(x=23\)
Vậy \(x=23\).
e) \(100-7\left(x-5\right)=58\)
\(7\left(x-5\right)=100-58\)
\(7\left(x-5\right)=42\)
\(x-5=42\div7\)
\(x-5=6\)
\(\Rightarrow x=6+5=11\)
Vậy \(x=11\).
j) \(24+5x=7^5\div7^3\)
\(24+5x=7^2\)
\(24+5x=49\)
\(5x=49-24\)
\(5x=25\)
\(\Rightarrow x=25\div5=5\)
Vậy \(x=5\).
541+(218-x)=735
(218-x)=735-541
(218-x)=194
x=218-194
x=24
vậy x=24
5 (x+35)=515
(x+35)=515:5
(x+35=103
x=103-35
x=68
vậy x=68
90-3 (x+1)=42
3 (x+1)=90-42
3 (x+1)=48
(x+1)=48:3
(x+1)=16
x=16-1
x=15
vậy x=15
12x-33=3^2.3^3
12x-33=3^5
12x-33=243
12x=243-33
12x=210
x=210:12
x=17,5
vậy x=17,5
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=1
vậy x=1
24+5x=7^5:7^3
24+5x=7^2
24+5x=49
5x=49-24
5x=25
x=25:5
x=5
vậy x=5
a,(3x+5)2=289
=>(3x+5)2=172
=>3x+5=17
=>3x=12
=>x=4
b,x50=x
=>x50-x=0
=>x=1
c,2x.7=224
=>2x=224:7
=>2x=32
=>2x=25
=>x=5
#Koo#
a, 2x=224:7
2x=32
2x=25
=> x=5
b, \(\left(3^x+5\right)^2=289\)
\(\left(3^x+5\right)^2=17^2\)
\(\Rightarrow3^x+5=17 \)
\(3^x=17-5\)
\(3^x=12\)
a,2x.7=224
=> 2x=32
Mà : 25=32
=> 2x=25
=> x=5
b,(3x+5)2=289
Ta có : 172=289
=> (3x+5)2=172
=> 3x+5=17
=> 3x=12
=> sai đề :v