2+4+6+8+...+(2x)=120

=> (1+2+3+4+...+x).2=120

=> 1+2+3+4+...+x = 120:2=60

=> (x+1).x : 2 = 60

=> (x+1).x=60.2 = 120

=> Không có x thỏa mãn

giúp tôi với tôi đang gấp , đi mà , năn nỉ đó . huhu

B

B

Câu 1:B

Câu 2:B

chịu

Chịu 

`@` `\text {Ans}`

`\downarrow`

`a,`

`(x-1954) \times 5=50`

`x-1954 = 50 \div 5`

`x-1954 = 10`

`x=10 + 1954`

`x= 1964`

`B.`

`48 - 40 \times x=8`

`40 \times x=48-8`

`40 \times x = 40`

` x=40 \div 40`

`x=1`

`C.`

`2+70 \div x = 3`

`70 \div x = 3-2`

`70 \div x = 1`

`x=70 \div 1`

`x=70`

`D. [3 \times (x+2) \div 7]  \times  4=120`

`3 \times (x+2) \div 7 = 120 \div 4`

`3 \times (x+2) \div 7 =30`

`3 \times (x+2) = 30 \times 7`

`3 \times (x+2)=210`

`x+2=210 \div 3`

`x+2=70`

`x=70-2`

`x=68`

`E.`

`7,2 \div [ (0,6 \times x+8) \div 20 + 59] = 0,12?` Thiếu ngoặc bạn nhé._.

\(a,\left(x-1954\right)\times5=50\\ x-1954=50:5\\ x-1954=10\\ x=1954+10\\ x=1964\\b.48-40\times x=8\\40\times x=48-8\\40\times x=40\\ x=40:40\\ x=1 \\ c.2+70:X=3\\ 70:x=3-2\\ 70:x=1\\ x=70:1\\ x=70\\ d,\left[3\times\left(x+2\right):7\right]\times4=120\\ 3\times\left(x+2\right):7-120:4\\ 3\times\left(x+2\right):7=30\\ \left(x+2\right):7=30:3\\ \left(x+2\right):7=10\\ x+2=10\times7\\ x+2=70\\ x=70-2\\ x=68\)

a, (x-35)-120=0

x-35=120+0

x-35=120

x=120+35

x=155

b,7x-8=713

7x=713+8

7x=721

x=721:7

x=103

c,4x:17=0

4x=0.17

4x=0

x=0:4

x=0

d,75+(131-x)=205

131-x=205-75

131-x=130

x=131-130

x=1

e,2x-36=4^6:4^3

2x-36=4^6-3

2x-36=4^3

2x-36=48

2x=48-36

2x=12

x=12:2

x=6

f, 2011^2.2011^x=2011^6

2011^x=2011^6: 2011^2

2011^x=2011^6-2

2011^x=2011^4

x=4

bn

\(\text{a/ ( x-35 ) - 120 = 0}\)

\(\Rightarrow\left(x-35\right)=120\)

\(\Rightarrow x=120+35\)

\(x=155\)

\(\text{b/ 7x - 8 = 713}\)

\(\Rightarrow7x=713+8=721\)

\(x=721:7=103\)

\(\text{c/ 4x : 17 = 0}\)

\(4x=0\)

\(x=0\)

\(\text{d/ 75+(131-x) = 205}\)

\(131-x=205-75=130\)

\(x=131-130=1\)

\(e.2x-36=4^6:4^3\)

\(2x-36=4^3=64\)

\(2x=64+36=100\)

\(x=100:2=50\)

\(f.2011^2.2011^x=2011^6\)

\(\text{Ta có công thức }:x^n.x^m=x^{n+m}\)

\(\Rightarrow x=6-2=4\)

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,(x+1)+(x+2)+(x+3)+....+(x+9)+(x+10)=240

Từ 1 đến 10 có số số hạng là:(10-1):1+1=10

Từ 1 đến 10 có tổng là:(10+1)x10:2=55

Ta có:(Xx10)+55=240

          Xx10        =240-55

          Xx10        =185

          X              =185:10

          X              =18,5