4x+3x=30-20÷10
x+3=12 4x+3x=30-20:10
\(x+3=12\\ \Rightarrow x=9\)
Vậy \(x=9\).
\(4x+3x=30-20:10\\ \Rightarrow x\left(4+3\right)=30-2\\ \Rightarrow x.7=28\\ \Rightarrow x=4\)
Vậy \(x=4.\)
Tìm x ∈ N, biết:
4x + 3x = 30 – 20 : 10
4x + 3x = 30 – 20 : 10
7x = 30 - 2
7x = 28
x = 28 : 7
x = 4
Tìm x ∈ N, biết:
b) 4x + 3x = 30 – 20 : 10
b)
4x + 3x = 30 – 20 : 10
7x = 30 - 2
7x = 28
x = 28 : 7
x = 4
tìm số tự nhiên x biết:
4x+ 3x = 30 - 20 : 10
\(4x+3x=30-20:10\)
\(4x+3x=30-2\)
\(4x+3x=28\)
\(\left(4+3\right)x=28\)
\(7x=28\)
\(x=28:7\)
\(x=4\)
Vậy \(x=4\)
Chúc cậu học tốt !!!
Tìm x ∈ N, biết:
a) (x - 3) : 2 = 5 14 : 5 12
b) 4x + 3x = 30 – 20 : 10
a) (x - 3) : 2 = 5 14 : 5 12
(x - 3) : 2 = 5 2
(x - 3) : 2 = 25
(x - 3) = 25.2
x = 50 + 3
x = 53
b) 4x + 3x = 30 – 20 : 10
7x = 30 - 2
7x = 28
x = 28 : 7
x = 4
Giải phương trình :
1 ) 5( x - 2 ) = 3x + 10
2 ) x2( x - 5 ) - 4x + 20 = 0
3 ) \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
4 ) \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
1) Ta có: \(5\left(x-2\right)=3x+10\)
\(\Leftrightarrow5x-10-3x-10=0\)
\(\Leftrightarrow2x-20=0\)
\(\Leftrightarrow2\left(x-10\right)=0\)
Vì 2>0
nên x-10=0
hay x=10
Vậy: x=10
2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy: x∈{-2;2;5}
3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)
\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)
\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)
\(\Leftrightarrow-5x=0\)
hay x=0
Vậy: x=0
4) ĐKXĐ: x≠5; x≠-5
Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow9x+45+14x-70-180=0\)
\(\Leftrightarrow23x-205=0\)
\(\Leftrightarrow23x=205\)
hay \(x=\frac{205}{23}\)(tm)
Vậy: \(x=\frac{205}{23}\)
1) 4x-(2x-5)=21
2) 14x+5=8x+10
3) 15+5x=3x+30
4) 2x-5=15-3x
5) 2(3x+5)+3(x+1)=6x+20
6) 4(x-3)+5(x-3)=18
1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
2)14x+5=8x+10
14x-8x=10-5
6x=5
x=5/6
3) 15+5x=3x+30
5x-3x=30-15
2x=15
x=15/2
Vậy x=15/2
|x +10| - (5 - 3x) = (4x - 10) - (x - 5)
|3x + 21| - ( 10 - 5x ) = 5x - |-20|
a) | x + 10 | - ( 5 - 3x ) = ( 4x - 10 ) - ( x - 5 )
=> | x + 10 | = ( 5 - 3x ) + ( 4x - 10 ) - ( x - 5 )
=> | x + 10 | = 5 - 3x + 4x - 10 - x + 5
=> | x +10 | = 0
=> x + 10 = 0
=> x = -10
Vậy...
b) Làm tương tự
Kết quả : | 3x + 21 | = -10 ( vô lí) ( vì |3x+21| >= 0 mà -10<0)
Vậy không tìm được x thỏa mãn bài toán
2) Tìm x \(\in\) N ,biết:
a) x+3=12
b) (x-3):2=\(5^{14}\):\(5^{12}\)
c) 4x+3x=30-20:10
d)2x-138=\(2^3\).\(3^2\)
giải chi tiết giúm mình với ạ
a) x+3=12
x=12-3
x=9
b)(x-3):2=514:512
=>(x-3):2=52
=>(x-3):2=25
=>x-3=25.2
=>x-3=50
=>x=50+3
=>x=53
c)4x+3x=30-20:10
=>x(4+3)=30-2
=>7x=28
=>x=28:7
=>x=4
d)2x-138=23.32
=>2x-138=8.9
=>2x-138=72
=>2x=72+138
=>2x=210
=>x=210:2
=>x=105
a) x + 3 = 12
x = 12 - 3
x = 9
b) ( x - 3 ) : 2 = 514 : 512
( x - 3 ) : 2 = 514-12
( x - 3 ) : 2 = 52
( x - 3 ) : 2 = 25
x - 3 = 50
x = 53
c) 4x + 3x = 30 - 20 : 10
7x = 28
x = 4
d) 2x - 138 = 23 x 32
2x - 138 = 8 x 9
2x - 138 = 72
2x = 210
x = 105