Tìm X :
-15 5 . (3X - 1)=25
Tìm x
X×4/5=15
6/5:x=1/3x:2/7=3/5
12/25×x=4/10
\(x\) \(\times\) \(\dfrac{4}{5}\) = 15
\(x\) = 15 : \(\dfrac{4}{5}\)
\(x\) = \(\dfrac{75}{4}\)
\(\dfrac{12}{25}\) \(\times\) \(x\) = \(\dfrac{4}{10}\)
\(x\) = \(\dfrac{4}{10}\) : \(\dfrac{12}{25}\)
\(x\) = \(\dfrac{5}{6}\)
\(\dfrac{6}{5}\): \(x\) = \(\dfrac{1}{3}\)\(x\) : \(\dfrac{2}{7}\) = \(\dfrac{3}{5}\) (xem lại đề bài)
Bài 2. Tìm x, biết :
a) 3x – 15 = 25 – 5x b) 3x - 17 = 2x – 7 c) 2x – 17 = – (3x – 18)
d) 3x – 14 = 2(x – 9) + 1 e) f) (x – 5)2 = 9
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
Bài 2. Tìm x, biết :
a) \(3x-15=25-5x\)
\(\Leftrightarrow8x=40\)
\(\Leftrightarrow x=5\)
Vậy x = 5
b) \(3x-17=2x-7\)
\(\Leftrightarrow x=10\)
Vậy x = 10
c) \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=18-3x\)
\(\Leftrightarrow5x=35\)
\(\Leftrightarrow x=7\)
Vậy x = 7
d) \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14=2x-17\)
\(\Leftrightarrow x=-3\)
Vậy x = -3
e) \(\left(x-5\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy x = {8; 2}
bài 1 tìm x
a) 5( x-7) = 0
b) 2015.( 3x-15)= 0
c) 2( x-5) -17= 25
a) 5(x - 7) = 0
x - 7 = 0 : 5 = 0
x = 0 + 7 = 7
b) 2015.( 3x - 15) = 0
3x - 15 = 0 : 2015 = 0
3x = 0 + 15 = 15
x = 15 : 3
x = 5
c) 2(x - 5) - 17 = 25
2(x - 5) = 25 + 17 = 42
x - 5 = 42 : 2
x - 5 = 21
x = 21 + 5
x = 26
Đúng cho mình nha bạn !!!
5(x-7) = 0
=> x - 7 = 0
=> x = 7
2015.(3x-15) = 0
=> 3x - 15 = 0
=> 3x = 15
=> x = 5
2(x-5)-17 = 25
=> 2(x-5) = 32
=> x - 5 = 16
=> x = 21
Bài 1: Tìm x [GIÚPPPPPPPPPPPPPPPPPPP]
1) 5 – (10 – x) = 7
2) - 32 - (x – 5) = 0
3) -12 + (2x – 9) + x= 0
4) 11 + (15 - x) = 1
5) 4 - (27 - 3) = x - (13 - 4)
6) 8 - (x - 10) = 23 - (- 4 +12)
7) 105 – 5(10 – 5x) = -20
8) (x -1)(8-2x)(3x+123) = 0
9) (x2 - 25)(x+ 10) = 0
10) x(x2+5) = 0
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
1)10-x=5-7
x=10-(-2)
x=12
2)x+5=0+32
x=32-5
x=27
3) -12+2x-9+x=0
-12+(2x+x-9)=0
3x-9=0+12
3x=12+9
x=21:3
x=7
4) 15-x=1-11
x=15-(-10)
x=25
5) 4-(27+3)=x-9
4-30=x-9
-26+9=x
x=-17
bài 19: tìm x
a) 5 . ( x - 7 ) = 0
b) 25 ( x - 4 ) = 0
c) ( 34 - 2x ) . ( 2x - 6 ) = 0
d) ( 2019 - x ) . ( 3x - 12 ) 0
e) 57 . ( 9x - 27 ) = 0
f) 25 + ( 15 - x ) = 30
g) 43 - ( 24 - x ) = 20
h) 2 . ( x - 5 ) - 17 = 25
i) 3 . ( x + 7 ) - 15 = 27
j) 15 + 4 . ( x - 2 ) = 95
k) 20 - ( x + 14 ) = 5
l) 14 + 3 . ( 5 - x ) = 27
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
Bài 1 ;Tìm x biết
1) (2x+1 )^3 - (2x+1)(4x^2-2x+1)-3(2x-1)=15
2) x(x-4)(x+4)-(x-5)(x^2 + 5x +25)=13
Bài 2 : Cmr các giá trị của biểu thức sau không thuộc vào giá ttij của biến :
A= (x+5)(x^2-5x+25)-(2x+1)^3-28x^3+3x(-11x+5)
B = (3x+2)^3 - 18x(3x+2)+(x-1)^3-28^3+3x(x-1)
C= (4x-1)(16x^2+4x+1)-(4x+1)^3+12(4x+1)^3+12(4x+1)-15
Tính giúp mình ạ ! Cảm ơn các cậu rất nhieeufuuuu :3
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
Cho biểu thức:
A=\(\left(\frac{3}{x+5}-\frac{3x-15}{2x-15}.\left(\frac{2x-15}{x^2-25}-2x+15\right)\right):\left(1-x\right)\)
Tìm x để biểu thức A xác định
diều kiện xác định là các mẫu phải khác o; số chia cũng khác o nhé:
ĐK: +) \(x+5\ne0\Rightarrow x\ne-5\)
+) \(2x-15\ne0\Rightarrow x\ne\frac{15}{2}\)
+) \(x^2-25\ne0\Rightarrow\left(x+5\right)\left(x-5\right)\ne0\Rightarrow x\ne\pm5\)
+) \(1-x\ne0\Rightarrow x\ne1\)
Vậy điều kiện xác đinh của A là : \(x\ne1;x\ne\frac{15}{2};x\ne\pm5\)
Tìm x
A. 31x - 3x -5x +25
B. 15(x -2) + 3(5-1) = -14
C. 4x (41 -52) -35x = -(7-2x)
D. 53 -(3x -17) + 218= 19 + 45x
Tìm x, biết: a) 121-(115+x)= 3x-(25-9-5x)-8
b)2x+2.3x+1.5x = 10800
c) (3|x-1/2) . (8/15-1/5)+2/3-1
d) x+1/2022 + x+2/2021= x+3/2020 + x+4/2019
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)