Tìm x:
1500 : [(30x + 40) : x] = 30
(2x - 15)5 = (2x - 15)3
32x + 2 = 9x + 3
tìm x
a , 3x(12x-4)-9x(4x-3)=30
b, x(5-2x)+2x(x-1)=15
bài 19: tìm x
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
nhanh nha, mik tick cho, ccau trình bày dễ hiểu, ko cần ''hoặc''
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
Tìm x, biết :
a) 3x.(12x - 4) - 9x.(4x - 3) = 30
b) x.(5 - 2x) + 2x.( x - 1) = 15
a, 3x.(12x-4)-9x(4x-3)=30
=>36x2-12x-36x2+27x=30
=>5x=30
=> x=6
b,x.(5-2x)+2x.(x-1)=15
=> 5x-2x2+2x2-2x=15
=>3x=15
=>x=5
tk mk nha bn
*****Chúc bạn học giỏi*****
a) 3x . (12x - 4) - 9x(4x - 3) = 30
3x . 12x - 12x - 9x.4x + 27x = 30
(3x . 12x - 9x . 4x) - (12x - 27x) = 30
(36x2 -36x2) + 15x = 30
=> 15x = 30
=> x = 30 : 15
=> x = 2
b) x.(5 - 2x) + 2x.(x - 1) = 15
5x - 2x2 + 2x2 - 2x = 15
(5x - 2x) - (2x2 - 2x2) = 15
=> 3x = 15
=> x = 15 : 3 = 5
Tìm x,biết
a)3x(12x-4)-9x(4x-3)=30
b)x(5-2x)+2x(x-1)=15
P(x) = 2x3 – 5x2 + 8x – 3
Nghiệm hữu tỷ nếu có của đa thức P(x) trên là:
(– 1); 1; (–1/2); 1/2 ; (–3/2); 3/2 ; –3…
Sau khi kiểm tra ta thấy x = 1/2 là nghiệm nên đa thức chứa nhân tử ( x – 1/2) hay (2x – 1). Do đó ta tìm cách tách các hạng tử của đa thức để xuất hiện nhân tử chung (2x – 1).
2x3 - 5x2 + 8x – 3 = 2x3- x2 – 4x2 + 2x + 6x – 3
= x2( 2x – 1) – 2x( 2x – 1) + 3(2x – 1)
= ( 2x – 1)(x2 – 2x + 3).
Hoặc chia P(x) cho (x – 1) ta được thương đúng là: x2 – 2x + 3
P(x) = 2x3 – 5x2 + 8x – 3 = ( 2x – 1)(x2 – 2x + 3)
Vậy P(x) = 2x3 – 5x2 + 8x – 3 = ( 2x – 1)(x2 – 2x + 3)
Tìm x, biết:
a) 3x(12x – 4) – 9x(4x – 3) = 30
b) x(5 – 2x) + 2x(x - 1) = 15
a) 3x(12x – 4) – 9x(4x – 3) = 30
36x2 – 12x – 36x2 + 27x = 30
15x = 30
x = 2
b) x(5 – 2x) + 2x(x – 1) = 15
5x – 2x2 + 2x2 – 2x = 15
3x = 15
x = 5
a) 3x(12x – 4) – 9x(4x – 3) = 30
36x2 – 12x – 36x2 + 27x = 30
15x = 30 x = 2
b) x(5 – 2x) + 2x(x – 1) = 15
5x – 2x2 + 2x2 – 2x = 15
3x = 15
x = 5
a) 3x(12x – 4) – 9x(4x – 3) = 30
36x2 – 12x – 36x2 + 27x = 30
15x = 30
x = 2
b) x(5 – 2x) + 2x(x – 1) = 15
5x – 2x2 + 2x2 – 2x = 15
3x = 15
x = 5
Các bn nhớ ủng hộ NTN Vlogs này nhá
Tìm x biết :
a) 3x(12x - 4 ) - 9x( 4x - 3 ) = 30
b) x( 5 - 2x ) + 2x ( x - 1 ) = 15
a) \(3x\left(12x-4\right)-9x\left(4x-3\right)=0\)
\(\Leftrightarrow3x\left(12x-4-12x+9\right)=0\)
\(\Leftrightarrow3x\cdot5=0\)
\(\Leftrightarrow x=0\)
b) \(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=15\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
a) 3x(12-4)-9x(4x-3)=0
=> 3x(12x-4-12x+9)=0
=>3x .5=0
=>x-0
b) 5x-2x^2+2x^2-2x=15
=>5x-2x=15
=>3x=15
=>x=5
a) 3x(12x-4)-9x(4x-3)=30
\(\Leftrightarrow\)3x(12x-4)-3.3x(4x-3)=30
\(\Leftrightarrow\)3x(12x-4-12x+9)=30
\(\Leftrightarrow\)3x.5=30
\(\Leftrightarrow\)3x=7
\(\Leftrightarrow\)x=\(\dfrac{7}{3}\)
b)x(5-2x)+2x(x-1)=15
\(\Leftrightarrow\)5x-2x2+2x2-2x=15
\(\Leftrightarrow\)3x=15
\(\Leftrightarrow\)x=5
Tìm x, biết:
a) 3x ( 12x – 4 ) – 9x ( 4x – 3 ) = 30
b) x ( 5 – 2x ) + 2x ( x – 1 ) = 15
a) 3x (12x – 4) – 9x (4x – 3) = 30
=> 36x2 – 12x – 36x2 + 27x = 30
=> 15x = 30
Vậy x = 2.
b) x (5 – 2x) + 2x (x – 1) = 15
=> 5x – 2x2 + 2x2 – 2x = 15
=> 3x = 15
=> x = 5
a) 3x (12x – 4) – 9x (4x – 3) = 30
=> 36x2 – 12x – 36x2 + 27x = 30
=> 15x = 30
Vậy x = 2.
b) x (5 – 2x) + 2x (x – 1) = 15
=> 5x – 2x2 + 2x2 – 2x = 15
=> 3x = 15
=> x = 5
a) 3x(12x - 4 ) -9x (4x -3 ) = 30
<=> 36x² - 12x - 36x²+27x = 30
<=> 15x = 30
<=> x=2
b) x(5-2x)+2x(x-1)=15
<=> 5x - 2x² + 2x² - 2x = 15
<=> 3x = 15
<=> x = 5.
1500 : [ ( 30x + 40 ) : x ] = 30
92 x 4 - 27 = x + 350 :x + 315
697 : 15x + 364 : x = 17
tìm \(x\)
a) \(1500:\left[\left(30x+40\right):x\right]=30\)
b)\(697:\frac{15x+364}{x}=17\)
c)\(3x=2x+10\)
d)\(4x=2x+100\)
e)\(\left(2x-1\right)+\left(4x-2\right)+\left(6x-3\right)+...+\left(400x-200\right)=5+10+15+...+1000\)
b) \(697:\frac{15x+364}{x}=17\)
\(\frac{15x+364}{x}=697:17\)
\(\frac{15x+364}{x}=41\)
\(15+\frac{364}{x}=41\)
\(\frac{364}{x}=41-15\)
\(\frac{364}{x}=26\)
\(\Rightarrow x=14\)