tìm x biết :
a,6x.(3x+5)-2x(9x-2) =17
b,2x.(3x-1)-3x(2x+11)-70=0
Tìm x biết
a) 6x(3x+5)-2x(9x-2)=17
b) 2x(3x-1)-3x(2x+11)-70=0
c) 5x(2x-3)-4(8-3x)=2(3+5x)
Giải chi tiết giúp mình nhé!!!!
a) 6x(3x +5)-2x(9x-2)=17
6x3x+6x5-2x9x-2x(-2)=17
\(18x^2\)+30x-\(18x^2\)+4x=17
\(18x^2-18x^2\)+ 34x=17
0 +34x=17
x=17:34
x=0.5
b)2x(3x-1)-3x(2x+11)-70=0
2x3x-2x1-3x2x+3x11-70=0
\(6x^2-2x-6x^2+33x-70=0\)
-2x+33x-70=0
31x-70=0
31x=0+70
31x=70
x=\(\frac{70}{31}\)
(trong câu c dấu . của mình là nhân nha)
c)5x(2x-3)-4(8-3x)=2(3+5x)
5x2x-5x3-4.8+4.3x=2.3+2.5x
\(10x^2-15x-32+12x=6+10x\)
\(10x^2-15x+12x-10x=6+32\)
\(10x^2-13x=38\)
tạm thời mình bí chổ này thông cảm nha bạn
Tìm x:
6x(3x+5)-2x(9x-2)=17
2x(3x-1)-3x(2x+11)-70=0
6x(3x + 5) - 2x(9x - 2) = 17
⇒ 18x2 + 30x - (18x2 - 4x) = 17
⇒ 18x2 + 30x - 18x2 + 4x = 17
⇒ 30x + 4x = 17
⇒ 34x = 17
⇒ x = 17 : 34
⇒ x = 1/2
2x(3x - 1) - 3x(2x + 11) - 70 = 0
⇒ 6x2 - 2x - (6x2 + 33x) - 70 = 0
⇒ 6x2 - 2x - 6x2 -33x = 70
⇒ -2x - 33x = 70
⇒ -35x = 70
⇒ x = -2
\(6x\left(3x+5\right)-2x\left(9x-2\right)=17\)
⇔\(18x^2+30x-18x^2+4x=17\)
⇔\(34x=17\)
⇔\(x=2\)
tìm x biết
a) (6x-3) (2x+4) + (4x-1) (5-3x) = -21
b) 6x (3x+5) - 2x (9x-2) + (17-x) (x-1) + x (x-18) =0
c) (15-2x) (4x+1) - (13-4x) (2x-3) - (x-1) (x+2) + x2=52
d) (8x-3) (3x+2) - (4x+7) (x+4) = (2x+1) (5x-1) - 33
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21
<=> 12x2 + 24x - 6x - 12 + 20x - 12x2 - 5 + 3x = -21
<=> 41x = -21 + 12 + 5
<=> 41x = -4
<=> x = -4/41
A. 6x . (3x+5) -2x .(9x-2) = 17
B . 2x . (3x-1)-3x.(2x+11) -70 = 0
C . 2x2 + 3 . ( x2 - 1 ) = 5x . x +1
Hu hu bạn nào giúp mik vs mùa hè nên quên tất 😭 Mik mong các b giúp đỡ cảm ơn .
Ta có : 6x(3x + 5) - 2x(9x - 2) = 17
<=> 18x2 + 30x - 19x2 + 4x = 17
<=> 34x2 = 17
=> x2 = 17 : 34
=> \(x^2=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
\(a,6x\cdot\left(3x+5\right)-2x\cdot\left(9x-2\right)=17\)
\(\Leftrightarrow18x^2+30x-18x^2+4x=17\)
\(\Leftrightarrow34x=17\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(b,2x\cdot\left(3x-1\right)-3x\cdot\left(2x+11\right)-70=0\)
\(\Leftrightarrow6x^2-2x-6x^2-33x-70=0\)
\(\Leftrightarrow-35x=70\)
\(\Leftrightarrow x=-2\)
Đầu bài ý c là j vậy ... mình k thấy zõ
Chúc bạn học giỏi
Kết bạn với mình nha
Tìm x biết:
a,(3x+2)(2x+9)-(x+2)(6x+1)=(x-1)-(x-6)
b,3(2x-1)(3x-1)-(2x-3)(9x-1)=0
. Tìm x, biết:
a) 6x.(x – 5) + 3x.(7 – 2x) = 18 b) 2x.(3x + 1) + (4 – 2x).3x = 7 c) 0,5x.(0,4 – 4x) + (2x + 5).x = -6,5 | d) (x + 3)(x + 2) – (x - 2)(x + 5) = 6 e) 3(2x - 1)(3x - 1) – (2x - 3)(9x - 1) = 0 |
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
tìm x
1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
2) (2x+1)^3-(3x+2)^2=(2x-5)(4x^2+10x+25)+6x(2x+1)-9x^2
Tìm x, biết:
a)2x*(6x-5)-4x*(3x+7)=7
b)-5x(2x+1)+3x(3x+2)+x(x-1/2)=0
c)3x*(6x-5)-2x(9x+7)=15
d)1/2x*(2x+4)-(x+3)=5
e)(-3x+2)*5x-5x*(2x+1)-5x=4
Mấy bạn giúp mk ik mk đang cần gấp!
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)