(x-1)(x+2)-x-2=0
x(3x+2)+(x+1)2-(2x-5)(2x+5)=0
(4x+1)(x-2) - (2x-3)(2x+1)=7
1) |2x - 1| = 5
2) |2x - 1| = |x + 5|
3) |3x + 1| = x - 2
4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
1)4x-20=0 ; 2) 5x+15=0 ; 3) 3x-5=7x+2 ; 4) 4x-(x-1)=2(1+x) ; 5) x2 -2x=0 ; 6) 2(3x-5)-3(x-2)=3(x+4) ; 7) (x+3)(2x-7)=0
8) 5x(x-3)+2x-6=0 ; 9) (3x-1)(2x-1)-(3x-1)(x+2)=0
10)|2x-1|+1=8 ; 11) |x-2|=3x+1 ; 12) |2x|=21-x
Giải các phương trình nha mọi người ^_^
Tìm x biết
1.(x+3)2-(x+2).(x-2)=4x+17
2.(2x+1)2-(4x-1).(x-3)-15=0
3.(2x+3).(x-1)+(2x-3).(1-x)=0
4.2(5x-8)-3(4x-5)=4(3x-4)+11
5.(3x-1).(2x-7)-(1-3x).(6x-5)=0
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
. 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}
Các bạn giúp mình nhé càng nhanh càng tốt nhà
(5x-1). (2x+3)-3. (3x-1)=0
x^3 (2x-3)-x^2 (4x^2-6x+2)=0
x (x-1)-x^2+2x=5
(3x+2)(x-1)-3 (5x+2)+5 (11-4x)=25
8 (x-2)-2 (3x-4)=25
(3x+4). (5x-1)+(5x+2). (1-3x)+2=0
(5x-1). (2x+7)-(2x-3). (5x+9)
4 (x-1). (X+5)-(x+5). (X+2)=3. (X-1)(x+2)
2x^2+3 (x-1). (X+1)=5x(x+1)
4. (18-5x)-12 (3x-7)=1825. (2x-16)-6 .(x+4)
1/2x. (2/5-4x)+(2x+5).x=-13/2
Nhiều các bạn giả đùm mình nha
Thanh nhiều
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
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
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35