1) 4x-10=0 2) 2x^3 + 6x^2 = x^2 + 3x 3) x - 5 = 3 - x 4) (-10x + 5 ) ( 2x - 8 ) = 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
C1.10x2=6x+8
C2.23x+10=23+13x
C3.9x-6=4x+1
C4.15x-12=11x+15
C5.21x+9=19-11x
C6.15+16x=8-3x
C7.19-4x=8x+23
C8.51-10x=3x-21
C9.8-6x=11-4x
C10.2(3x+4)-3(1-2x)=8x+10
C11.5(3-4x)-4(2x-5)=9-10x
C12.3(5x-6)-2(2x-5)=11x-10
C13.10x+5(3x-2)=25-10x
C14.6(2x-3)+3(3-5x)=8x-9
C15.3(4x-2)+2(6-2x)=10-6x
C16.5(3-6x)-4(2-2x)=4x-9
B2:tìm cặp số nguyên x, y thỏa mãn
X y+2x+y=0
nhiều quá bạn ơi , mk nghĩ bạn nên tách ra rồi hãy đăng lên
Bài 1:
16:
=>15-30x-8+8x=4x-9
=>-22x+7=4x-9
=>-26x=-16
=>x=8/13
15: \(\Leftrightarrow12x-6+12-4x=10-6x\)
=>8x+6=10-6x
=>14x=4
=>x=2/7
14: \(\Leftrightarrow12x-18+9-15x=8x-9\)
=>-3x-9=8x-9
=>x=0
13: \(\Leftrightarrow10x+15x-10=25-10x\)
=>25x-10=25-10x
=>35x=35
=>x=1
12: \(\Leftrightarrow15x-18-4x+10=11x-10\)
=>11x-8=11x-10(loại)
1. 6x(x - 10) - 2x+20=0 6. 3x2 - 6x+3=0
2. 3x2(x - 3) + 3(3 - x)=0 7. 4x2 - 10x+2=0
3. x2 - 8x+16=2(x -4) 8. x2 - 12x -18=0
4. x2 - 16 + 7x ( x+4)=0 9. 3x2 - 10x+3=0
5. x2 - 13x - 14=0 10. 5x2 - 10x+10=0
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
Giải các phương trình tích sau:
1.a)(3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0
c)(4x + 2)(x2 + 1) = 0 d) (2x + 7)(x – 5)(5x + 1) = 0
2. a)(3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
b)x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
c)2x(x – 3) + 5(x – 3) = 0 d)(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
3.a)(2x – 5)2 – (x + 2)2 = 0 b)(3x2 + 10x – 8)2 = (5x2 – 2x + 10)2
c)(x2 – 2x + 1) – 4 = 0 d)4x2 + 4x + 1 = x2
4. a) 3x2 + 2x – 1 = 0 b) x2 – 5x + 6 = 0
c) x2 – 3x + 2 = 0 d) 2x2 – 6x + 1 = 0
e) 4x2 – 12x + 5 = 0 f) 2x2 + 5x + 3 = 0
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
bài 2:
a, (3x+2)(x^2-1)=(9x^2-4)(x+1)
(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)
(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0
(3x+2)(x+1)(1-2x)=0
b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0
x(x^2-9)-(x^3+8)=0
x^3-9x-x^3-8=0
-9x-8=0
tự tìm x nha
Tìm x:
1) -3.(1-2x) - 4.(1+3x) = -5x + 5
2) 3.(2x - 5) - 6.(1 - 4x) = -3x + 7
3) (1 - 3x) - 2.(3x - 6) = -4x - 5
4) x.(4x - 3) - 2x.(2x - 1) = 5x - 7
5) 3x.(2x - 1) - 6x.(x + 2) = -3x + 4
6) (1 - 2x).3 - 4.(6x - 1) = 7x - 5
7) 6x - 3.(1 - 4x) - 5.(x + 1) = 2x + 7
8) 6.(1 - 3x) - 3.(2x + 5) = -10x + 7
9) 3x.(1 - 2x) + 6x^2 - 7x = 8.(1 - 2x) - 9
10) 2x.(1 + 3x) - 3x.(4 + 2x) = 3x - 4
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Phân tích đa thức thành nhân tử
a) x³-3x²+3x-1-8y³
b) x⁴-4x³+8x²-16x+16
Giải pt
a) 6(x-3) +(x-1) ²-(x+1) ²=2x
b) (x+4) ²-(x+8) (x-8) =96
c) 4x²-1=(2x+1) (3x-5)
d) 2x²-x=3-6x
e) 2x³+5x²-3x=0
f) x(2x-7) -4x+14=0
g) (2x-5) ²-(x+2) ²=0
h) (3x+1) (7x+3) =(5x-7) (3x+1)
i) x²+10x+25-4x(x+5) =0
k))(4x-5) ²-2(16x²-25) =0
l) (4x+3) ²=4(x²-2x+1)
m) x²-11x+28=0
n) 3x³-3x²-6x=0
o) x²-9x+20=0
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
\(m,x^2-11x+28=0\)
\(\Leftrightarrow x^2-4x-7x+28=0\)
\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)
\(l,\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)
Bài 3
1.(x-1)(x+2)+5x-5=0
2.(3x+5)(x-3)-6x-10=0
3.(x-2)(2x+3)-7x2+14x=0
4.(x+1)(x-3)-15+5x=0
5.5(2x-1)(x+3)+5x-10x2=0
Bài4
1.3x-6+(x+1)(x-2)=0
2.4x2+6x-(2x+3)(3x-x)=0
3.5x-10-(2-x)(4+x)=0
4.10-10x+(x-1)(x-3)=0
5.20x2+30x-2(x-5)(2x+3)=0
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
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