( 3x^3 - 2x^2 + x + 2 )(5x^2)
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Bài 1 làm tính nhân
2x.(x^2-7x-3)
(-2x^3+y^2-7xy).4xy^2
(-5x^3).(2x^2+3x-5)
(2x^2-xy+y^2).(-3x^3)
(x^2-2x+3).(x-4)
(2x^3-3x-1).(5x+2)
Bài 2 Thực hiện phép tính
A,(2x+3y^2)
B, (5x-y)^2
C, (2x+y^2)^3
D, ( 3x^2-2y)^3
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)
Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số:
a.3x-5 >15-x b.3(x-2).(x+2)<3x^2+x
c.(2x+1)^2+(1-x).3x<hoặc=(x+2)^2
d.5x-20/3 - 2x^2+x/2 > x.(1-3x)/3 -5x/4
e.4-2x <hoặc= 3x-6
f.(x+4).(5x-1)>5x^2+16x+2
g)x.(2x-1)-8<5-2x(1-x)
h)3x-1/4 - 3.(x-2)/8 - 1>5-3x/2
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
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
Bài:Chia 1 biến đã sắp xếp 1)(2x^3+11x^2+18x-3):(2x+3) 2)(2x^3+11x^2+18x-3):(3x+3) 3)(2x^3+9x^2+5x+41):(2x^2-x+9) 4)(13x+41x^2+35x^3-14):(5x-2) 5)(5x^2-3x^3+15-9x):(5-3x) 6)(-4x^2+x^3-20+5x):(x-4)
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
Bài : Rút gọn
1) (3x - 5).(7-5x)-(5x+2).(2-3x)
2) (3x2+5x-2).(2x2-4x+3)
3) (3x2+5x-2).(2x2-4x+3)
4)4(2x3+x)-(2x-1)(4x2+3x+1)
5) 2x(x+3)-(x-5).(7+2x)
Nhanh + đúng auto tick :Đ!
Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số:
a.3x-5 >15-x b.3(x-2).(x+2)<3x^2+x
c.(2x+1)^2+(1-x).3x<hoặc=(x+2)^2
d.5x-20/3 - 2x^2+x/2 > x.(1-3x)/3 -5x/4
e.4-2x <hoặc= 3x-6
f.(x+4).(5x-1)>5x^2+16x+2
g)x.(2x-1)-8<5-2x(1-x)
h)3x-1/4 - 3.(x-2)/8 - 1>5-3x/2
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
a) ( 2x^4-5x^2+x^3-3-3x):(x^2-3)
b)(2x^3+5x^2+3):(2x^2-x+1)
c) (2x+4y)^2 : (x+2y)-(9x^3-12x^2-3x):(-3x)-3(x^2+3)
p(x)=3x^2+7+2x^4-3x^2-4-5x+2x^3
Q(x)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4
tính p(x) + Q(x)
Mình thu gọn 2 đa thức trước r mới cộng nhé
\(P\left(x\right)=3x^2+7+2x^4-3x^2-4-5x+2x^3\)
\(P\left(x\right)=\left(3x^2-3x^2\right)+\left(7-4\right)+2x^4-5x+2x^3\)
\(P\left(x\right)=2x^4+2x^3-5x+3\)
\(Q\left(x\right)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)
\(Q\left(x\right)=\left(-3x^3+x^3\right)+2x^2+\left(-x^4+5x^4\right)+\left(x+4x\right)-2\)
\(Q\left(x\right)=-2x^3+4x^4+2x^2+5x-2\)
\(P\left(x\right)+Q\left(x\right)=2x^4+2x^3-5x+3-2x^3+4x^4+2x^2+5x-2\)
\(P\left(x\right)+Q\left(x\right)=\left(2x^4+4x^4\right)+\left(2x^3-2x^3\right)+\left(-5x+5x\right)+\left(3-2\right)+2x^2\)
\(P\left(x\right)+Q\left(x\right)=6x^4+1+2x^2\)