Bài 9: Tìm x biết:
d) (2x - 1) . (3x + 1) + (3x + 4) . (3 - 2x) = 5
e) 0,6x (x - 0,5) - 0,3x (2x + 1,3) = 0,38
f) (c - 3x2)(x + 6) + x(3x2 + 17x) = 24
Tìm x
a) 6x(5x + 3) + 3x(1 – 10x) = 7 b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
c) (x + 1)(x + 2)(x + 5) – x2(x + 8) = 27
d) 5x(12x + 7) – 3x(20x – 5) = - 100
e) 0,6x(x – 0,5) – 0,3x(2x + 1,3) = 0,138
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
d) 5x(12x + 7) – 3x(20x – 5) = - 100
⇒60x2+35x-60x2+15=-100
⇒35x+15=-100
⇒35x=-100-15
⇒35x=-115
⇒x=\(\dfrac{-115}{35}\)
⇒x=\(\dfrac{-23}{7}\)
a) 5x(12x+7) - 3x(20x-5)= -100
b) 0,6x(x-0,5) - 0,3x(2x+1,3)= 0,138
c) 4x(3x-7) - 6(2x2 - 5x + 1) =12
a) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
b) 0,6x(x - 0,5) - 0,3x(2x + 1,3) = 0,138
<=> 0,6x2 - 0,3x - 0,6x2 - 0,39x = 0,138
<=> -0,69x = 0,138
<=> x = -0,2
c) 4x(3x - 7) - 6(2x2 - 5x + 1) = 12
<=> 12x2 - 28x - 12x2 + 30x - 6 = 12
<=> 2x - 6 = 12
<=> 2x = 18
<=> x = 9
Bài 1: Thực hiện phép tính:
a) 2x.(3x + 3) b) 5x.(3x2-2x + 1) c) 3x2(2x +4)
d) 5x2.(3x2 + 4x – 1) e) (x-1).(2x +3) f) (x+2).(3x-5)
Bài 2: Tìm x, biết:
a) 3x(x+1) – 3x2 = 6
b) 3x(2x+1) – (3x +1).(2x-3) = 10
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
1, Chứng tỏ rằng các đa thức sau ko phụ thuộc vào biến:
a,x(2x+1)-x2(x+2)+(x3-x+3)
b,4(x-6)-x2(2+3x)+x(5x-4)+3x2(x-1)
2,Tìm x
a,5x(12x+7)-3x(20x-5)=-100
b,0,6x(x-0,5)-0,3x(2x+1,3)=0,138
1)
a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^3-x+3=3\)
=>đpcm
b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2=-24\)
=>đpcm
2,
a) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
\(\Leftrightarrow x=-2\)
b) \(0,6x\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)
\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(\Leftrightarrow-0,69x=0,138\)
\(\Leftrightarrow x=-0,2\)
Câu 1:
a)\(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^2-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^2-x+3\)
\(=x^3+3\)(ko thể CM)
b)\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)(đpcm)
1 a ) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^3-x+3\)
\(=3\)
=> đpcm
b ) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)
=> đpcm
2 ) Tìm x :
a ) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=100\)
\(\Leftrightarrow50x=100\)
\(\Leftrightarrow x=2\)
b ) \(0,6c\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)
\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(\Leftrightarrow-0,69x=0,138\)
\(\Leftrightarrow-0,2\)
bài 1 Tim x a, 0,6x(x-0,5)-0,3x(2x+1,3)=0,138. , b, 3x2-3x(-2+x)=36. c, 2.(5x-8)-3(4x-5)=4(3x-4)+11 d, 1/3x2-4x+2x(2-3x)=0. e, 5x-3{4x-2(5x-2)]}=182 bài 2 thực hiện phép tính a, (-2x3-1/4y-4yz)8xy2 c,-6x+3(7+2x) e, 1/3xz(-9xy+15yz)+3x2(2yz2-yz)
Bài 1: Thực hiện phép tính:
a) x(3x2 – 2x + 5) b) 1/3 x2 y2 (6x + 2/3x2 – y)
c) ( 1/3x + 2)(3x – 6) d) ( 1/3x + 2)(3x – 6)
e) (x2 – 3x + 1)(2x – 5) f) ( 1/2x + 3)(2x2 – 4x + 6)
Bài 2: 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
Bài 3: Chứng tỏ rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của biến: a) A = x(2x + 1) – x2 (x + 2) + x3 – x + 3
b) B = (2x + 11)(3x – 5) – (2x + 3)(3x + 7) + 5
Bài 4: Tính giá trị của biểu thức
a) A = 2x( 1/2x2 + y) – x(x2 + y) + xy(x3 – 1) tại x = 10; y = – 1 10
b) B = 3x2 (x2 – 5) + x(–3x3 + 4x) + 6x2 tại x = –5
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
Bài 4:
b: Ta có: \(B=3x^2\left(x^2-5\right)+x\left(-3x^3+4x\right)+6x^2\)
\(=3x^4-15x^2-3x^3+4x^2+6x^2\)
\(=-5x^2\)
\(=-5\cdot25=-125\)
Tìm x, biết:
a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1)-3x2=54
b)(x-3)3-(x-3)(x2+6x+9)+6(x+1)2+3x2=-33
\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)
Tìm x biết
a. 5x.(12x + 7) - 3x(20x - 5) = - 100
b. 0,6x(x - 0,5) - 0,3x(2x + 1,3) = 0,138
a. 5x.(12x + 7) - 3x(20x - 5) = - 100
↔ 60x2 + 35x - 60x2 + 15x = - 100
↔ 50x = - 100
→ x = - 2
b. 0,6x(x - 0,5) - 0,3x(2x + 1,3) = 0,138
↔ 0,6x2 - 0,3x - 0,6x2 - 0,39x = 0,138
↔ - 0,6x = 0,138
↔ x = 0,138 : (- 0,6)
↔ x = - 0,2
Bài 1: Tìm x:
1) (x-3)3 -( x-3)(x2+ 3x+9) +6( x+1)2+ 3x2 = -33
2) (X-3)( X2+ 3X+9) - X(X-2)( 2+X) = 1
3) (X+2)(X2 - 2X+4) – X(X-3)(X+3) = 26
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)