(X - 1 )x ( X + 2 ) - X x ( X - 2 )- 3X.
Bài1:Rút gọn
a,(4x-5)(3x+2)-(7-3x)(x+2)
b,(-2x+1)(x-5)-3(x-2)(x+1)
c,(x^2-7)(x-5)+(3x^2+5)(2x-4)
d,(x^2+3x-2)(x+4)-4x(x-5)
Bài2:Tìm xbiết
a,(x-4)(x+3)-(x+1)(x-5)=8
b,(3x-2)(x+1)-3x(x+7)=13
c,(x+5)(x-5)-x(x+2)=9
d,(x-1)(x^2+x+1)-x(x^2-3)=1
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
Cho R(x) = 2x 2 + 3x - 1; M(x) = x 2 - x 3 thì R(x) - M(x)=
A.-3x 3 + x 2 + 3x – 1 B. -3x 3 - x 2 + 3x – 1
B. 3x 3 - x 2 + 3x – 1 D. x 3 + x 2 + 3x + 1
R(x) = 2x2 + 3x - 1
- M(x) = -x3 + x2
x3 + x2 + 3x - 1
Vậy R(x) - M(x) = x3 + x2 + 3x - 1
\(1, (-x+5)(x-2)+(x-7)(x+7)=(3x+1)^2-(3x-2)(3x+2)\)
\(2, (5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(x^2-x)\)
3, \((x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2 +(x-2)^2-x^2\)
1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)
\(=-x^2+2x+5x-10+x^2-49=7x-59\)
\(B=\left(3x+1\right)^2-\left(3x-2\right)\left(3x+2\right)\)
\(=9x^2+6x+1-9x^2+4=6x+5\)
=>7x-59=6x+5
=>x=64
2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)
\(=5x^2+5x-x-1-2x^2+12x-9\)
\(=3x^2+16x-10\)
\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)
\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)
\(=3x^2-4x-18\)
=>16x-10=-4x-18
=>20x=-8
hay x=-2/5
a, (3x+2)2 - (3x-2)2 =5x+38 b, 3(x-2)2 +9(x-1) =3(x2+x-3)
c, (x+3)3 -(x-3)2 -(x-3)2 =6x+18 d, (x-1)3-x(x+1)2=5x(2-x)-11(x+2)
e, (x+1)(x2-x+1)-2x=x(x-1)(x+1) f, (x-2)3+(3x-1)(3x+1)=(x+1)3
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
Bài1 thực hiện phép tính
a 5/2x^2+6x - 4-3x^2/x^2-9 -3
b , 3x^2+5x+14/x^3+1 + x-1/x^2-x+1 - 4/x+1
c, x-6/x^2+1 × 3x^2-3x+3/x^2-36 + x-6/x^3+1 × 3x/x^3-36
d,x^2+1/3x ÷ x^2+1/x-1 ÷x^3-1/x^2+x ÷ x^2+2x+1/x^2+x+1
1.Tìm x
a) (x - 5)(x + 5) - (x + 3)^2 + 3 (x - 2)^2 = (x + 1)^2 - (x + 4)(x - 4) +3x^2
b) (2x + 3)^2 + (x - 1)(x + 1) = 5 (x + 2)^2 - (x - 5)(x + 1) + (x + 4)^2
c) (-x + 5)(x - 2) + (x - 7)(x + 7) = (3x + 1)^2 - (3x - 2)(3x + 2)
d) (5x - 1)(x + 1) - 2(x - 3)^2 = (x + 2)(3x - 1) - (x + 4)^2 + (x^2 - x)
2.Rút gọn :
a) A= 3 (x - 1)^2 - (x + 1)^2 + 2(x - 3)(x + 3) - (2x + 3)^2 - (5 - 20x)
b) B= 5x (x - 7)(x + 7) - x (2x - 1)^2 - (x^3 + 4x^2 - 246x) - 175
c) C = -2x (3x + 2)^2 + (4x + 1)^2 + 2 (x^3 + 8x + 3x - 2 ) - (5 - x)
Tìm x biết:
1. (x-2)^2-(x-3)(x+3)=6
2. 4(x-3)^2-(2x-1)(2x+1)=10
3. (x-4)^2-(x-2)(x+2)=6
4.9(x+1)^2-(3x-2)(3x+2)=10
5. 3x +2(5-x)=0
6.x(2x-1)(x+5)-(2x^2+1)(x+4,5)=3,5
7, 3x^2-3x(x-2)=36
8. (3x^2-x+1)(x-1) +x^2(4-3x)=5/2
Tìm x :( bài 14 trang 11 sách bồi dưỡng năng lực tự học toán 8)
Câu 2 : (2x+3)2+(x-1)*(x+1)=5*(x+2)2-(x-5)*(x+1)+(x+4)2
Câu 3 : (-x+5)*(x-2)+(x-7)*(x+7)=(3x+1)2-(C)*(3x+2)
Câu 4 : (5x-1)*(x+1)-2(x-3)2=(x+2)*(3x-1)-(x+4)2+(x2-x)
Câu 5 : (4x-1)2-(3x+2)*(3x-2)=(7x-1)*(x+2)+(2x+1)2-(3x+2)
Câu 6 : (2x+3)2-(5x-4)*(5x+4)=(x+5)2-(3x-1)*(7x+2)-(x2-1+1)
Câu 7 : (1-3x)2-(x-2)*(9x+1)=(3x-4)*(3x+4)-9(x+3)2
Câu 8 : (3x+4)*(3x-4)-(2x+5)=(x-5)+(2x+1)2-(x2-2x)+(x-1)2
Câu 9 : (x-7)*(x+1)-(x-3)2=(3x-5)*(3x+5)-(3x+1)+(x-2)2-x2
Câu 10 : -5(x+3)2+(x-1)*(x+1)+(2x-3)=(5x-2)2-5x(5x+3)
1/Tìm x,biết:
a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555
b)1+2+3+4+...+x=820
c)3(x+1)=9.27
d)x+2x+3x+...+99x+100x=15150
e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550
f)3x+3x+1+3x+2=351
a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555
=> 101x +5050 = 5555
=> 101x = 505
=> x = 505 : 101 = 5
Vậy, x = 5
b)1+2+3+4+...+x=820
=> ( x+1) x :2 = 820
=> (x+1)x = 1640
Mà 1640 = 40 . 41
=> x = 40 ( vì {x+1} - x = 1)
Vậy, x = 40
c) 3x+1 = 9.27=243
=> 3x+1 = 35
=>x + 1 = 5
=> x = 4
Vậy, x=4
d) x+2x+3x+...+99x+100x=15150
=> [( 100 + 1) x 100 :2 ] x = 15150
=> 5050x = 15150
=> x = 15150:5050 = 3
Vậy, x =3
e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050= 200500
=> x = 200500 : 100 = 2005
Vậy, x = 2005
f)3x+3x+1+3x+2=351
=> 3x + 3x . 3 + 3x x 9 = 351
=> 3x ( 1+3+9) = 351
=> 3x . 13 = 351
=> 3x = 351 :13=27 mà 27 = 33
=> x=3
Vậy, x=3
a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)
\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)
\(\Rightarrow101\cdot x+5050=5555\)
\(\Rightarrow101\cdot x=5555-5050\)
\(\Rightarrow101\cdot x=505\)
\(\Rightarrow x=505:101\)
\(\Rightarrow x=5\)
b) \(1+2+3+4+...+x=820\)
\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)
\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)
\(\Rightarrow\left(x+1\right)\cdot x:2=820\)
\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)
\(\Rightarrow x\cdot\left(x+1\right)=1640\)
Ta thấy: \(40\cdot41=1640\)
Vậy: \(x=40\)
a. 3[x^2 - 2x +1]+x[2-3x]=7
b. 5[x-2] +2[x+3]=10
c. [x+1][-3]+5[x-4]=-3
d. 2[x-1]-x[3-x]=x^2
đ. 3x[x+5]-2[x+5]=3x^2
e. 4x[x+2]+x[4-x]=3x^2 +12
f. 1/3x [3x+6]-x[x-5]=9
a) \(3\left(x^2-2x+1\right)+x\left(2-3x\right)=7\)
\(\Rightarrow3x^2-6x+3+2x-3x^2=7\)
\(\Rightarrow-4x+3=7\)
\(\Rightarrow-4x+3-7=0\)
\(\Rightarrow-4x-4=0\)
\(\Rightarrow-4\left(x+1\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
b) \(5\left(x-2\right)+2\left(x+3\right)=10\)
\(\Rightarrow5x-10+2x+6=10\)
\(\Rightarrow7x-4=10\)
\(\Rightarrow7x=10+4=14\)
\(\Rightarrow x=\dfrac{14}{7}=2\)
c) \(\left(x+1\right)\left(-3\right)+5\left(x-4\right)=-3\)
\(\Rightarrow-3x-3+5x-20=-3\)
\(\Rightarrow2x-23=-3\)
\(\Rightarrow2x=-3+23=20\)
\(\Rightarrow x=\dfrac{20}{2}=10\)
d) \(2\left(x-1\right)-x\left(3-x\right)=x^2\)
\(\Rightarrow2x-2-3x+x^2=x^2\)
\(\Rightarrow-x-2+x^2-x^2=0\)
\(\Rightarrow-x-2=0\)
\(\Rightarrow-x=2\)
\(\Rightarrow x=-2\)
đ) \(3x\left(x+5\right)-2\left(x+5\right)=3x^2\)
\(\Rightarrow3x^2+15x-2x-10=3x^2\)
\(\Rightarrow3x^2-3x^2+13x-10=0\)
\(\Rightarrow13x-10=0\)
\(\Rightarrow13x=10\)
\(\Rightarrow x=\dfrac{10}{13}\)
e) \(4x\left(x+2\right)+x\left(4-x\right)=3x^2+12\)
\(\Rightarrow4x^2+8x+4x-x^2=3x^2+12\)
\(\Rightarrow3x^2+12x=3x^2+12\)
\(\Rightarrow3x^2+12x-3x^2-12=0\)
\(\Rightarrow12\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
f) \(\dfrac{1}{3}x\left(3x+6\right)-x\left(x-5\right)=9\)
\(\Rightarrow x^2+2x-x^2+5x=9\)
\(\Rightarrow7x=9\)
\(\Rightarrow x=\dfrac{9}{7}\)