Khai triển HDT:
1. ( 2x + y2 )
2. ( 7 + 3x - y) ( 7 - 3x + y )
3. ( 6/5 - 5x )2
Bài 1: Khai triển hằng đẳng thức:
a, ( x - y + 2z )2
b, ( 2x-3 ). ( 2x+3 ) . ( 4x2+9 )
Bài 2: Rút gọn biểu thức:
a, ( 5x+2 ).( 2-5x ) - ( 3x+2 ).( 2x+5 )2
b, ( -2x-3 )2 + 2(2x+1).( 2x+5 ) + ( 2x+5 )2
Bài 1:
a, \(\left(x-y+2z\right)^2=x^2+y^2+4z^2-2xy-4yz+4zx\)
b, \(\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)=\left(4x^2-9\right)\left(4x^2+9\right)=16x^4-81\)
Triển khai giúp mình ạ
a) ( 2 + xy )^2
b) ( 5 - x^2 ) . ( 5 + x^2 )
c ) ( 2x - y ) . ( 4x^2 + 2xy + y^2 )
d) ( 5 - 3x )^2
e) ( 5x - 1 ) ^3
f) ( x + 3 ) . ( x^2 - 3x + 9)
g) -x^3 + 3x^2 - 3x + 1
h)( 2x^2 + 3y)^2
a. \(\left(2+xy\right)^2=x^2y^2+4xy+4\)
b. \(\left(5-x^2\right)\left(5+x^2\right)=25+5x^2-5x^2-x^4=-x^4+25\)
c. \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3\)
\(=8x^3-y^3\)
d. \(\left(5-3x\right)^2=25-30x+9x^2\)
e. \(\left(5x-1\right)^3=125x^3-75x^3+15x-1\)
f. \(\left(x+3\right)\left(x^2-3x+9\right)=x^3-3x^2+9x+3x^2-9x+27=x^3+27\)
h. \(\left(2x^2+3y\right)^2=4x^4+12x^2y+9y^2\)
a) (2+xy)2 = 22+4xy+(xy)2 = 4 + 4xy +x2y2
b) ( 5 - x^2 ) . ( 5 + x^2 ) = 52-x4=25-x4
c) ( 2x - y ) . ( 4x^2 + 2xy + y^2 ) = 8x3-y3
d)(5-3x)2=52-2.5.3x+9x2=25-30x+9x2
e) (5x-1)3=(5x)3-3.(5x)2.1+3.5x.1-1 =125x3-75x2+15x-1
f) (x+3)(x2-3x+9)=(x+3)(x2-3x+32)=x3+27
g) -x3+3x2-3x+1 =(−x+1)(x−1)(x−1)= -(x-1)3
h) (2x2+3y)2=4x4+2.2x2.3y+9y2=4x4+12x2y+9y2
d) (3x – 5)(7 – 5x) – (5x + 2)(2 – 3x) = 4 g) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) =0 j) (2x – 1)(3x + 1) – (4 – 3x)(3 – 2x) = 3 k) (2x + 1)(x + 3) – (x – 5)(7 + 2x) = 8 m) 2(3x – 1)(2x + 5) – 6(2x – 1)(x + 2) = - 6
g: 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
a) 3x(x+1)-x(3x+2)
b) 2x(x2-5x+6)+(x-1)(x+3)
c) (x2-xy+y2)-(x2+2xy+y2)
d) (2/5xy+x-y)-(3x+4y)-2/5xy
e) 2xy(x2-4xy+4y2)
f) (x+y)(xy+5)
g) (x3-2x2-x+2):(x-1)
h) (2x2+3x-2):(2x-1)
Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2
Vậy phương trình có nghiệm là x = 2
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:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Chứng tỏ rằng các đa thức sau ko phụ thuộc vào biến
A=(3x-5)(2x+11)-(2x+3)(3x+7)
B=(x-5)(2x+3)-2x(x-3)+x+7
C= 4(x-6)-x^2(2+3x)+x(5x-4)+3x^2(x-1)
D=x(y+z-yz)-y(z+x-zx)+z(y-x)
Chứng tỏ rằng các đa thức sau ko phụ thuộc vào biến
a) Ta có: \(A=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21\)
=-74
Vậy: Đa thức A không phụ thuộc vào biến(đpcm)
b) Ta có: \(B=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)
\(=2x^2+3x-10x-15-2x^2+6x+x+7\)
\(=-8\)
Vậy: Đa thức B không phụ thuộc vào biến(đpcm)
c) Ta có: \(C=4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)
Vậy: Đa thức C không phụ thuộc vào biến(đpcm)
d) Ta có: \(D=x\left(y+z-yz\right)-y\left(z+x-zx\right)+z\left(y-x\right)\)
\(=xy+xz-xyz-yz-xy+xyz+zy-zx\)
=0
Vậy: Đa thức D không phụ thuộc vào biến(đpcm)
Bài 1 : Tìm x,y,z
1) 3x+2/5x+7 = 3x-1/5x+1
2) x+1/2x+1 = 0,5x+2/x+3
3) x/3=y/4:y/5 = z/7
4) x/3=y/4; y/3 = z/5 và 2x-3y+z=36
5) x-1/2 = y-2/3 = z-3/4 và 2x+3y-z= 50
6)2x+3/5x+2 = 4x+5/10x+2
CÁC BẠN BIẾT CÂU NÀO THÌ GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
Bài 2: Thực hiện các phép tính sau:
a)(x3+5x2-2x+1)(x-7)
b)(2x2-3xy+y2)(x+y)
c)(x-2)(x2-5x+1)-x(x2+11)
d)x(1-3x)(4-3x)-(x-4)(3x+5)
\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)