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7(x - 3) - x(3 - x)
= (x - 3)(7 + x)
chỉ bt có v mà k bt có đúng k
1 ) 7 ( x - 3 ) - x ( 3 - x )
= 7 ( x - 3 ) + x ( x - 3 )
= ( x - 3 ) ( 7 + x )
2 ) 4x2 - 6x + 3 - 2x
= 4x2 - 2x - 6x + 3
= 2x ( 2x - 1 ) - 3 ( 2x - 1 )
= ( 2x - 1 ) ( 2x - 3 )
3 ) ( 4 - x ) - 4x + x2
= ( 4 - x ) - x ( 4 - x )
= ( 4 - x ) ( 1 - x )
4 ) x2 - 2xy + y2
= ( x - y )2
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
Làm ngắn gọn thôi nhé :v
\(A=\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)
\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)
\(A=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x-3\right)}{\left(x-1\right)\left(x-1\right)\left(x-3\right)\left(x-3\right)}\)
\(A=\frac{x+2}{x-3}\)
\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)
\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)
\(B=\frac{\left(-x-2\right)\left(x+2\right)\left(x+2\right)\left(x-2\right)}{\left(-x-2\right)\left(x-2\right)\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{x+2}{x-2}\)
\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{10x}{-x^2+9}\)
\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)
\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)
\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)
\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)
\(D=\frac{51x-15}{2x^3-18x}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\left(\frac{3x-2}{x^2+2x+1}\right)\)
\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)
\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(E=\frac{10x^2+10}{x^4-2x+1}\)
a )\(2x\left(xy-3\right)+3xy\left(x+1-y\right)+3x\left(y^2-1\right)=2x^2y-6x+3x^2y+3xy-3xy^2+3xy^2-3x=5x^2y-9x+3xy\)
=> Phụ thuộc vào giá trị của biến
b) \(\left(x+2y\right)\left(x-2y\right)-x\left(x+4y^2\right)+5=x^2-4y^2-x^2-4xy^2+5=-4y^2-4xy^2+5\)
=> Phụ thuộc vào giá trị của biến
c) \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)=27x^3+8-9x^2+4=27x^3-9x^2+12\)
=> Phụ thuộc vào giá trị của biến
a: Ta có: \(2x\left(xy-3\right)+3xy\left(x-y+1\right)+3x\left(y^2-1\right)\)
\(=2x^2y-6x+3x^2y-3xy^2+3xy+3xy^2-3x\)
\(=5x^2y+3xy-9x\)
c: Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)\)
\(=27x^3+8-9x^2+4\)
\(=27x^3-9x^2+12\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
4) (3x-2)(x-3)= 3x(x-3)-2(x-3)
=3x.x+3x.(-3)-2.x-2.(-3)
=\(3x^2\)-9x-4x+6
=\(3x^2\)+(-9x-4x)+6
=\(3x^2\)-13x+6
5) (2x+1)(x+3)=2x(x+3)+1(x+3)
=2x.x+2x.3+1.x+1.3
=\(2x^2\)+6x+1x+3
=\(2x^2\)+(6x+1x)+3
=\(2x^2\)+7x+3
6) (x-3)(3x-1)=x(3x-1)-3(3x-1)
=x.3x+x.(-1)-3.3x-3.(-1)
=\(3x^2\)-1x-9x+3
=\(3x^2\)+(-1x-9x)+3
=\(3x^2\)-10x+3
rút gọn biểu thức
A) \(x^2\)-(x+4)(x-1)=\(x^2\)- x(x-1)-4(x-1)
=\(x^2\)-x.x-x.(-1)-4.x-4.(-1)
=\(x^2\)-\(x^2\)+1x-4x+4
=(\(x^2-x^2\))+(1x-4x)+4
= -3x+4
B) x(x+2)-(x-2)(x+4)=x.x+x.2-x(x+4)+2(x+4)
=\(x^2+2x\)-x.x-x.4+2.x+2.4
=\(x^2+2x-x^2-4x+2x+8\)
=(\(x^2-x^2\))+(2x-4x+2x)+8
=8
tính giá trị biểu thức
A=3(x-2)-(2+x)(x-3)
=3.x+3.(-2)-2(x-3)-x(x-3)
=3x-6-2.x-2.(-3)-x.x-x(-3)
=3x-6-2x+6-\(x^2\)+3x
=(3x-2x+3x)+(-6+6)\(-x^2\)
=4x - \(x^2\)
thay x=-8 vào biểu thức thu gọn ta được:
4.(-8)- (-8)\(^2\)
= - 32 +64
= 32
B= x(3-x)-(1+x)(1-x)
=x.3+x.(-x)-1(1-x)-x(1-x)
=3x -\(x^2\)-1.1-1 .(-x)-x.1-x.(-x)
=3x\(-x^2\)-\(1^2\)+1x-1x+\(x^2\)
=(3x+1x-1x)+(\(-x^2+x^2\))-1
=3x-1
thay x=-5 vào biểu thức thu gọn ta được:
3.(-5)-1
=-15-1
=-16
Thu gọn biểu thức
4) (3x - 2) (x - 3)
= ( 3x2 - 2x ) - ( 3x x 3 - 2 x 3 )
= 3x2 - 2x - 3x x 3 + 2 x 3
= 3x2 - 2x - 9x + 6
= 3x2 - 11x + 6
5) (2x + 1) (x + 3)
= ( 2x2 + 1x ) + ( 6x + 3 )
= 2x2 + 1x + 6x + 3
= 2x2 + 7x + 3
6) (x - 3) (3x - 1)
= ( 3x2 - 9x ) - ( x - 3 )
= 3x2 - 9x - x + 3
= 3x2 - 10 + 3
Rút gọn biểu thức
A) x^2 - (x + 4) (x - 1)
= x2 - ( x2 + 4x ) - ( x + 4 )
= x2 - x2 - 4x - x - 4
= -5x - 4
B) x (x + 2) - (x - 2) (x + 4)
= x2 + 2x - ( x2 - 2x ) + ( 4x - 8 )
= x2 + 2x - x2 + 2x + 4x - 8
= 8x - 8
Tính giá trị biểu thức
A = 3 (x - 2) - (2 + x) (x - 3) tại x = - 8
Thế x = -8 vào, ta có :
= 3 ( -8 -2 ) - ( 2 + -8 ) ( -8 - 3 )
= 3 x ( -10 ) - ( - 6 ) ( -11 )
= -30 - 66
= -96
B = x (3 - x) - (1 + x) ( 1 - x) tại x = - 5
Thế x = - 5 vào, ta có :
= -5 ( 3 - -5 ) - ( 1+ -5 ) ( 1 - -5 )
= -5 x 8 - (-4) x 6
= - 40 - -24
= -40 + 24
= -16
100% đúng
hok tốt nha
a)xy(x2+2y)=xy.x2+xy.2y
=x3y+2xy2
b)-4(6x2-xy)=-4.6x2+4.xy
=-24x2+4xy
c)4x[x2+6x-1/2]
=4x.x2+4x.6x-4x.1/2
=4x3+24x2-2x
a) \(xy\times\left(x^2+2y\right)=x^3y+2xy^2\)
b) \(-4\times\left(6x^2-xy\right)=-24x^2+4xy\)
c)\(4x\times\left(x^2+6x-\frac{1}{2}\right)=4x^3+24x^2-2x\)