Rút gọn:
a) A=(5-x)(5+x)-x(4-x)-25
b) B=(x2+1)(x+1)-(x+1)3
c) C=(x+y-2)2-2(x+y-2)(y+x)+(x+y)2
Rút gọn:
a) A= \(\dfrac{x}{x-y}+\dfrac{2y^2}{x^2-y^2}-\dfrac{x}{x+y}\)
b) B= \(\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}\)
c) C= \(\dfrac{5}{x+1}-\dfrac{10}{-x^2+x-1}-\dfrac{15}{x^3+1}\)
a) \(\dfrac{x}{x-y}+\dfrac{2y^2}{x^2-y^2}-\dfrac{x}{x+y}=\dfrac{x\left(x+y\right)+2y^2-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x^2+xy+2y^2-x^2+xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y^2+2xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y}{x-y}\)
b) \(B=\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}=\dfrac{x\left(x+2\right)-4x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c) \(\dfrac{5}{x+1}-\dfrac{10}{-x^2+x-1}-\dfrac{15}{x^3+1}=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3+1}=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)
Bài 1: Chứng minh rằng :
cho ab=2;a+b=-3 tính giá trị biểu thức a^3 + b^3
Bài 2: rút gọn:
a, 2(x-y)×(x+y)+(x+y)^2(x-y)^2
b, x(x+4)×(x-4)-(x^2+1)×(x^2-1)
c, (a+b-c)-(a-c)^2-2ab+2ab
Bài 2:
b: Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-4x-x^4+1\)
\(=-x^4+x^3-4x+1\)
c: Ta có: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ab\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)\)
\(=b\left(2a+b-2c\right)\)
\(=2ab+b^2-2bc\)
\(a + b = -3\)
\(ab = 2\)
Từ \(ab = 2\), ta có thể giải ra được \(a = \frac{2}{b}\) hoặc \(b = \frac{2}{a}\).
Đặt \(a = \frac{2}{b}\) vào \(a + b = -3\) ta được:
\(\frac{2}{b} + b = -3\)
\(2 + b^2 = -3b\)
\(b^2 + 3b + 2 = 0\)
\((b + 1)(b + 2) = 0\)
\(b = -1\) hoặc \(b = -2\).
Khi \(b = -1\), ta có \(a = -2\). Khi \(b = -2\), ta có \(a = -1\).
Vậy giá trị của biểu thức \(A = a^3 + b^3\) khi \(a = -2, b = -1\) hoặc khi \(a = -1, b = -2\).
Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
a: \(=\dfrac{x-z}{2}\)
b: \(=\dfrac{3x}{4y^3}\)
Rút gọn:
a) A= \(\dfrac{x+y}{x-y}-\dfrac{x}{x+y}+\dfrac{2y^2}{x^2-y^2}\)
b) B= \(\dfrac{x}{x-2}-\dfrac{10}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-1}{x+3}\)
c) C= \(\dfrac{1}{x-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x^3-1}\)
a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)
Rút gọn:
a)2x.(3x-1)-(x-3).(6x+2)
b)(2x-3)2-(1+2x).(2x-1)+3.(2x-3)
c)(x+y-1)2-2.(x+y-1).(x+y)+(x+y)2
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)
\(=4x^2-12x+9-4x^2+1+6x-9\)
\(=-6x+1\)
c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2\)
=1
a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)
b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)
c) \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)
Bài 2: Rút gọn biểu thức:
a/ A = (3x–1)2 + (x+3)(2x–1)
b/ B = x(x–y) + y(x–y)
e/ C = (x–2)(x2+2x+ 4) – x(x2 –2)
f/ D = (x+y)2– (x–y)2
\(a.\left(3x-1\right)^2+\left(x+3\right)\left(2x-1\right)\)
\(=9x^2-6x+1-2x^2+x-6x+3\)
\(=7x^2-11x+4\)
bài 1 chứng minh giá trị của biểu thức ko phụ thuộc vào giá trị của x
a,A=3(x-1)2-(x+1)2+2(x-3)(x+3)-(2x+3)2-(5-20x)
b,B=-x(x+2)2+(2x+1)2+(x+3)(x2-3x+9)-1
bài 2 rút gọn biểu thưc
a,27(1-x)(x2+x+1)+81(x-1)
b,y[x2+x(x-y)+(x-y)2 ]+(x-y)3
Rút gọn:
a) A=(4-x)(16+4x+x2)-(4-x)3
b) B=(3x+2)(9x2-6x+4)-(9x2+6x+4)(3x-2)
c) C=(x+1)(x2-x+1)-x(x+1)2
a) Ta có: \(A=\left(4-x\right)\left(16+4x+x^2\right)-\left(4-x\right)^3\)
\(=64-x^3+\left(x-4\right)^3\)
\(=64-x^3+x^3-12x^2+48x-64\)
\(=-12x^2+48x\)
b) Ta có: \(B=\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(=27x^3+8-27x^3+8\)
=16
c) Ta có: \(C=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)^2\)
\(=x^3+1-x\left(x^2+2x+1\right)\)
\(=x^3+1-x^3-2x^2-x\)
\(=-2x^2-x+1\)
Rút gọn các biểu thức sau:
a) ( x + y)2 + (x - y)2 b) ( x + y)2 + (x - y)2 + 2( x+ y) ( x- y)
c) (2+3y)2-(2x-3y)2-12xy d) ( 3x + 1)2 - (3x - 1)2
e)(x+1)(x2-x+1)-(x-1)(x2+x+1)
a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)
b: \(=\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)
d: \(=9x^2+6x+1-9x^2+6x-1=12x\)