Những câu hỏi liên quan
Chứng minh đẳng thức:
a, (x^2-2x/2x^2+8-2x^2/8-4x+2x^2-x^3)(1-1/x-2/x^2)=x+1/2x
b, [2/3x-2/x+1(x+1/3x-x-1)]:x-1/x=2x/x-1
c, [2/(x+1)^3(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
Bài 1: CM hằng đẳng thức:
(a+b)4=a4+4a3b+6ab2+4ab2+b4
Bài 2:Rút gọn:
a) (x-2)(x2+2x+4)-(x-1)+7
b) 8(x-1)(x2+x+1)-(2x-1)(4x2+2x-1)
Bài 1 :
Ta có : \(VP=\left(a+b\right)^4=\left(a+b\right)\left(a+b\right)^3\)
\(=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
=> HĐT ko đc CM
Bài 2 :
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)
\(=x^3+2x^2+4x-2x^2-4x-8-x+1+7=x^3-x=x\left(x^2-1\right)\)
Sửa đề : b, \(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=8\left(x^3-1\right)-8x^3+1=8x^3-8-8x^3+1=-7\)
Xin phép chủ nahf cho mjnh sửa đề:D
\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
a,\(\left(a+b\right)^4\)
\(=\left[\left(a+b\right)^2\right]^2\)
\(=\left(a^2+2ab+b^2\right)^2\)
\(=\left[\left(a^2+2ab\right)+b^2\right]^2\)
\(=\left(a^2+2ab\right)^2+2\left(a^2+2ab\right)b^2+b^4\)
\(=a^4+4a^3b+4a^2b^2+2a^2b^2+4ab^3+b^4\)
\(=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
Bài 2:
a,\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)
\(=\left(x^3-8\right)-\left(x-1\right)+7\)
b,\(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x-1\right)\)
\(=8\left(x^3-1\right)-\left(8x^3-1\right)\)
\(=8x^3-8-8x^3+1\)
\(=-7\)
cho biểu thức
P=(\(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\)).(1-\(\dfrac{1}{x}-\dfrac{2}{x^2}\)) ( x≠0; x≠2)
rút gọn biểu thức P
tính giá trị biểu thức P với x=1/2
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(=\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x^2-x-2\right)}{x^2}\)
\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:
\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)
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Cho biểu thức
(\(\dfrac{x^2-2x}{2x^2+8}\) - \(\dfrac{2x^2}{8-4x+2x^2-x^3}\)).(1-\(\dfrac{1}{x}\)-\(\dfrac{2}{x^2}\))(x≠0;x≠2)
rút gọn biểu thức
tính giá trị biểu thức với x1/2
rút gọn biểu thức M=((x2-2x)/(2x2+8)-2x/(8-4x+2x2-x3))*(1-1/x-2/x2)
Bài 3: Rút gọn biểu thức (Dùng hằng đẳng thức)1, (x+y)^2-(x-y)^22, (x+y)^3-(x-y)^3-2y^33,(x+y)^2-2(x+y)(x-y)+(x-y)^24,(2x+3)^2-2(2x+3)(2x+5)+(2x+5)^25, 9^8. 2^8-(18^4+1)(18^4-1)
Đọc tiếp
Bài 3: Rút gọn biểu thức (Dùng hằng đẳng thức)
1, (x+y)\(^2\)-(x-y)\(^2\)
2, (x+y)\(^3\)-(x-y)\(^3\)-2y\(^3\)
3,(x+y)\(^2\)-2(x+y)(x-y)+(x-y)\(^2\)
4,(2x+3)\(^2\)-2(2x+3)(2x+5)+(2x+5)\(^2\)
5, 9\(^8\). 2\(^8\)-(18\(^4\)+1)(18\(^4\)-1)
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
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1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
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cho biểu thức M=(x^2-2x/2x^2+8-2x^2/8-4x+2x^2-x^3)(1-1/x-2/x^2)
rút gọn M
tính gtrij của M sao cho x=1/2
Chứng minh đẳng thức :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x^3}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b) left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3}-x-1right)right]:dfrac{x-1}{x}dfrac{2x}{x-1}
c) left[dfrac{2}{left(x+1right)^3}.left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
Đọc tiếp
Chứng minh đẳng thức :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b) \(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3}-x-1\right)\right]:\dfrac{x-1}{x}=\dfrac{2x}{x-1}\)
c) \(\left[\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
A=(x ² - 2x/ 2x ² +8 +2x ²/x ³ -2x ² -4x-8) × (1-1/2-2/x ²) a.rút gọn