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bài 1 : điền vào chỗ chấm để đk khẳng định đúng :
a) (.x..+2y...)2=x2+..4y.+4y2
b) (.a..-.3b..)2=a2-6ab+.9b2..
c) (.m..+.\(\frac{1}{2}\)..)2=.m2..+m+1/4
d) 25a2-..\(\frac{1}{4}b\).=(.5a..+1/2b)(..5a..-1/2b)
e)(.2x...+.1..)^2 = 4x^2 +.4x..+1
g)(2-x)(.4..+.2x..+.x2..)=8-x^3
h) 16a^2 - ..9. = (..4a.+3)(..4a.-3)
f)25 - ..30y.+9y^2=(..5.+...3y.)^2
Bài 4:
a: \(=1-\left(x-y\right)^2\)
\(=\left(1-x+y\right)\left(1+x-y\right)\)
b: \(=a^2-2ab+b^2-c^2+2cd-d^2\)
\(=\left(a-b\right)^2-\left(c-d\right)^2\)
\(=\left(a-b-c+d\right)\left(a-b+c-d\right)\)
d: \(=x^2\left(x^6-64\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
Phân tích đa thức thành nhân tử:
a, \(36a^2-\left(a^2+9\right)^2\)
\(=\left(6a\right)^2-\left(a^2+9\right)^2\)
\(=\left(6a-a^2-9\right)\left(6a+a^2+9\right)\)
b, \(\left(a+3b\right)^2-\left(a^2+9\right)^2\)
\(=\left(a+3b-a^2-9\right)\left(a+3b+a^2+9\right)\)
c, \(9\left(2a-x\right)^2-4\left(3a-x\right)^2\)
\(=\left[3\left(2a-x\right)\right]^2-\left[2\left(3a-x\right)\right]^2\)
\(=\left(6a-3x\right)^2-\left(6a-2x\right)^2\)
\(=\left(6a-3x-6a+2x\right)\left(6a-3x+6a-2x\right)\)
\(=\left(-x\right)\left(12a-5x\right)\)
e, \(x^4+x^3+x+1\)
\(=\left(x^4+x^3\right)+\left(x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x^3+1\right)\left(x+1\right)\)
A=(a+1)(a+2)(a^2+4)(a-1)(a^2+1)(a-2)
A =(a+1)(a-1)(a+2)(a-2)(a^2+4)(a^2+1)
A =(a^2-1)(a^2+1)(a^2-4)(a^2+4)
A =(a^4-1)(a^4-16)
A =\(a^{16}-16\cdot a^4-a^4+16\)
A =\(a^{16}-17\cdot a^4+16\)
B=(a+2b-3c-d)(a+2b+3c+d)
B=[(a+2b)^2 - (3c +d)^2]
B=[a^2+4ab+4b^2-(9c^2+6cd+d^2)]
B=a^3+4ab+4b^2 - 9c^2 - 6cd - d^2
C=(1-x-2x^3+3x^2)(1-x+2x^3-3x^2)
C=[(1-x)^2-(2x^3-3x^2)^2]
C=[(1-2x+x^2) - (4x^6-12x^5+9x^4)]
C=[1-2x-x^2-4x^6+12x^5-9x^4]
C=-4x^6+12x^5-9x^4-x^2-2x+1
D=(a^6-3a^3+9)(a^3+3)
D=a^9+27
a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0
\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)
\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)
\(-5x-8=0\)
\(x=-\frac{8}{5}\)
Mai mik làm mấy bài kia sau
e: \(\left(a^2-1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)
\(=\left(a^3-1\right)\left(a^3+1\right)\)
\(=a^6-1\)
lm hết giúp mk vs