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\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)
a) \(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
c) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
a.\(\left(x^2+x\right)^2+3\left(x^2+x\right)+2=\left(x^2+x\right)^2+2\left(x^2+x\right)+\left(x^2+x+2\right)\)
\(=\left(x^2+x\right)\left(x^2+x+2\right)+\left(x^2+x+2\right)=\left(x^2+x+2\right)\left(x^2+x+1\right)\)
b. \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)(1)
Đặt \(t=x^2+3x\)
(1) \(\Leftrightarrow t\left(t+2\right)+1\)
\(=t^2+2t+1=\left(t+1\right)^2\)(2)
Thay \(t=x^2+3x\)vào (2) t/có:
\(\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)
c. dài lắm mình lười làm, bn bấm thử mạng tìm ik nhớ tíck cho mình nha thanks
ta có :
\(a^3+c^3=\left(a+c\right)^3-3ac\left(a+c\right)\)
nên \(a^3+c^3-b^3+3abc=\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)\)
\(=\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2-3ac\right]=\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)\)
b. tương tự ta có :
\(a^3-b^3-c^3-3abc=a^3-\left(b+c\right)^3+3bc\left(b+c-a\right)\)
\(=\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2-3bc\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
c. ta có : \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=\left(x-z+z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+3\left(x-z\right)\left(z-y\right)\left(x-y\right)+\left(z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=3\left(x-z\right)\left(z-y\right)\left(x-y\right)\)
bài 2 nè
a+b+c = 0
=>(a+b+c)^3 = 0
a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c) = 0
vì a+b = -c
a+c = -b
b+c = -a
thay vào => a^3 + b^3 + c^3 - 3abc = 0
=> a^3 + b^3 + c^3 = 3abc
Áp dụng hằng đẳn thức này: (a+b)^ 3 = a^3 + 3a^2b+3ab^2+b^3 = a^3 + b^3 +3ab(a+b)
a/. Có: a3+b3 +c3-3abc = (a+b)3-3ab(+b)+c3-3abc
= (a+b)3+c3-3ab(a+b) - 3abc= (a+b+c)[(a+b)2-(a+b)c+c2] - 3ab(a+b+c)
=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2 - 3ab)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
b/. tương tự a. khi nhóm thì nhóm (a^3 - c^3) trước
c/. 6x^4 - 11x^2 + 3 = 6t^2 -11t + 3 (Với t = x^2 >=0)
=6t^2 - 2t - 9t +3 = (6t^2 -2t) -(9t - 3) = 2t(3t - 1) - 3(3t-1) = (3t-1)(2t-3)
bạn tham khảo nha
a) 85 .12,7 + 5.3.12,7 = 12,7 . ( 85 + 5.3 ) = 12,7 . (85+15) = 12,7 . 100 = 1270
b) 52.143 - 52.39 - 8.26 = 52.143 - 52.39 - 4.2.26 = 52.143 - 52.39 - 4.52 = 52.( 143 -39 - 4 ) = 52 . 100 = 5200
phân tích :
5x - 20y = 5 .(x - 4y)
5x(x-1)-3x.(x-1) = (x-1).(5x-3x) = 2x(x-1) hoặc 5x(x-1)-3x(x-1) = x(x-1)(5-3) = 2x(x-1)
x.(x+y)-5x-5y = x(x+y) - 5 (x+y) = (x+y)(x-5)
\(a,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-40\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)=\left(x^2+6x\right)\left(x^2+6x+13\right)\\ b,=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\\ =a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)-b^2c^2\left(c-a\right)+c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(a^2b^2-b^2c^2\right)-\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)-c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(a-c\right)\left[b^2\left(a+c\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(a-c\right)\left(a^2b+b^2c-b^2c+a^2c\right)\\ =a^2\left(a-b\right)\left(a-c\right)\left(b+c\right)\)
\(c,=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)