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Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
Nếu ol thì tham khảo nah nguoiemtinhthong.
1.1
2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1
⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)
Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0
pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0
a=2ba=2b v a=13ba=13b
Các bạn tự giải quyết tiếp nhé.
1.2
TXĐ D=[1;+∞)D=[1;+∞)
đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0
pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0
⇔a=b⇔a=b v a=23ba=23b
...
1.3
D=[3;+∞)D=[3;+∞)
Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0
pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2
⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0
⇒a=5b⇒a=5b
...
1.4
ĐK
⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)
⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)
Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)
⇔2a2+2b2=3ab
1.5
Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)
⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x
⇔t2−t−4x2+2x=0t2−t−4x2+2x=0
Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2
⇒t=1−2xt=1−2x hoặc t=2xt=2x
1.1
2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1
2(.2+x+1)+3(x-1)
3a+b=11a2-19b2
tóm tắt
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
a) Ta có: \(4x^2-6x\)
\(=2x\left(2x-3\right)\)
b) Ta có: \(9x^4y^3+3x^2y^4\)
\(=3x^2y^3\left(3x^2+y\right)\)
c) Ta có: 3(x-y)-5x(y-x)
=3(x-y)+5x(x-y)
=(x-y)(3+5x)
d) Ta có: \(x^3-2x^2+5x\)
\(=x\left(x^2-2x+5\right)\)
e) Ta có: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=\left(x+3y\right)\left(5-15x\right)\)
\(=5\left(x+3y\right)\cdot\left(1-3x\right)\)
f) Ta có: \(2x^2\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+4\right)\)
\(=2\left(x+1\right)\left(x^2+2\right)\)
nhiều
bt mà bn
Phạm Minh Hiền 10 ý
bn rảnh thì lm đi bn
a, x2-6x+8
=x2-2x-4x+8
=x(x-2)-2(x-2)
=(x-2)2
b, x2+4x+3
=x2+x+3x+3
=x(x+1)+3(x+1)
=(x+1)(x+3)
x, 4x2+4x-3
=4x2-2x+6x-3
=2x(2x-1)+3(2x-1)
=(2x-1)(2x-3)
d, x2-x-12
=x2+3x-4x-12
=x(x+3)-4(x+3)
=(x+3)(x-4)
e, x2-5x+6
=x2-2x-3x+6
=x(x-2)-3(x-2)
=(x-2)(x-3)
f, x2-5x-14
=x2+2x-7x-14
=x(x+2)-7(x+2)
=(x+2)(x-7)
g, 4x2-3x-1
=4x2-4x+x-1
=4x(x-1)+(x-1)
=(x-1)(4x+1)
h, 3x2-2x-5
=3x2+3x-5x-5
=3x(x+1)-5(x+1)
=(x+1)(3x-5)
i, 9x2-6x+1
=9x2-3x-3x+1
=3x(3x-1)-(3x-1)
=(3x-1)(3x-1)
=(3x-1)2
\(\text{a) }x^2-6x+8\\ =x^2-2x-4x+8\\ =\left(x^2-2x\right)-\left(4x-8\right)\\ =x\left(x-2\right)-4\left(x-2\right)\\ =\left(x-4\right)\left(x-2\right)\\ \)
\(\text{b) }x^2+4x+3\\ =x^2+x+3x+3\\ =\left(x^2+x\right)+\left(3x+3\right)\\ =x\left(x+1\right)+3\left(x+1\right)\\ =\left(x+3\right)\left(x+1\right)\\ \)
\(\text{c) }4x^2+4x-3\\ 4x^2+6x-2x-3\\ =\left(4x^2+6x\right)-\left(2x+3\right)\\ =2x\left(2x+3\right)-\left(2x+3\right)\\ =\left(2x-1\right)\left(2x+3\right)\\ \)
\(\text{d) }x^2-x-12\\ =x^2-4x+3x-12\\ =\left(x^2-4x\right)+\left(3x-12\right)\\ =x\left(x-4\right)+3\left(x-4\right)\\ =\left(x+3\right)\left(x-4\right)\\ \)
\(\text{Chữa đề: }\\\text{e) }x^2-5x-6\\ =x^2-6x+x-6\\ =\left(x^2+x\right)-\left(6x+6\right)\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x-6\right)\left(x+1\right)\\ \)
\(\text{f) }x^2-5x-14\\ =x^2-7x+2x-14\\ =\left(x^2-7x\right)+\left(2x-14\right)\\ =x\left(x-7\right)+2\left(x-7\right)\\ =\left(x+2\right)\left(x-7\right)\\ \)
\(\text{g) }4x^2-3x-1\\ =4x^2-4x+x-1\\ =\left(4x^2-4x\right)+\left(x-1\right)\\=4x\left(x-1\right)+\left(x-1\right)\\ =\left(4x+1\right)\left(x-1\right)\\ \)
\(\text{h) }3x^2-2x-5\\ =3x^2-5x+3x-5\\ =\left(3x^2+3x\right)-\left(5x+5\right)\\ =3x\left(x+1\right)-5\left(x+1\right)\\ \left(3x-5\right)\left(x+1\right)\\ \)
\(\text{ i) }9x^2-6x+1\\ =\left(3x\right)^2-2\cdot3x\cdot1+1^2\\ =\left(3x-1\right)^2\)
Lần sau đăng từ từ thôi nhé.