(x2+7x+10).(x2+11x+28)=12
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NN
1
12 tháng 2 2023
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
=>1/x+2-1/x+6=1/8
=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>x^2+8x+12=32
=>x^2+8x-20=0
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
6 tháng 6 2017
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
DH
6 tháng 6 2017
a,\(x^3-7x+6\)
\(=x^3-2x^2+2x^2-4x-3x+6\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)
\(=x^2.\left(x-2\right)+2x.\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)
\(=\left(x-2\right).\left[x.\left(x-1\right)+3.\left(x-1\right)\right]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
b,\(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)
\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)
\(=\left(x-8\right).\left(x^2-x-2\right)\)
\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)
\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)
\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)
\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)
c,\(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2.\left(x-5\right)-x.\left(x-5\right)-6.\left(x-5\right)\)
\(=\left(x-5\right).\left(x^2-x-6\right)\)
\(=\left(x-5\right).\left(x^2+2x-3x-6\right)\)
\(=\left(x-5\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)
\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)
\(=\left(x-5\right).\left(x+2\right).\left(x-3\right)\)
Chúc bạn học tốt!!!
DH
6 tháng 6 2017
d,\(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2.\left(2x+1\right)-x.\left(2x+1\right)+3.\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x^2-x+3\right)\)
e, \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)
\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)
Chúc bạn học tốt!!!
HV
1
13 tháng 7 2023
a: =3x^2-3x-8x+8=(x-1)(3x-8)
b: =x^2-x-5x+5=(x-1)(x-5)
c: =x^2-6x+2x-12=(x-6)(x+2)
4 tháng 1 2020
7 x 2 + 28 + 14 x 2
= 14 + 28 + 14 x 2
= 42 + ( 14 x 2 )
= 42 + 28
= 70
CM
20 tháng 11 2017
a) x 2 – 7 x + 12 = 0
Có a = 1; b = -7; c = 12
⇒ Δ = b 2 – 4 a c = ( - 7 ) 2 – 4 . 1 . 12 = 1 > 0
⇒ Phương trình có hai nghiệm phân biệt x 1 ; x 2 thỏa mãn:
Vậy dễ dàng nhận thấy phương trình có hai nghiệm là 3 và 4.
b) x2 + 7x + 12 = 0
Có a = 1; b = 7; c = 12
⇒ Δ = b2 – 4ac = 72 – 4.1.12 = 1 > 0
⇒ Phương trình có hai nghiệm phân biệt x1; x2 thỏa mãn:
Vậy dễ dàng nhận thấy phương trình có hai nghiệm là -3 và -4.
D
1
18 tháng 1 2022
\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)
\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)
\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)
\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)x\left(x+1\right)=0.\Leftrightarrow\left(x-1\right)\left(x+1\right)^2x=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)