Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
Rình mãi ms được 1 câu!
Bài 3:
\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(A=\left[\left(x+1\right).\left(x+7\right)\right].\left[\left(x+3\right).\left(x+5\right)\right]+15\)
\(A=\left(x^2+7x+x+7\right).\left(x^2+5x+3x+15\right)+15\)
\(A=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)
Đặt \(t=x^2+8x+7\Rightarrow t+8=x^2+8x+15\)
\(\Rightarrow A=t.\left(t+8\right)+15\)
\(A=t^2+8t+15=t^2+3t+5t+15\)
\(A=\left(t^2+3t\right)+\left(5t+15\right)=t.\left(t+3\right)+5.\left(t+3\right)\)
\(A=\left(t+3\right).\left(t+5\right)\)
Vì \(t=x^2+8x+7\) nên
\(A=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)
\(A=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)
\(A=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)
\(A=\left(x^2+8x+10\right).\left[\left(x^2+2x\right)+\left(6x+12\right)\right]\)
\(A=\left(x^2+8x+10\right).\left[x.\left(x+2\right)+6.\left(x+2\right)\right]\)
\(A=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)
Chúc bạn học tốt!!!
a. \(=x^3+2^3+1^3-x^3\)
\(=\left(x^3-x^3\right)+8+1\)
\(=0+8+1\)
\(=9\)
Bài 1 :
a) ( x + 2 )( x2 - 2x + 4 ) + (1 - x)(1+x+ + x2 )
= ( x3 - 8 ) + ( 1 - x3 )
= x3 - 8 + 1 - x3
= 7
b) 7x( 4x - 2) - ( x - 3)( x+1 ) + 16x
= 28x2 - 14x - x2 - x + 3x + 3 + 16x
= 27x2 + 3
Bài a) nhóm thành 2 nhóm; nhóm thứ nhất gồm số hạng đầu và cuối
bài b) dùng hằng đẳng thức là đc rồi
\(a,\)\(x^3-13x-12\)
\(=x^3-x-12x-12\)
\(=x\left(x^2-1\right)-12\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
a) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen
c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha
Giải ra dài lắm nên cho đáp án nè
a/ B = (z - x - y)(z - x + y)(z + x - y)(z + x + y)
b/ Nó là 3 cạnh tam giác nên
(z - x - y ) < 0
(z - x + y) > 0
(z + x - y) > 0
(z + x + y) > 0
Nên B < 0
Bài 1:
a, \(x^2-x-12\)
\(=x^2-4x+3x-12=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x.\left(x-4\right)+3.\left(x-4\right)=\left(x-4\right).\left(x+3\right)\)
b, \(x^2+8x+15\)
\(=x^2+3x+5x+15=\left(x^2+3x\right)+\left(5x+15\right)\)
\(=x.\left(x+3\right)+5.\left(x+3\right)=\left(x+3\right).\left(x+5\right)\)
c, \(x^{16}+x^8-2\)
\(=x^{16}-x^8+2x^8-2=\left(x^{16}-x^8\right)+\left(2x^8-2\right)\)
\(=x^8.\left(x^8-1\right)+2.\left(x^8-1\right)=\left(x^8-1\right)\left(x^8+2\right)\)
d, \(x^2+7x+12\)
\(=x^2+3x+4x+12=\left(x^2+3x\right)+\left(4x+12\right)\)
\(=x.\left(x+3\right)+4.\left(x+3\right)=\left(x+3\right).\left(x+4\right)\)
Chúc bạn học tốt!!!
1,2,4 sử dụng Casio