(4a+2).(-3a+1)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
OY
19 tháng 8 2021
(3a+1).(3a+2)
Ta có: nếu a là số lẻ thì 3a+1 là số chẵn
⇒(3a+1).(3a+2)⋮2 (thỏa mãn)
Ta có: nếu a là số chẵn thì 3a+2 là số chẵn
⇒(3a+1).(3a+2)⋮2 (thỏa mãn)
Vậy với mọi a thì (3a+1).(3a+2)⋮2
17 tháng 4 2021
Ta có:
\(VT=\left[\dfrac{16a-a^2-\left(3+2a\right)\left(a+2\right)-\left(2-3a\right)\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}\right]:\dfrac{a-1}{a^3+4a^2+4a}\)
\(=\dfrac{16a-a^2-3a-6-2a^2-4a-2a+4+3a^2-6a}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}\)
\(=\dfrac{a-2}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}=\dfrac{a\left(a+2\right)}{a-1}\left(a\ne\pm2;a\ne1\right)\)
\(=a-\dfrac{a\left(a+2\right)}{a-1}=\dfrac{a^2-a-a^2-2a}{-1}=\dfrac{-3a}{a-1}=\dfrac{3a}{1-a}=VP\left(đpcm\right)\)
18 tháng 7 2021
a) = 9x2 - ( y2 - 10y + 25y2 ) = ( 3x )2 - ( y - 5 )2 = ( 3x - y + 5 )( 3x + y - 5 )
b) = ( x3 - 8 ) - ( x2 - 4x + 4 ) = ( x - 2 )( x2 + 2x + 4 ) - ( x - 2 )2 = ( x - 2 )( x2 + x + 6 )
c) = ( 4a2 - 4a + 1 ) - ( b2 - 2bc + c2 ) = ( 2a - 1 )2 - ( b - c )2 = ( 2a - b + c - 1 )( 2a + b - c - 1 )
d) = ( a3 + 3a2 + 3a + 1 ) - 27b3 = ( a + 1 )3 - ( 3b )3 = ( a - 3b + 1 )( a2 + 9b2 + 3ab + 3b )
NM
Nguyễn Minh Quang
Giáo viên
18 tháng 7 2021
a. \(9x^2-\left(y^2-10y+25\right)=9x^2-\left(y-5\right)^2=\left(3x-y+5\right)\left(3x+y-5\right)\)
b.\(x^3-8-x^2+4x-4=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)^2=\left(x-2\right)\left(x^2+x+6\right)\)
c.\(\left(4a^2-4a+1\right)-\left(b^2-2bc+c^2\right)=\left(2a-1\right)^2-\left(b-c\right)^2=\left(2a-1+b-c\right)\left(2a-1-b+c\right)\)
d.\(\left(a^3+3a^2+3a+1\right)-27b^3=\left(a+1\right)^3-\left(3b\right)^3=\left(a+1-3b\right)\left[\left(a+1\right)^2+3b\left(a+1\right)+9b^2\right]\)
TA
1
18 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}3a>=0\\a+1>=0\end{matrix}\right.\)
=>a>=0
\(\sqrt{a+1}-4a^2=\sqrt{3a}-1\)
=>\(\sqrt{a+1}-\sqrt{3a}=4a^2-1\)
=>\(\dfrac{a+1-3a}{\sqrt{a+1}+\sqrt{3a}}=\left(2a-1\right)\left(2a+1\right)\)
\(\Leftrightarrow\dfrac{-2a+1}{\sqrt{a+1}+\sqrt{3a}}-\left(2a-1\right)\left(2a+1\right)=0\)
=>\(\dfrac{2a-1}{\sqrt{a+1}+\sqrt{3a}}+\left(2a-1\right)\left(2a+1\right)=0\)
=>\(\left(2a-1\right)\cdot\left(\dfrac{1}{\sqrt{a+1}+\sqrt{3a}}+2a+1\right)=0\)
=>2a-1=0
=>a=1/2(nhận)
HH
0
TN
0
TD
1
11 tháng 11 2023
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
\(\left(4a+2\right)\left(-3a+1\right)\)
\(=-12a^2+4a-6a-2\)
\(=-12a^2-2x-2\)
Sửa dòng cuối giúp mình thành:
\(=-12a^2-2a+2\)