a^2-2a-24=0
Cm các đẳng thức sau
a) (a-1)*(a-2)+(a-3)*(a-4)-2a2+5a-24=-7a+24
b) (a+c)*(a-c)-b*(2a-b)-(a+b+c)*(a-b-c)=0
bai1:CMR
a, (a-1).(a-2)+(a-3).(a+4)-(2a^2+5a-34)=-7a+24
b,(a+c)(a-c)-b(2a-b)-(a-b+c).(a-b-c)=0
b. \(\left(a+c\right)\left(a-c\right)-b\left(2a-b\right)-\left(a-b+c\right)\left(a-b-c\right)\)
=\(\left(a^2-c^2\right)-2ab+b^2-\left(a-b\right)^2+c^2\)
=\(a^2-c^2-2ab+b^2-a^2+2ab-b^2+c^2\)
=0=VP=> đpcm
a. \(\left(a-1\right)\left(a-2\right)+\left(a-3\right)\left(a+4\right)-\left(2a^2+5a-34\right)\)
=\(\left(a^2-3a+2\right)+\left(a^2+a-12\right)-\left(2a^2+5a-34\right)\)
=\(a^2-3a+2+a^2+a-12-2a^2-5a+34\)
=-7a+24
=VP => đpcm
chứng minh rằng:
a) (a-1)(a-2) + (a-3)(a+4) - (2a^2 + 5a - 34)= -7a +24
b) (a+c)(a-c) - b(2a-b) - (a-b+c)(a-b-c) = 0
c) (a - b)(a^2 +ab+b^2) - (a+b)(a^2-ab+b^2) = - 2b^3
m.n giúp mk vs, mk đang rất gấp...tks trc nạ.!
Chứng minh:
a. (a - 1)(a - 2) + (a - 3)(a + 4) - 2(2a2 + 5a - 34) = -7a + 24
b. (a +c)(a - c) - b(2a - b) - (a - b + c)(a - b - c) = 0
c. (a -b)(a2 + ab + b2) - (a +b)(a2 - ab + b2) = -2b3
Giải các bất phương trình sau:
a) (a2 - 2)2 - 3(a 2- 2)(a2 - 2a + 2) - 10(a2 - 2a + 2)2=0
b)(a2 - a -6)2 + (a2 + 4)2 - 24(a2 + 14) + 144=0
c)(x2 - 4x + 36)2 - (x2 - 4x +36)(x2 + x - 3) = 12(x2 + x - 3)2
Rút gọn : \(\left(\frac{1+2a}{4+2a}-\frac{a}{3a-6}+\frac{2a^2}{12-3a^2}\right)\div\frac{13a+6}{24-12a}\)
\(=\left(\dfrac{2a+1}{2\left(a+2\right)}-\dfrac{a}{3\left(a-2\right)}-\dfrac{2a^2}{3\left(a-2\right)\left(a+2\right)}\right):\dfrac{13a+6}{24-12a}\)
\(=\dfrac{3\left(2a+1\right)\left(a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}:\dfrac{13a+6}{-12\left(a-2\right)}\)
\(=\dfrac{3\left(2a^2-3a-2\right)-2a\left(a+2\right)-4a^2}{6\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-12\left(a-2\right)}{13a+6}\)
\(=\dfrac{6a^2-9a-6-2a^2-4a-4a^2}{a+2}\cdot\dfrac{-2}{13a+6}\)
\(=\dfrac{-\left(13a+6\right)}{a+2}\cdot\dfrac{-2}{13a+6}=\dfrac{2}{a+2}\)
Chứng minh rằng:
1) (2n – 3)^2 – 9 chia hết cho 4 với mọi số nguyên n
2) a^4 - 2a^3 – a^2 + 2a chia hết cho 24 với a là số nguyên
\(1,\left(2n-3\right)^2-9=\left(2n-3-3\right)\left(2n-3+3\right)=\left(2n-6\right)2n=4n\left(n-3\right)⋮4\)
\(2,=a^3\left(a-2\right)-a\left(a-2\right)=\left(a-2\right)\left(a^3-a\right)=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\)
Vì đây là tích 4 số nguyên lt nên chia hết cho \(1\cdot2\cdot3\cdot4=24\)
4 giờ trước (10:24)cho a/b=c/d\nCMR: 2a-3c/2b-3d=2a+3c/2a+3d
(2a+1)+(2a+2)+....+ (2a+2015)=0
tim a