Cho a+b+c=1, \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) abc khác 0.Tính A = a2 + b2 + c2
Cho a+b+c=1, \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) abc khác 0.Tính A = a2 + b2 + c2
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}\)
\(\Rightarrow\frac{bc+ac+ab}{abc}=0\)
\(\Rightarrow bc+ca+ab=0\)
\(\Rightarrow2bc+2ac+2ab=0\)
Đặt \(B=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Rightarrow B=\left(a+b+c\right)^2=1^2=1\) ( áp dụng hằng đẳng thức )
\(\Rightarrow B=a^2+b^2+c^2+0=1\)
\(\Rightarrow A=a^2+b^2+c^2=1-0=1\)
Vậy \(A=1\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\frac{bc+ac+ab}{abc}=0\)
=> ac + ab + bc = 0
a2 + b2 + c2
= (a + b + c)2 - 2(ab + ac + bc)
= 12 - 2 . 0
= 1
(x^3+8y^3):(x+2y)=
\(\left(x^3+8y^3\right):\left(x+2y\right)\)
\(=\left[x^3+\left(2y\right)^3\right]:\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right):\left(x+2y\right)\)
\(=x^2-2xy+4y^2\)
chứng minh rằng với mọi số tự nhiên n ta đều có:
A=n4-14n3+71n2-154n+120 chia hết cho 24
B= (n^4 - 14n^3 + 49n^2) + 22n^2 -154n +120
= n^2(n^2 -14n +49) + 22n(n-7) +120
= (n(n-7))^2 +10n(n-7) + 12n(n-7) + 10*12
= n(n-7)[n(n-7) + 10] + 12[n(n-7) +10]
= [n(n-7) +10] * [n(n-7) + 12]
= (n^2 - 7n + 10)(n^2 - 7n +12)
= (n-2)(n-5)(n-3)(n-4)
= (n-5)(n-4)(n-3)(n-2)
B là tích của 4 số tự nhiên liên tiếp
=> B chia hết cho 2, 3, 4 mà 2, 3, 4 nguyên tố cùng nhau
Suy ra: B chia hết 2x3x4
Hay B chia hết cho 24.
Bn chịu khó đọc nha!
chứng minh rằng với mọi số tự nhiên n ta đều có: A= n4+6n3+11n2+6n chia hết cho 24
@Tuấn Anh Phan Nguyễn Copy không nhìn hả :vvv đề bài n4 + 6n3 + 11n2 + 6n biến thành n4 + 6n3 + 11n2 + 30n - 24 luôn kìa. Hơn nữa với pp quy nạp cần xét n = 1 :vvvv
Chứng minh :
a) ( n^3 - n ) chia hết cho 6 với mọi số nguyên n.
b) ( 55^n+1 - 55^n ) chia hết cho 54 với mọi số nguyên n.
a) n3 - n
= n.(n2 - 1)
= n.(n - 1).(n + 1)
Vì n.(n - 1).(n + 1) là tích 3 số nguyên liên tiếp
=> n.(n - 1).(n + 1) chia hết cho 2 và 3
Mà (2;3)=1 => n.(n - 1).(n + 1) chia hết cho 6
=> n3 - n chia hết cho 6 (đpcm)
b) 55n+1 - 55n
= 55n.55 - 55n
= 55n.(55 - 1)
= 55n.54 chia hết cho 54 (đpcm)
Hình thang có , . Khi đó ...độ
AB // CD
=> B + C = 1800
mà B - C = 300
=> B = (1800 + 300) : 2 = 1050
AB // CD
=> A + D = 1800 (2 góc trong cùng phía)
mà A = 3D => D = A/3
=> A + A/3 = 1800
4/3A = 1800
A = 1800 . 3/4
A = 1350
=> A + B = 1350 + 1050 = 2400
Nghiệm của đa thức là
f(x) có nghiệm khi 2x - 8 = 0
2x = 8
x = 8/2
x = 4
Vậy x = 4 là nghiệm của f(x) = 2x - 8
\(f\left(x\right)=2x-8\)
\(=>2x-8=0\)
\(=>2x=8\)
\(=>x=4\)
Tìm x :
a) 3x( x - 10 ) = x - 10
b) x( x + 7 ) - ( 4x + 28 ) = 0
c) x( x - 4 ) = 2x - 8
d) ( 2x + 3 )( x - 1 ) + ( 2x - 3 )( x - 1 ) = 0
3x(x - 10) = x - 10
(x - 10)(3x - 1) = 0
Th1:
x - 10 = 0
x = 10
TH2:
3x - 1 = 0
3x = 1
x = 1/3
Vậy x = 10 hoặc x = 1/3
x(x + 7) - (4x + 28) = 0
x(x + 7) - 4(x + 7) = 0
(x + 7)(x - 4) = 0
Th1:
x + 7 = 0
x = - 7
Th2:
x - 4 = 0
x = 4
Vậy x = - 7 hoặc x = 4
x(x - 4) = 2x - 8
x(x - 4) - 2(x - 4) = 0
(x - 2)(x - 4) = 0
Th1:
x - 2 = 0
x = 2
Th2:
x - 4 = 0
x = 4
Vậy x = 2 hoặc x = 4
(2x + 3)(x - 1) + (2x - 3)(x - 1) = 0
(x - 1)(2x + 3 + 2x - 3) = 0
4x(x - 1) = 0
Th1:
x = 0
Th2:
x - 1 = 0
x = 1
Vậy x = 0 hoặc x = 1
a)
\(3x\left(x-10\right)=x-10\)
\(\Rightarrow3x\left(x-10\right)-\left(x-10\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(x-10\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-1=0\\x-10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=10\end{array}\right.\)
b)
\(x\left(x+7\right)-\left(4x+28\right)=0\)
\(\Rightarrow x\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=-7\end{array}\right.\)
c)
\(x\left(x-4\right)=2x-8\)
\(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)
d)
\(\left(2x+3\right)\left(x-1\right)+\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow2\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x+3=0\\x-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=1\end{array}\right.\)
Tìm x :
a) x(x - 1 ) = 0
b) 3x^2 - 6x = 0
c) x( x - 6 ) + 10( x - 6 ) = 0
d) x^3 - x = 0
a) x(x - 1) = 0
=> \(\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
b) 3x2 - 6x = 0
=> 3x.(x - 2) = 0
=> x.(x - 2) = 0
=> \(\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
c) x(x - 6) + 10(x - 6) = 0
=> (x - 6)(x + 10) = 0
=> \(\left[\begin{array}{nghiempt}x-6=0\\x+10=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=6\\x=-10\end{array}\right.\)
d) x3 - x = 0
=> x.(x2 - 1) = 0
=> x.(x - 1).(x + 1) = 0
=> \(\left[\begin{array}{nghiempt}x=0\\x-1=0\\x+1=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=0\\x=1\\x=-1\end{array}\right.\)
a)
\(x\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
Vậy x=0 ; x =1
b)
\(3x^2-6x=0\)
\(\Rightarrow3x\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
Vậy x=0 ; x =2
c)
\(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-6=0\\x+10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=6\\x=-10\end{array}\right.\)
Vậy x=6 ; x = -10
d)
\(x^3-x=0\)
\(\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\\x+1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\\x=-1\end{array}\right.\)
Vậy x = 0 ; x = 1 ; x= - 1
a,\(x\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
b,\(3x^2-6x=0\)
\(\Rightarrow3x\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-2=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
c,\(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Rightarrow\left(x+10\right)\left(x-6\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-6=0\\x+10=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=6\\x=-10\end{array}\right.\)
d,\(x^3-x=0\)
\(\Rightarrow x^2\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x^2=0\\x-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
Phân tích đa thức thành nhân tử :
a) 4a^2b^2 + 36a^2b^3 + 6ab^4
b) 4a^2b^3 - 6a^3b^2
4a2b2 + 36a2b3 + 6ab4
= 2ab2(2a + 18ab + 3b2)
4a2b3 - 6a3b2
= 2a2b2(2b - 3a)