Chứng minh rằng \(\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(b-c\right)\left(b-d\right)\left(c-d\right)⋮12\)
Chứng minh \(\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(b-c\right)\left(b-d\right)\left(c-d\right)⋮12\)
Với 4 số nguyên a, b, c, d bất kì, chứng minh rằng:
\(\left(a-b\right)\left(a-c\right)\left(a-d\right)\left(b-c\right)\left(b-d\right)\left(c-d\right)\) chia hết cho 12
a. Cho \(A\subset C\) và \(B\subset D\), chứng minh rằng \(\left(A\cup B\right)\subset\left(C\cup D\right)\)
b. Chứng minh rằng A\ \(\left(B\cap C\right)=\left(A\B\right)\cup\left(A\C\right)\)
c. Chứng minh rằng A\ \(\left(B\cup C\right)=\left(A\B\right)\cap\left(A\C\right)\)
Cho a,b,c,d dương thỏa mãn \(a^2+b^2+c^2+d^2=4.\)Chứng minh:
\(16\left(2-a\right)\left(2-b\right)\left(2-c\right)\left(2-d\right)\ge\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)\)
Chứng minh đẳng thức:
a) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)=\left(a-c\right)\left(d-b\right)\)
b) \(\left(a-c\right)\left(b+d\right)-\left(a-d\right)\left(b+c\right)=\left(a+b\right)\left(d-c\right)\)
a) Vế trái = a.(c + d) + b.( c+ d) - a.(b + c) - d.(b + c)
= a.[(c+ d) - (b + c)] + [b(c+d) - d.(b + c)]
= a.(d - b) + (bc + bd - db - dc) = a.(d - b) + c.(b - d) = a.(d - b) - c.(d - b) = (a - c).(d - b) = Vế phải
Vậy....
b) làm tương tự:
a) (a+b) (c+d) - (a+d) (b+c) = (ac + ad + bc + bd) - (ab + ac +bd + cd) = ac + ad + bc + bd - ab -ac - bd - cd
và bằng ad + bc - ab - cd = a( d-b ) + c( b-d ) = a (d-b) - c (d-b) = (a-c)(d-b) (dpcm)
p/s: ý B chứng minh tương tự.
Cho 5 số thực khác nhau a,b,c,d,x.Chứng minh :
\(\frac{b+c+d}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}+\frac{a+c+d}{\left(a-b\right)\left(c-b\right)\left(d-b\right)\left(x-b\right)}+\frac{a+b+d}{\left(a-c\right)\left(b-c\right)\left(d-c\right)\left(x-c\right)}+\)
\(\frac{a+b+c}{\left(a-d\right)\left(b-d\right)\left(c-d\right)\left(x-d\right)}=\frac{a+b+c+d-x}{\left(a-x\right)\left(b-x\right)\left(c-x\right)\left(d-x\right)}\)
Chứng minh với a; b; c; d > 0
\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\) \(\ge\) \(\left(a+b\right)\left(c+d\right)\)
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\ge\sqrt{\left(ac+bc\right)^2}=ac+bc\)
CMTT : \(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ad+bd\)
Ta có :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)
Áp dụng BĐT Bunhiacopxki:
CMTT :
Ta có :
1/ cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng:
a) \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\frac{a.d}{c.b}=\frac{\left(a+b\right).\left(a-b\right)}{\left(c+d\right).\left(c-d\right)}\)
2/ cho a.b=c2 chứng minh: \(\frac{a}{b}=\frac{\left(2.a+3.c\right)^2}{\left(2.c\right)+\left(3.b\right)^2}\)
Chứng minh rằng nếu a + b + c + d = 0 thì
a)\(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
b)\(\left(b+d\right)\left(ac-bd\right)=\left(b+c\right)\left(cd-bc\right)\)