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Giải:
\(a+b+c+d=0\)
\(\Leftrightarrow a+c=-b-d\)
\(\Leftrightarrow a+c=-\left(b+d\right)\)
Ta có:
\(\left(a+c\right)^3=-\left(b+d\right)^3\)
\(\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-\left(b^3+3b^2d+3bd^2+d^3\right)\)
\(\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-3b^2d-3bd^2-d^3\)
\(\Leftrightarrow a^3+3ac\left(a+c\right)+c^3=-b^3-3cd\left(b+d\right)-d^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bd\left(b+d\right)-3ac\left(a+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bd\left(b+d\right)+3ac\left(b+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
Vậy ...
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\left(\frac{c}{d}\right)\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau có :
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Mà \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}\)
\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
Vậy ...
Câu hỏi của ✰✰ βєsէ ℱƐƝƝIƘ ✰✰ - Toán lớp 8 - Học toán với OnlineMath
a) Ta có: (a + b + c + d)(a - b - c +d )=( (a + d) + (b + c) )( (a + d) - (b + c) )
=(a + d )2 - (b +c )2 (1)
(a - b + c - d)(a + b - c - d)=(a - d)2 - (b - c)2 (2)
Từ (1) và (2) => a2 + 2ad + d2 - b2 - 2bc - c2=a2 - 2ad + d2 - b2 + 2bc - c2
4ad=4bc => ad=bc <=> \(\frac{a}{c}=\frac{b}{d}\) (đpcm)
cau 1 ne:
a^2 + b^2 + c^2 + 3
theo bat dang thuc cosi ban se co
a^2 + a + 1 >= 3a
b^2 + b + 1 >= 3b
c^2 + c + 1 >= 3c
cong 3 ve bat dang thuc lai voi nhau ban se co
a^2 + b^2 + c^2 + (a + b + c) + 3>= 3(a + b + c)
=> a^2 + b^2 + c^2 + 3 >= 2(a + b + c)
dau = xay ra <=> a= b= c = 1
ma theo de bai ta lai co a^2 + b^2 + c^2 + 3 = 2(a + b + c)
=> a = b = c = 1 (dpcm)
b) (a - b)^2 + (b-c)^2 + (c - a)^2 = (a + b - 2c)^2 + (b + c - 2a)^2 + (c + a - 2b)^2
hay (a + b - 2b)^2 + (b + c - 2c)^2 + (c + a - 2a)^2 = (a + b - 2c)^2 + (b + c - 2a)^2 + (c + a - 2b)^2
dat. a + b = A
b + c = B
c + a = C
=> ban se co:
(A - 2b)^2 + (B - 2c)^2 + (C - 2a)^2 = (A - 2c)^2 + (B - 2a)^2 + (C - 2b)^2
tu day ban nhan pha ra roi rut gon 2 ve cho nhau ban se co
Ab + Bc + Ca = Ac + Ba + Cb
hay (a + b)b + (b + c)c + (c + a)a = (a + b)c + (b + c)a + (c + a)b
hay ab + b^2 + bc + c^2 + ac + a^2 = 2ab + 2bc + 2ac
hay a^2 + b^2 + c^2 - ab - bc - ac = 0
hay 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0
hay (a-b)^2 + (b-c)^2 +(c - a)^2 = 0
dau = xay ra <=> a = b = c (dpcm)
c) a^3 + b^3 + c^3 + d^3 = (a + b)(a^2 -ab +b^2) + (c+d)(c^2 - cd + d^2) (**)
ban nhan thay a + b + c + d = 0
=> a + b = - c - d
thay vao pt (**) ban se co
-(c + d)(a^2 - ab + b^2) + (c + d)(c^2 - cd + d^2)
(c + d)(c^2 - cd + d^2 -a^2 + ab - b^2)
hay (c + d)(ab - cd + (c^2 + d^2 - a^2 - b^2)) (***)
ban co a + b = - c - d
hay (a + b)^2 = (c + d)^2
hay a^2 + b^2 + 2ab = c^2 + d^2 + 2cd
hay c^2 + d^2 - a^2 - b^2 = 2ab - 2cd
thay vao pt (***) ban se co
(c + d)(ab - cd + 2ab - 2cd)
hay (c +d)(3ab - 3cd) = 3(c+d)(ab - cd) (dpcm)
b1: ta có: a^2+b^2 >0 ; b^2 +c^2>0 ; c^2 +a^2>0
=> \(a^2+b^2\ge2\sqrt{a^2.b^2}\) (BĐT cau chy)
\(b^2+c^2\ge2\sqrt{b^2.c^2}\) (BĐT cau chy)
\(c^2+a^2\ge2\sqrt{c^2.a^2}\)(BĐT cauchy)
=>\(\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\ge8a^2.b^2.c^2\)
Dấu '= xảy ra khi a=b=c (đpcm)
Bài 1:
\(P=\left(5x-1\right)^2+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(1-5x+5x+4\right)^2\)
\(=5^2=25\)
Bài 2:
a: \(\left(a+b+c\right)^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)\cdot c^2+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3a^2c+6abc+3b^2c+3ac+3bc+c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+2abcd+a^2d^2-2abcd+b^2c^2\)
\(=\left(a^2c^2+b^2c^2\right)+\left(b^2d^2+a^2d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
Lời giải:
Sử dụng điều kiện $a+b+c+d\Rightarrow a+c=-b-d$. Khi đó ta có:
\(a^3+b^3+c^3+d^3=(a^3+c^3)+(b^3+d^3)\)
\(=(a+c)^3-3ac(a+c)+(b+d)^3-3bd(b+d)\)
\(=(-b-d)^3-3ac(a+c)+(b+d)^3-3bd(b+d)\)
\(=-3ac(a+c)-3bd(b+d)=3ac(b+d)-3bd(b+d)\)
\(=3(b+d)(ac-bd)\)
Ta có đpcm.