Cho \(a^2+b^2+c^2=a^3+b^3+c^3=1\) Tính \(S=a^2+b^9+c^{2016}\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Bài 1: Cho a,b,c thỏa mãn (a+b-c)/c=(b+c-a)/a=(c+a-b)/b
tính P=(1+b/a)*(1+c/b)*(1+a/c)
Bài 2: Cho a+b+c=0
tính B=((a^2+b^2-c^2)*(b^2+c^2-a^2)*(c^2+a^2-b^2))/(10*a^2*b^2*c^2)
Bài 3: cho a^3*b^3+b^3*c^3+c^3*a^3=3*a^3*b^3*c^3
tính M(1+a/b)*(1+b/c)*(1+c/a)
Bài 4: cho 3 số a,b,c TM a*b*c=2016
tính P=2016*a/(a*b+2016*a+2016) + b/(b*c+b+2016) + c/(a*c+c+1)
Bài 5: cho a+b+c=0
tính Q=1/(a^2+b^2-c^2) + 1/(b^2+c^2-a^2) + 1/(a^2+c^2-b^2)
Cho \(a^2+b^2+c^2=a^3+b^3+c^3=1\). Tính \(S=a^{2016}+b^{2017}+c^{2018}\)
từ giả thiết => a;b;c<=1
\(a\le1\\ \Rightarrow a^3\le a^2\)
tt b^3<=b^2;c^3<=c^2
=>a^3+b^3+c^3\(\le\)a^2+b^2+c^2
dấu = xảy ra <=> a=0hoặc a=1 tt với b;c và a^2+b^2+c^2=a^3+b^3+c^3=1
=>S=1
a2 + b2 + c2 = a3 + b3 + c3 = 1
\(\Rightarrow\)a2 ( a - 1 ) + b2 ( b - 1 ) + c2 ( c - 1 ) = 0 ( 1 )
a2 + b2 + c2 = 1 ; a2,b2,c2 \(\ge\)0 \(\Rightarrow\)a2,b2,c2 \(\le\)1
\(\Rightarrow\)a \(\le\)1,b \(\le\)1, c \(\le\)1 \(\Rightarrow\)1 - a \(\ge\)0 ; 1-b \(\ge\)0 ; 1 - c \(\ge\)0
\(\Rightarrow\)a2 ( a - 1 ) + b2 ( b - 1 ) + c2 ( c - 1 ) \(\le\)0 ( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)a2 ( a - 1 ) = b2 ( b - 1 ) = c2 ( c - 1 ) = 0
\(\Rightarrow\)a = b = 0 ; c = 1 hoặc b = c = 0 ; a = 1 hoặc a = c = 0 ; b = 1
\(\Rightarrow\)S = 1
a^2+b^2+c^2=a^3+b^3+c^3=1. tinh S=a^2+b^2016+c^2017
cho a^3+b^3=c(3ab-c^2) và a+b+c=3 tính giá trị của: K=675(a^2016+b^2016+c^2016)+1
a)Tính A = (a/b+c)+(b/a+c)+(c/a+b) Biết a+b+c =2016 và (1/a+b)+(1/b+c)+(1/a+c)=1/2016
b)Tìm n thuộc Z sao cho 2n-3 chia hết cho a+1
c)tìm số có 3 chữ số biết rằng số đó chia hết cho 9 và các chữ số tỉ lệ với 2:3:4
d) tìm x,y,z biết rằng 2x=3y;5y=3z và 4x2 + 2y2 - z2=7
giải gấp nha các bạn
? nghĩa là sao
giúp mk vs các bạn ơi, mk cần gấp lắm
cho a2+b2+c2=a3+b3+c3=1. tính S=a2+b2016+c2017
Cho a^2 + b^2 + c^+ =1 và a^3 + b^3 + c^3 =1. Tính S = a^2 + b^9 + c^1945. .Giup mình nhanh nhé 😢😢😢😢😢
Vì \(a^2+b^2+c^2=1\)
\(\Rightarrow-1\le a,b,c\le1\)
\(\Rightarrow a-1\le0;b-1\le0;c-1\le0\)
Lây cai xau trừ cai trươc được
\(\left(a^3+b^3+c^3\right)-\left(a^2+b^2+c^2\right)=0\)
\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)
Ta co \(VT\le0\)
Dâu = xảy ra khi: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)
\(\Rightarrow S=1\)
Cho a + b + c = 1; a2 + b2 + c2 = 1; a3 + b3 + c3 = 1. Tính giá trị của biểu thức:
D = a2016 + b2016 + c2016
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Ta có
(m+n+p)^q >= m^q+n^q+p^q
=>a+b+c=1
=>(a+b+c)^2016=1 >= a2016 + b2016 + c2016
Mà a2016 + b2016 + c2016 >=0
=> a2016 + b2016 + c2016=1
Cho a,b,c>0 sao cho ab+bc+ac=3. CMR
1/(a^2+b^2+2016)+1/(b^2+c^2+2016)+1/(c^2+a^2+2016) >= 3/2018