Ta có: a+b+c=0

nên a+b=-c

Ta có: \(a^2-b^2-c^2\)

\(=a^2-\left(b^2+c^2\right)\)

\(=a^2-\left[\left(b+c\right)^2-2bc\right]\)

\(=a^2-\left(b+c\right)^2+2bc\)

\(=\left(a-b-c\right)\left(a+b+c\right)+2bc\)

\(=2bc\)

Ta có: \(b^2-c^2-a^2\)

\(=b^2-\left(c^2+a^2\right)\)

\(=b^2-\left[\left(c+a\right)^2-2ca\right]\)

\(=b^2-\left(c+a\right)^2+2ca\)

\(=\left(b-c-a\right)\left(b+c+a\right)+2ca\)

\(=2ac\)

Ta có: \(c^2-a^2-b^2\)

\(=c^2-\left(a^2+b^2\right)\)

\(=c^2-\left[\left(a+b\right)^2-2ab\right]\)

\(=c^2-\left(a+b\right)^2+2ab\)

\(=\left(c-a-b\right)\left(c+a+b\right)+2ab\)

\(=2ab\)

Ta có: \(M=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)

\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

\(=\dfrac{a^3+b^3+c^3}{2abc}\)

Ta có: \(a^3+b^3+c^3\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-cb+c^2\right)-3ab\left(a+b\right)\)

\(=-3ab\left(a+b\right)\)

Thay \(a^3+b^3+c^3=-3ab\left(a+b\right)\) vào biểu thức \(=\dfrac{a^3+b^3+c^3}{2abc}\), ta được: 

\(M=\dfrac{-3ab\left(a+b\right)}{2abc}=\dfrac{-3\left(a+b\right)}{2c}\)

\(=\dfrac{-3\cdot\left(-c\right)}{2c}=\dfrac{3c}{2c}=\dfrac{3}{2}\)

Vậy: \(M=\dfrac{3}{2}\)

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\left(đpcm\right)\)

Thay b² = ac vào biểu thức trên, ta có:

\(\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)

\(\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)

Câu hỏi của Hattory Heiji - Toán lớp 8 - Học toán với OnlineMath

tvbobnokb' n

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  ni;bv nn0

Ta có: \(a^2+b^2=\left(a+b-c\right)^2\)

=>\(a^2+b^2=\left(a+b\right)^2-2c\left(a+b\right)+c^2\)

=>\(a^2+b^2+2ab-2ac-2bc+c^2=a^2+b^2\)

=>\(c^2=2ac+2bc-2ab\)

Đặt \(A = a^2 + (a - c)^2\)

\(=a^2+a^2-2ac+c^2=2a^2-2ac+c^2\)

=>\((b - c) \cdot A = (b - c)(2a^2 - 2ac + c^2)\)

\(= 2a^2b - 2a^2c - 2abc + 2ac^2 + bc^2 - c^3\)

\(= a(2ab - 2ac) - 2abc + 2ac^2 + bc^2 - c^3\)

=>\((b - c) \cdot A = a(2bc - c^2) - 2abc + 2ac^2 + bc^2 - c^3\)

\(= 2abc - ac^2 - 2abc + 2ac^2 + bc^2 - c^3\)

\(=ac^2+bc^2-c^3=c^2\left(a+b-c\right)\)

Đặt \(B = b^2 + (b - c)^2\)

\(=b^2+b^2-2bc+c^2=2b^2-2bc+c^2\)

=>\((a - c) \cdot B = (a - c)(2b^2 - 2bc + c^2)\)

\(= 2ab^2 - 2b^2c - 2abc + 2bc^2 + ac^2 - c^3\)

\(= b(2ab - 2bc) - 2abc + 2bc^2 + ac^2 - c^3\)

=>\((a - c) \cdot B = b(2ac - c^2) - 2abc + 2bc^2 + ac^2 - c^3\)

\(= 2abc - bc^2 - 2abc + 2bc^2 + ac^2 - c^3\)

\(=ac^2+bc^2-c^3=c^2\left(a+b-c\right)\)

Do đó: A(b-c)=B(a-c)

=>\(\frac{a-c}{b-c}=\frac{A}{B}=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)

\(a,a^2+b^2=\left(a+b\right)^2-2ab=9^2-2\cdot20=41\\ b,a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=41^2-2\left(ab\right)^2\\ =1681-2\cdot400=881\\ c,\left(a-b\right)^2=a^2+b^2-2ab=41-2\cdot20=1\\ \Rightarrow a-b=1\\ \Rightarrow C=a^2-b^2=\left(a-b\right)\left(a+b\right)=9\cdot1=9\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{8a}{8c}=\frac{9b}{9d}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{8a}{8c}=\frac{9b}{9d}=\frac{8a+9b}{8c+9d}=\frac{8a-9b}{8c-9d}\left(dpcm\right)\)

b) xem lại đề nha b

Do a+b+c= 0

<=> a+b= -c 

=> (a+b)²= c² 

Tương tự: (c+a)²= b², (c+b)²= a²   

Ta có: \(A=\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}\)

\(=\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{c^2+a^2-\left(c+a\right)^2}+\frac{1}{a^2+b^2-\left(a+b\right)^2}\)

\(=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}\)

\(=\frac{a+b+c}{-2abc}=0\)

a)10

b)13.33333(3)

c)36

d)-5

a)=10

b)=13,3 rút gọn số thập phân trong máy tính thì =13 nha

c)=36

d)=-5

help me

a, False

b, True

c, False

d, True