ta có: \(\frac{a^2+c^2}{b^2+a^2}\)do \(a^2=bc\)

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

vậy \(\frac{a^2+c^2}{b^2+a^2}=\frac{c}{b}\)

\(\text{Ta có : }\frac{a^2+c^2}{b^2+a^2}\text{ do }a^2=bc\)

\(\Rightarrow\frac{a^2+c^2}{b^2+a^2}=\frac{b.c+c.c}{b.b+b.c}=\frac{c.\left(b+c\right)}{b.\left(b+c\right)}=\frac{c}{b}\)

\(\text{Vậy }\frac{a^2+c^2}{b^2+a^2}=\frac{c}{b}\)

b, Ta có \(m=a+b+c\)

          \(\Rightarrow am+bc=a\left(a+b+c\right)+bc=a\left(a+b\right)+ac+bc=\left(a+c\right)\left(a+b\right)\)

CMTT \(bm+ac=\left(b+c\right)\left(b+a\right)\);\(cm+ab=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

\(4,VT=-a+b+c-a+b-c+a-b-c=-a+b-c=-\left(a-b+c\right)=VP\\ 5,M=-a+b-b-c+a+c-a=-a\\ M>0\Rightarrow-a>0\Rightarrow a< 0\)

Lời giải:

a.

$(a-b)-(c-d)+(b+c)=a-b-c+d+b+c=(a+d)+(-b+b)+(-c+c)$

$=a+d+0+0=a+d$

b.

$(a+b-c)-(a-b+c)=a+(-b-a+c)$

$a+b-c-a+b-c=a-b-a+c$

$(a-a)+(b+b)-(c+c)=(a-a)-b+c$

$2b-2c=-b+c$

$2b+b=2c+c$

$3b=3c$

$b=c$ (đpcm)

chuyển vế đổi dấu

hk lớp 9 rùi mà ko biết

Ta có: (a-b-c)+(-a+b-c)=-(a-b+c)

          a-b-c-a+b-c=-(a-b+c)

          -2c=-a+b-c

          -2c-(-a+b-c)=0

          -2c+a-b+c=0

           a-b-c=0

           a-(b+c)=0

           a=b+c

Theo định lí cos ta có: \({a^2} = {b^2} + {c^2} - 2bc\;\cos A\)

\( \Rightarrow {b^2} + {c^2} - {a^2} = 2bc\;\cos A\)(1)

a) Nếu góc A nhọn thì \(\cos A > 0\)

Từ (1), suy ra \({b^2} + {c^2} > {a^2}\)

b) Nếu góc A tù thì \(\cos A < 0\)

Từ (1), suy ra \({b^2} + {c^2} < {a^2}\)

c) Nếu góc A vuông thì \(\cos A = 0\)

Từ (1), suy ra \({b^2} + {c^2} = {a^2}\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

\(\frac{a}{b}=1\Rightarrow a=b\)

\(\frac{b}{c}=1\Rightarrow b=c\)

\(\frac{c}{a}=1\Rightarrow c=a\)

Vay : \(\Rightarrow a=b=c\)