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Đặt \(\frac{x}{a+2b+c}\)=\(\frac{y}{2a+b-c}\)=\(\frac{z}{4a-4b+c}\)=k
=>x=ak+2bk+ck; y=2ak+bk-ck; z=4ak-4bk+ck
=> \(\frac{a}{x+2y+c}\)=\(\frac{a}{ak+2bk+ck+4bk+2bk-2ck+4ak-4bk+ck}\)=\(\frac{a}{9ak}\)=\(\frac{1}{9k}\)
Tương tự => \(\frac{a}{x+2y+c}\)=\(\frac{b}{2x+y-z}\)=\(\frac{c}{4x-4y+z}\)=\(\frac{1}{9k}\)
Đặt \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=A\)
Áp dụng TC DTSBN ta có :
\(A=\frac{x+2y+z}{a+2b+c+2\left(2a+b-c\right)+4a-4b+c}=\frac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}\)
\(=\frac{x+2y+z}{9a}=\frac{1}{9}.\frac{x+2y+z}{a}\) (1)
\(A=\frac{2x+y+z}{2\left(a+2b+c\right)+2a+b-c+4a-4b+c}=\frac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b-c}\)
\(=\frac{2x+y-z}{9b}=\frac{1}{9}.\frac{2x+y-z}{b}\) (2)
\(A=\frac{4x-4y+z}{4\left(a+2b+c\right)-4\left(2a+b-c\right)+4a-4b+c}=\frac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}\)
\(=\frac{4x-4y+z}{9c}=\frac{1}{9}.\frac{4x-4y+z}{c}\)(3)
Từ (1);(2);(3) \(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y+z}=\frac{c}{4x-4y+z}\) (đpcm)
Ta có: \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x}{2a+4b+2c}=\dfrac{2y}{4a+4b-2c}=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+y+z}{\left(a+2b+c\right)+\left(2a+b-c\right)+\left(4a-4b+c\right)}=\dfrac{x+2y+z}{9b}\left(1\right)\)
\(\dfrac{2x}{2a+2b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-z}{\left(2a+2b+2c\right)+\left(2a+b-c\right)-\left(4a-4b+c\right)}=\dfrac{2x+y-z}{9a}\left(2\right)\)
\(\dfrac{4x}{4a+4b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{\left(4a+8b+4c\right)-\left(8a+4b-4c\right)+\left(4a-4b+c\right)}=\dfrac{4x-4y+z}{9c}\left(3\right)\)
Từ (1), (2), (3) \(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y+z}{9b}=\dfrac{4x-4y+z}{9b}\)
\(\Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\left(đpcm\right)\)
Chúc bạn học tốt!
#)Giải :
b)Ta có :
\(\left(-2a^2b^3\right)^{10}+\left(3b^2c^4\right)^{15}=0\)
\(\Leftrightarrow2^{10}.a^{20}.b^{30}+3^{15}.b^{30}.c^{60}=0\)
\(\Leftrightarrow\hept{\begin{cases}a^{20}.b^{30}=0\\b^{30}.c^{60}=0\end{cases}\Leftrightarrow\hept{\begin{cases}a.b=0\\b.c=0\end{cases}}\Leftrightarrow b=0;a,b\in Z}\)
Cmr
(-2a2 -b3) n *0n . 3b*n
=2a -3b15 =0*
=15*0=b *ab x bn
=0
suy ra ta có:
ab +14-0
15-0--2ab -n
kq:
abn=15
=-0
hk tốt