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áp dụng t/c dãy tỉ số = nhau ta đc
\(+)\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\)(do a+b+c=1)
=> \(x+y+z=\frac{x}{a}\Leftrightarrow\left(x+y+z\right)^2=\frac{x^2}{a^2}\left(1\right)\)
+) \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=>\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)(do a^2 +b^2 +c^2 =1)
\(\Leftrightarrow x^2+y^2+z^2=\frac{x^2}{a^2}\left(2\right)\)
từ (1) zà (2)
=>\(\left(x+y+z\right)^2=x^2+y^2+z^2\left(dpcm\right)\)
Có \(a+b+c=a^2+b^2+c^2=1\) và \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(a;b;c\ne0\right)\left(1\right)\)
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\left(\frac{x}{a}\right)^2=\left(\frac{y}{b}\right)^2=\left(\frac{z}{c}\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}\left(2\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=\frac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}\). Theo \(\left(1\right)\)
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\). Theo \(\left(2\right)\)
Có \(a+b+c=a^2+b^2+c^2=1\Leftrightarrow\left(a+b+c\right)^2=1^2=1\).
Từ các đẳng thức trên, ta suy ra : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=\frac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(=\frac{x+y+z}{1}=\frac{\left(x+y+z\right)^2}{1}=\frac{x^2+y^2+z^2}{1}\Leftrightarrow1\left(x+y+z\right)^2=1\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\Leftrightarrowđpcm\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(2\right)\)
Mà \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow2ab=c\left(a+b\right)\left(1\right)\)
Nhận thấy ( 1 )=( 2 ) => đpcm
\(\frac{1}{c}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b})\)
\(\Rightarrow\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{a}{ab}+\frac{b}{ab}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=(a+b)\cdot c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b(a-c)=a(c-b)\)
\(\frac{a}{c}=\frac{a-c}{c-b}(đpcm)\)
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}.\left(\frac{a+b}{ab}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow ac+cb=2ab\Rightarrow ac-ab=-cb+ba\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
bn ghi sai đề kìa :v
\(ac=bb=>\frac{a}{b}=\frac{b}{c}=\frac{2012b}{2012c}\)
áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{2012b}{2012c}=\frac{a+2012b}{b+2012c}\)
\(=>\left(\frac{a}{b}\right)^2=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)
vì \(\frac{a}{b}=\frac{b}{c}=>\left(\frac{a}{b}\right)^2=\frac{a.b}{b.c}=\frac{a}{c}\)
\(=>\frac{a}{c}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\left(dpcm\right)\)
1.
Ta có : \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
\(\Rightarrow\frac{a.\left(2bz-3cy\right)}{a^2}=\frac{2b.\left(3cx-az\right)}{4b^2}=\frac{3c.\left(ay-2bx\right)}{9c^2}\)
\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)
Áp dụng tính chất của dãy tỉ số bằng hau ta có :
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)
\(=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz=3cy\\3cx=az\\ay=2bx\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{x}{3c}\left(đpcm\right)\)
Chúc bạn học tốt !!!
1. Sửa lại dòng cuối
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
1) \(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\)
-->\(\frac{a}{b}=\frac{a-c}{b-d}\left(đpcm\right)\)
2) ta có \(\frac{a}{b}=\frac{c}{d}\)
đặt a=kb và c=kd
\(\frac{a+b}{a-b}=\frac{kb+b}{kb-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)
\(\frac{c+d}{c-d}=\frac{kd+d}{kd-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)
từ (1) và (2) --> \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
Xét \(a+b+c=0\) thì \(\hept{\begin{cases}a+2b=c\\b+2c=a\\c+2a=b\end{cases}}\)\(\Rightarrow P=\frac{\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)}{abc}=1\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(a+b+c=\frac{a+2b-c}{c}=\frac{b+2c-a}{a}+\frac{c+2a-b}{b}=\frac{a+2b-c+b+2c-a+c+2a-b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+2b=3c\\b+2c=3a\\c+2a=3b\end{cases}}\)\(\Rightarrow P=\frac{3a.3b.3c}{abc}=27\)
Có a+2b-c/c=b+2c-a/a=c+2a-b/b
suy ra a+2b-c/c=b+2c-a/a=c+2a-b/b=a+2b-c+b+2c-a+c+2a-b/a+b+c=2a+2b+2c/a+b+c=2
suy ra a+2b-c=2c suy ra a+2b=3c
b+2c-a=2a suy ra b+2c=3a
c+2a-b=2b suy ra c+2a=3b
Có P=(2+a/b)(2+b/c)(2+c/a)=(2b+a/b)(2c+b/c)(2a+c/a)=(3c/b)(3a/c)(3b/a)=27abc/abc=27