\(\frac{x-2}{7}+\frac{x-7}{2}=\frac{7}{x-2}-\frac{2}{x-7}\)ĐKXĐ: \(x\ne2\)và  \(x\ne7\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{14}+\frac{7\left(x-7\right)}{14}=\frac{\left(x-7\right)7}{\left(x-7\right)\left(x-2\right)}-\frac{\left(x-2\right)2}{\left(x-7\right)\left(x-2\right)}\)

\(\Leftrightarrow2\left(x-2\right)+7\left(x-7\right)=\left(x-7\right)7-\left(x-2\right)2\)

\(\Leftrightarrow2x-4+7x-49=7x-49-2x-4\)

\(\Leftrightarrow\left(2x-4+7x-49\right)-\left(7x-49-2x-4\right)=0\)

\(\Leftrightarrow2x-4+7x-49+7x+49+2x+4=0\)

\(\Leftrightarrow4x-49x-8-98=0\)

\(\Leftrightarrow x\left(4-49\right)-106=0\)

\(\Leftrightarrow x\left(-45\right)=106\)

\(\Leftrightarrow x=\frac{-106}{45}\)

nhân ra ko đc à em

nhân ra sau là tính đc mà

hok tốt

\(A=\left(x^n+1\right)\left(x^n-2\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2013\)

\(A=x^n\cdot x^n-x^n\cdot2+x^n-2-x^{n-3}\cdot x^{n+3}+x^{n-3}\cdot x^3+20\)

\(A=x^{2n}-2x^n+x^n-x^{n-3+n+3}+x^{n-3+3}+20-2\)

\(A=x^{2n}-2x^n+x^n-x^{2n}+x^n+18\)

\(A=\left(x^{2n}-x^{2n}\right)-\left(2x^n-x^n-x^n\right)+18\)

\(A=0-0+18\)

\(A=18\)

Với mọi x thì A luôn bằng 18

Vậy giá trị của A không phụ thuộc vào biến x ( đpcm )

b.2,4

Sử dụng tam giác đòng dạng

Chỉ cần k!

Nếu muốn giải đc..

Phải k cho mk,mk sẽ giải cho!

hay quá thế là đỡ cho mình

Tam giác ABC có DE//BC=>\(\frac{AB}{AD}=\frac{AC}{AE}=>\frac{AE}{AD}=\frac{AC}{AB}\left(1\right)\)

TA có AC^2=AB.AD=>\(\frac{AC}{AB}=\frac{AD}{AC}\)mà (1)=>\(\frac{AD}{AC}=\frac{AE}{AD}\)=> AC.AE=AD^2

Mặt khác CD^2=AC.AE

=>AD=CD

Đặt A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)

     B = \(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

\(\Rightarrow\)A . B  = 9

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

Nhân abc với A ta được:

 A_abc = \(\frac{abc\left(a-b\right)}{c}+\frac{abc\left(b-c\right)}{a}+\)\(\frac{abc\left(c-a\right)}{b}\)

A_abc =  ab.( a - b ) + bc.( b - c ) + ac.( c - a )

A_abc = ab.( a - b ) + bc.( a - c + b - a ) + ac.( a - c )

A_abc = ab.( a - b ) - bc.( a - b ) - bc.( c - a ) + ac.(c - a )

A_abc = b.( a - b ).( a - c ) - c.( a - b ).(c - a ) 

A_abc= ( a - b ).( a - c ).( b - c )

A =  \(\frac{\left(a-b\right).\left(a-c\right).\left(b-c\right)}{abc}\)

Xét a + b + c = 0 \(\Rightarrow\) a + b = - c ; c + a = -b ; b + c = -a

Nhân ( a - b ).( c - a ).( b - c ) với B ta được :

B( a - b).( c - a ).( b - c ) = \(\frac{c\left(a-b\right).\left(c-a\right).\left(b-c\right)}{a-b}\)+  \(\frac{a\left(a-b\right).\left(b-c\right).\left(c-a\right)}{b-c}\)+ \(\frac{b\left(a-b\right).\left(b-c\right).\left(c-a\right)}{c-a}\)

B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) + a.( b - c ).( c - a ) + b.( a - b ).( b - c)

B( a - b ).( c - a ) .( b - c ) = c.( c - a ).( b - c ) + ( a - b ).( -b - c ).( c - a ) + b.( a - b ).( b - c )

B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) - b.( a - b ).( c- a ) + b.( a - b ).(b - c ) - c.( a - b ).( c - a )

B( a - b ).( c - a ).( b - c ) = c.( c - a ).( -a + 2b - c ) + b.( a - 2c +b).(a - b )

B( a - b).( c - a ).( b - c ) = -3bc.( b + c - 2a )

B( a - b ).( c - a ).( b - c ) = -9abc

B = \(\frac{9abc}{\left(a-b\right).\left(c-a\right).\left(b-c\right)}\)

NHÂN A VỚI B :

\(\frac{\left(a-b\right).\left(b-c\right).\left(a-c\right)}{abc}\)\(.\)\(\frac{9abc}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)= 9

\(\Rightarrow\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right).\)\(\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)

MÌNH CŨNG KHÔNG CHẮC LẮM !