a=24 b=18c=5

ta có \(\frac{52}{a-20}=\frac{39}{b-15}=\frac{13}{c-5}\)

=> \(\frac{a-20}{52}=\frac{b-15}{39}=\frac{c-5}{13}\)

=> \(\frac{a}{52}-\frac{20}{52}=\frac{b}{39}-\frac{15}{39}=\frac{c}{13}-\frac{5}{13}\)

=>\(\frac{a}{52}-\frac{5}{13}=\frac{b}{39}-\frac{5}{13}=\frac{c}{13}-\frac{5}{13}\)

=> \(\frac{a}{52}=\frac{b}{39}=\frac{c}{13}\)

=>\(\frac{a^2}{52^2}=\frac{b^2}{39^2}=\frac{c^2}{13^2}=\frac{bc}{39.13}=\frac{3}{3.13.13}=\frac{1}{13^2}\)

=>\(\hept{\begin{cases}a^2=\frac{1}{13^2}.52^2=4^2\\b^2=\frac{1}{13^2}.39^2=3^2\\c^2=\frac{1}{13^2}.13^2=162\end{cases}}\)

=>\(\hept{\begin{cases}a=4;-4\\b=3;-3\\c=1;-1\end{cases}}\)

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Ta có : \(\frac{a-20}{52}=\frac{b-15}{39}=\frac{c-5}{13}\)

=> \(\frac{a}{52}-\frac{20}{52}=\frac{b}{39}-\frac{15}{39}=\frac{c}{13}-\frac{5}{13}\)

=> \(\frac{a}{52}-\frac{5}{13}=\frac{b}{39}-\frac{5}{13}=\frac{c}{13}-\frac{5}{13}\)

=> \(\frac{a}{52}=\frac{b}{39}=\frac{c}{13}\)

Đặt \(\frac{a}{52}=\frac{b}{39}=\frac{c}{13}=k\)

=> \(\hept{\begin{cases}a=52k\\b=39k\\c=13k\end{cases}}\)

=> b.c = 39k.13k

=> 507k² = 3

=> k = \(\pm\frac{1}{13}\)

=> \(\hept{\begin{cases}a=52\cdot\frac{1}{13}=4\\b=39\cdot\frac{1}{13}=3\\c=13\cdot\frac{1}{13}=1\end{cases}}\)hoặc \(\hept{\begin{cases}a=52\cdot\left(-\frac{1}{13}\right)=-4\\b=39\cdot\left(-\frac{1}{13}\right)=-3\\c=13\cdot\left(-\frac{1}{13}\right)=-1\end{cases}}\)

Tại hạ xin test cách khác :))

\(\frac{52}{a-20}=\frac{39}{b-15}=\frac{13}{c-5}\)

\(=>\frac{a-20}{52}=\frac{b-15}{39}=\frac{c-5}{13}\)

\(=>\frac{a}{52}-\frac{5}{13}=\frac{b}{39}-\frac{5}{13}=\frac{c}{13}-\frac{5}{13}\)

\(=>\frac{a}{52}=\frac{b}{39}=\frac{c}{13}\)

\(=>\frac{a^2}{52^2}=\frac{b^2}{39^2}=\frac{c^2}{13^2}=\frac{bc}{39.13}=\frac{3}{13^2.3}=\frac{1}{13^2}\)

\(=>\hept{\begin{cases}\frac{a^2}{52^2}=\frac{1}{13^2}=>a^2=\frac{1}{13^2}.52^2=4^2\\\frac{b^2}{39^2}=\frac{1}{13^2}=>b^2=\frac{1}{13^2}.39^2=3^2\\\frac{c^2}{13^2}=\frac{1}{13^2}=>c^2=\frac{1}{13^2}.13^2=1^2\end{cases}}\)

\(=>\hept{\begin{cases}a=\pm4\\b=\pm3\\c=\pm1\end{cases}}\)

1.

a) 5/8 x 4/10 + 2/3 =

= 1/4+ 2/3 = 11/12

b)5/12 x 4/7+5/12 x3/7

=5/12 x (4/7 +3/7)

=5/12 x1 = 5/12

c)(4/5 + 3/10 - 1/5 ) x 6 : 4/7

= ( 8/10 + 3/10 + 2/10) x 6 x 7/4

=13/10 x 21/2

=273/20

2.

5/8 và 3/2

ta có 5/8 =10/16    ;        3/2 =24 /16 

vì 24 /16 >10 /16 nên 3/2 > 5/8

b. tương tự như câu a nha

c 418/417 và 925 /926

418/417 > 1     ; 925 /926 < 1

vì 418 /417 >1 mà 925/926 < 1 nên 418 / 417 > 925 /926

chúc bạn học tốt nha !

mình làm sai chỗ nào à

GIẢ SỬ \(\frac{A}{B}=\frac{C}{D}\)

ĐẶT\(\frac{A}{B}=\frac{C}{D}=T\)=>A = BT , C = DT 

TA CÓ\(\frac{\left(A^2+B^2\right)}{\left(C^2+D^2\right)}=\frac{\left(\left(B\cdot T\right)^2+B^2\right)}{\left(\left(D\cdot T\right)^2+D^2\right)}=\frac{\left(B^2\cdot\left(T^2+1\right)\right)}{\left(D^2\cdot\left(T^2+1\right)\right)}=\frac{B^2}{D^2}=\left(\frac{B}{D}\right)^2\left(1\right)\)

LẠI CÓ\(\frac{\left(A\cdot B\right)}{\left(C\cdot D\right)}=\frac{\left(B\cdot T\cdot B\right)}{\left(D\cdot T\cdot D\right)}=\frac{B^2}{D^2}=\left(\frac{B}{D}\right)^2\left(2\right)\)

TỪ (1) VÀ (2) \(\Rightarrow\frac{\left(A^2+B^2\right)}{\left(C^2+D^2\right)}=\frac{\left(A\cdot B\right)}{\left(C\cdot D\right)}\)( THÕA ĐỀ )

=> ĐIỀU GIẢ SỬ ĐÚNG => DPCM

sao ban ko k cho minh

a) -2 /3 x + 1/5 = 3/10

 -2/3x =1/10 

 x = -3/20 

 vậy x = -3/20

b) 25/9 - 12/13x = 7/

12/13x = 2

x = 13/6

c) (x) - 3/4 =5/3 

(x) = 29/12

x = 29/12 ; -29/-12

 d)  x = 11/2