Bài 1: 

a: Ta có: 5x=4y+2x

\(\Leftrightarrow3x=4y\)

hay \(\dfrac{x}{4}=\dfrac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{-56}{7}=-8\)

Do đó: x=-32; y=-24

a) đặt \(\frac{a}{b}=\frac{c}{d}=k\) => \(a=bk;c=dk\) thay vào hai vế

VT=\(\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\) (1)

thay c=dk vào VP

\(VP=\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)

từ(1)(2)=> VT=VP(dpcm)

b) làm tương tự thay a=bk

\(VT=\frac{7\left(bk\right)^2+3\left(bk\right)b}{11\left(bk\right)^2-8b^2}=\frac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\) (3)

thay c=dk vào VP

\(VP=\frac{7\left(dk\right)^2+3\left(dk\right)d}{11\left(dk\right)^2-8d^2}=\frac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\) (4)

từ (3)(4)=> VT=VP

bài 2:

\(\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}\)

=> \(\frac{x}{8}=\frac{y}{64}=\frac{z}{216}=k\)

=> \(x=8k;y=64k;z=216k\)

thay vào điều kiện

\(\Rightarrow2\left(8k\right)^2+2\left(64k\right)^2+\left(216k\right)^2=1\)

\(2\cdot64k^2+2\cdot4096k^2+46656k^2=1\)

\(128k^2+8192k^2+46656k^2=1\)

\(54976k^2=1\)

\(k=\pm\frac{1}{234}\)

TH1: \(k=\frac{1}{234}\)

=> \(x=8\cdot\frac{1}{234}=\frac{4}{117}\)

\(y=64\cdot\frac{1}{234}=\frac{32}{117}\)

\(z=216\cdot\frac{1}{234}=\frac{12}{13}\)

TH2: \(k=-\frac{1}{234}\)

=> \(x=-\frac{4}{117}\)

\(y=-\frac{32}{117}\)

\(z=-\frac{12}{13}\)

bài 3:

ta có: \(\frac{\left(2x+1\right)}{5}=\frac{\left(4y-5\right)}{9}=\frac{\left(2x+4y-4\right)}{14}\) ( tính chất dãy tỉ số bằng nhau)

CM: \(\frac{\left(2x+4y-4\right)}{14}=\frac{\left(2x+4y-4\right)}{7x}\)

TH1: 2x+4y-4=0

=> 2x+1=0

=>x=\(\frac{-1}{2}\) thay vào biểu thức cầm CM trên

=> \(2\left(-\frac12\right)+4y-4=0\)

=> \(y=\frac54\left(TM\right)\)

TH2: 7x=14

=>x=2

thay vào phân số đầu tiên

\(\frac{2\cdot2+1}{5}=\frac55=1\)

=> \(\frac{4y-5}{9}=1\)

=>\(y=\frac72\)

bài 4:

=> \(\left(\frac{a}{a^{,}}+\frac{b^{,}}{b}\right)\cdot\frac{b}{b^{,}}=1\cdot\frac{b}{b^{,}}\)

=> \(\frac{a\cdot b}{a^{,}\cdot b^{^{,}}}+\frac{b^{,}\cdot b}{b\cdot b^{,}}=\frac{b}{b^{,}}\)

=> \(\frac{ab}{a^{,}b^{,}}+1=\frac{b}{b^{,}}\left(5\right)\)

ta có: \(\frac{b}{b^{,}}+\frac{c^{,}}{c}=1\Rightarrow\frac{b}{b^{,}}=1-\frac{c^{,}}{c}\left(6\right)\)

thay (6) vào (5)

=> \(\frac{ab}{a^{,}b^{,}}+1=1-\frac{c^{,}}{c}\)

=> \(\frac{ab}{a^{,}b^{,}}=-\frac{c^{,}}{c}\)

=> abc=\(-a^{,}b^{,}c^{,}\)

=> \(abc+a^{,}b^{,}c^{,}=0\left(đpcm\right)\)

cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k=> a=bk; c=dk

a. Vế trái =\(\frac{5a+3b}{5a-3b}\)=\(\frac{5bk+3b}{5bk-3b}\)=\(\frac{b\left(5k+3\right)}{b\left(5k-3\right)}\)=\(\frac{\left(5k+3\right)}{\left(5k-3\right)}\)(1)

Vế phải =\(\frac{5c+3d}{5c-3d}\)=\(\frac{5dk+3d}{5dk-3d}\)=\(\frac{d\left(5k+3\right)}{d\left(5k-3\right)}\)=\(\frac{\left(5k+3\right)}{\left(5k-3\right)}\)(2)

Từ (1) và (2) ta có\(\frac{5a+3b}{5a-3b}\)=\(\frac{5c+3d}{5c-3d}\)

b. Vế trái=\(\frac{7a^2+3ab}{11a^2-8b^2}\)=\(\frac{7b^2k^2+3b.k.b}{11b^2.k^2-8b^2}\)=\(\frac{b^2.k\left(7k+3\right)}{b^2\left(11k^2-8\right)}\)=\(\frac{k\left(7k+3\right)}{\left(11k^2-8\right)}\)(1)

Vế phải =\(\frac{7c^2+3cd}{11c^2-8d^2}\)=\(\frac{7d^2k^2+3d.k.d}{11d^2.k^2-8d^2}\)=\(\frac{d^2.k\left(7k+3\right)}{d^2\left(11k^2-8\right)}\)=\(\frac{k\left(7k+3\right)}{\left(11k^2-8\right)}\)(2)

Từ (1) và (2) ta có: \(\frac{7a^2+3ab}{11a^2-8b^2}\)=\(\frac{7c^2+3cd}{11c^2-8d^2}\)

giups mình với cảm ơn

đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

a) thay \(a=bk;c=dk\) ta có

\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)

\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)

từ (1);(2)\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

b) thay \(a=bk;c=dk\) ta có

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7(bk)^2+3bkb}{11(bk)^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}\)

\(=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(3)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3dkd}{11\left(dk\right)^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}\)

\(=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(4)

từ (3);(4)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Ta có:

\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7b^2k^2+3\cdot bk\cdot b}{11b^2k^2-8b^2}=\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(1\right)\)

\(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7d^2k^2+3dk\cdot d}{11d^2k^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrowđpcm\)

Mấy bài khác tương tự

à quên 
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{5a}{5c}=\frac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số = nhau 
\(\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)

LÀM CÁCH GIẢI LUÔN ĐI!!~~
 

Bài này bạn chỉ cần đặt k rồi thế k vào là làm được à, dễ lắm