Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k$

$\Rightarrow a=bk, c=dk$

Khi đó:

$\frac{2a+3b}{3a-5b}=\frac{2bk+3b}{3bk-5b}=\frac{b(2k+3)}{b(3k-5)}=\frac{2k+3}{3k-5}(1)$

$\frac{2c+3d}{3c-5d}=\frac{2dk+3d}{3dk-5d}=\frac{d(2k+3)}{d(3k-5)}=\frac{2k+3}{3k-5}(2)$

Từ $(1); (2)$ ta có đpcm.

a) \(...=P\left(x\right)=2x^4-x^4+3x^3+4x^2-3x^2+3x-x+3\)

\(P\left(x\right)=x^4+3x^3+x^2+2x+3\)

\(...=Q\left(x\right)=x^4+x^3+3x^2-x^2+4x+4-2\)

\(Q\left(x\right)=x^4+x^3+2x^2+4x+2\)

b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)

\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)

\(P\left(x\right)-Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)-\left(x^4+x^3+2x^2+4x+2\right)\)

\(\)\(\Rightarrow P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+3-x^4-x^3-2x^2-4x-2\)

\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)

\(\dfrac{-3}{26}+2\dfrac{4}{69}\)

\(=\dfrac{-3}{26}+\dfrac{142}{69}\)

\(=\dfrac{-3.69}{26.69}+\dfrac{142.26}{26.69}\)

\(=\dfrac{-207+3692}{1794}\)

\(=\dfrac{3485}{1794}\)

-3/26 + 2 và 4/69 = 3485/1794

Bài 5 :

a) \(\dfrac{y}{4}=\dfrac{9}{y}\)

\(\Rightarrow y^2=36\left(y\ne0\right)\)

\(\Rightarrow y=\pm6\)

b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)

\(\Rightarrow\left(y+7\right)^2=100=10^2\)

\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)

c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)

\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)

\(\Rightarrow20-25y=3y+6\)

\(\Rightarrow28y=14\)

\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)

Bài 4 :

\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)

\(x+y-2xy=4\)

\(\Rightarrow\left(\sqrt[]{x}-\sqrt[]{y}\right)^2-2^2=0\)

\(\Rightarrow\left(\sqrt[]{x}-\sqrt[]{y}-2\right)\left(\sqrt[]{x}-\sqrt[]{y}+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x}-\sqrt[]{y}-2=0\\\sqrt[]{x}-\sqrt[]{y}+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x}-\sqrt[]{y}=2\\\sqrt[]{x}-\sqrt[]{y}=-2\end{matrix}\right.\) \(\left(x;y\ge0\right)\)

\(TH1:\sqrt[]{x}-\sqrt[]{y}=2\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(4;0\right);\left(9;1\right);\left(16;4\right);...\right\}\left(x;y\inℕ\right)\)

\(TH2:\sqrt[]{x}-\sqrt[]{y}=-2\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;4\right);\left(1;9\right);\left(4;16\right);...\right\}\left(x;y\inℕ\right)\)

Đính chính mình nhầm sorry

\(x+y-2xy=4\)

\(\Rightarrow2x+2y-4xy=8\)

\(\Rightarrow2x-4xy+2y=8\)

\(\Rightarrow2x\left(1-2y\right)-\left(1-2y\right)=8-1\)

\(\Rightarrow\left(2x-1\right)\left(1-2y\right)=7\)

\(\Rightarrow\left(2x-1\right);\left(1-2y\right)\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;4\right);\left(1;-3\right);\left(-3;1\right);\left(4;0\right)\right\}\)

\(B=\dfrac{2n+6}{n-5}=\dfrac{2n-10+16}{n-5}=\dfrac{2\left(n-5\right)+16}{n-5}=2+\dfrac{16}{n-5}\)

Để \(B=2+\dfrac{16}{n-5}\inℤ\)

\(\Rightarrow n-5\in\left\{-1;1;-2;2;-4;4;-8;8;-16;16\right\}\)

\(\Rightarrow n\in\left\{4;6;3;7;1;9;-3;13;-11;21\right\}\)

Có (3x - \(\dfrac{1}{6}\))² ≥ 0 ∀ x; |2y-6| ≥ 0 ∀ y

=> (3x - \(\dfrac{1}{6}\))² + |2y-6| ≥ 0 ∀x,y

Mà (3x - \(\dfrac{1}{6}\))² + |2y-6| ≤ 0

=> (3x - \(\dfrac{1}{6}\))² = 0; |2y - 6| = 0

=> x = \(\dfrac{1}{18}\); y = 3;

=> A = \(\left(\dfrac{1}{18}\right)^2\) + 3² = \(9\dfrac{1}{324}\)