Ta có 

7y chia hết cho 7 112 chia hết cho 7

=> 5x phải chia hết cho 7

=> x chia hết cho 7

=> x=7k 

     y=(112-35k)/7

\(5x+7y=112\) (*)

\(\Leftrightarrow5x=112-7y=7\left(16-y\right)\)

Vì \(7\left(16-y\right)⋮7\) nên để \(5x=7\left(16-y\right)\Leftrightarrow5x⋮7\) Mà \(\left(5;7\right)=1\Rightarrow x⋮7\)(1)

Vì \(x;y\in N^{\text{*}}\) nên từ (*) \(\Rightarrow0< 5x< 112\Rightarrow0< x< \frac{112}{5}=22,4\)(2)

Từ (1) ;(2) \(\Rightarrow x=\left\{7;14;21\right\}\)

Với x = 7 thì \(5.7+7y=112\Rightarrow y=11\left(TM\right)\)

Với x = 14 thì \(5.14+7y=112\Rightarrow y=6\left(TM\right)\)

Với x = 21 thì \(5.21+7y=112\Rightarrow y=1\left(TM\right)\)

Vậy \(\left(x;y\right)=\left\{\left(7;11\right);\left(14;6\right);\left(21;1\right)\right\}\)

vô ib mk chỉ cho

\(a,1-sin^2\alpha=cos^2\alpha\)

\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)

\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)

\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)

\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)

\(=1+2sin^2\alpha.cos^2\alpha\)

\(\sqrt{3+\sqrt{x}}=3\)

\(3+\sqrt{x}=3^2=9\)

\(\sqrt{x}=9-3=6\)

\(x=6^2=36\)