a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

a)x-2/5=3/8

(x-2).8=5.3

x-2=5.3:8

x-2=1

=>x=3

b,c )tương tự

\(\frac{5}{6}\div\frac{3}{8}\times\frac{3}{4}\)\(=\frac{5}{3}\)

\(\frac{5}{6}:\frac{3}{8}x\frac{3}{4}=\frac{5}{6}x\frac{8}{3}x\frac{3}{4}=\frac{5x8x3}{6x3x4}=\frac{5}{3}\)

5 phần 3

\(\left(\frac{1}{4}+\frac{3}{8}\right)\times\frac{8}{7}\)

\(=\frac{5}{8}\times\frac{8}{7}\)

\(=\frac{5}{7}\)

#H

\(\left(\frac{1}{4}+\frac{3}{8}\right)\cdot\frac{8}{7}=\frac{5}{8}\cdot\frac{8}{7}=\frac{5}{7}\)

1/4 + 3/8 x 8/7

= ( 1/4 x 3/8 ) x 8/7

=         5/8      x 8/7 

= 40/56 .

#Songminhtử

1) \(M=\frac{x^2+y^2+7}{x^2+y^2+5}=1+\frac{2}{x^2+y^2+5}\)

Ta có: \(x^2+y^2\ge0,\forall x;y\)

=> \(x^2+y^2+5\ge5\) với mọi x; y 

=> \(\frac{2}{x^2+y^2+5}\le\frac{2}{5}\)

=> \(M\le1+\frac{2}{5}=\frac{7}{5}\)

Dấu "=" xảy ra <=> x = y = 0 

Vậy max M = 7/5 đạt tại x = y = 0 

2) \(f\left(x-1\right)=x^2-3x+5=x^2-x-2x+2+3\)

\(=x\left(x-1\right)-2\left(x-1\right)+3=x\left(x-1\right)-\left(x-1\right)-\left(x-1\right)+3\)

\(=\left(x-1\right)\left(x-1\right)-\left(x-1\right)+3\)

=> \(f\left(x\right)=x.x-x+3=x^2-x+3\)