\(7.PTK_{CuSO_4}=7.\left(64+32+16.4\right)=7.160=1120\)

7CuSO_{4 \(=7.64+32+16.4=448+32+64=544\left(đvC\right)\)}

(Nhớ k cho mình với nhé!)

Ta có:\(x+y=1\)\(\Rightarrow x=1-y\)

Khi đó: \(P=\left(1-y\right)^3+y^3+\left(1-y\right)y\)

               \(=1-3y+3y^2-y^3+y^3+y-y^2\)

                \(=2y^2-2y+1\)

                 \(=2\left(y^2-y+\frac{1}{4}\right)-\frac{1}{2}+1\)

                  \(=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu ''='' xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

= 6( 1.2 + 32 + 16.4) = 588

6H₂SO_{4 \(=6.1.2+32+16.4=12+32+64=10\left(đvC\right)\)}

(Nhớ k cho mình với nhé!)

\(x^4+2x^2+1=\left(x^2+1\right)^2\) (Nhớ k cho mình với nhé!)

\(6x^2-\left(2x+5\right)\left(3x-2\right)=-12.\)

\(\Leftrightarrow6x^2-6x^2-11x+10+12=0\)

\(\Leftrightarrow11x=22\Leftrightarrow x=2\)

\(6x^2-2x\left(3x-2\right)-5\left(3x-2\right)\)= -12

\(6x^2-6x^2+4x-15x+10=-12\)

\(-11x+10=-12\)

\(x=2\)

a) 9x⁴+22+6x²+y²+2y

= (3x²)²+2.3x².1+1+y²+2y+1+20

=(3x²+1)² + (y+1)²+2²+4²

b)x⁴+4+4y²+5x²+4xy

=x⁴+5x²+4+4y²+4xy

=x⁴+4x²+4+4y²+4xy+x²

=(x²)²+2x²2+2²+(2y)²+2.2yx+x²

=(x²+2)²+(2y+x)²

c)z²+y²-6z+2y+10

=z²-6z+9+y²+2y+1

=z²-2.z.3+9+(y+1)²

=(z-3)²+(y+1)²

d)x²+4y²+m²+4mn+4xy+4n²

=x²+4xy+4y²+4n²+4mn+m²

=x²+2x2y+(2y)²+(2n)²+2.2nm+m²

=(x+2y)²+(2n+m)²

e)x²+y²-6nx+9n²+4my+4m²

=x²-6nx+9n²+y²+4my+4m²

=x²-2x3n+(3n)²+y²+2y2n+(2m)²

=(x-3n)²+(y+2m)²

f)4x²-4xm+2m²+4mn+4n²

=4n²-4xm+m²+4n²+4mn+m²

=(2n)²-2.2xm+m²+(2n)²+2.2nm+m²

=(2n-m)²+(2n+m)²

g) Ghi thiếu đề,đề đúng :

9x²-12xy+5y²+2y+1

=9x²-12xy+4y²+y²+2y+1

=(3x)²-2.3x2y+(2y)²+(y+1)²

=(3x-2y)²+(y+1)²