b) \(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]:2\)

\(=\left[\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\cdot2+\dfrac{8}{3}\right]:2\)

\(=\left[\left(\dfrac{1}{4}-\dfrac{2}{4}\right)\cdot2+\dfrac{8}{3}\right]:2\)

\(=\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right):2\)

\(=\left(-\dfrac{1}{2}+\dfrac{8}{3}\right):2\)

\(=\dfrac{13}{6}:2\)

\(=\dfrac{13}{12}\)

d) \(\left(-0,75\right)-\left[\left(-2\right)+\dfrac{3}{2}\right]:1,5+\left(-\dfrac{5}{4}\right)\)

\(=\left(-0,75\right)-\left[\left(-2\right)+1,5\right]:1,5+\left(-\dfrac{5}{4}\right)\)

\(=\left(-0,75\right)-\left(-0,5\right):1,5+\left(-1,25\right)\)

\(=\left(-0,75\right)-\left(-\dfrac{1}{3}\right)-1,25\)

\(=-2+\dfrac{1}{3}\)

\(=-\dfrac{5}{3}\)

25.(-1/5)^3+1/5-2.(-1/2)^2-1/2

=25.-1/125+1/5-2.1/4-1/2

=1/5+1/5-1/2-1/2

=2/5-1/2-1/2

=4/10-5/10-5/10

=-1/10-5/10

=-6/10

\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\\ =2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}-2\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

\(=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

\(=2-\sqrt{2}+3\sqrt{2}-2\)

\(=2\sqrt{2}\)

\(B=2x^2+10x-1\)

=> \(B=2\left(x^2+5x\right)-1\)

=> \(B=2\left(x^2+2.x\frac{5}{2}+\frac{25}{4}\right)-\frac{27}{2}\)

=> \(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)

Có \(2\left(x+\frac{5}{2}\right)^2\ge0\)với mọi x

=> \(2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\)

Dấu "=" xảy ra <=> \(\left(x+\frac{5}{2}\right)^2=0\)<=> \(x+\frac{5}{2}=0\)<=> \(x=\frac{-5}{2}\)

KL: B_min = \(\frac{-27}{2}\)<=> \(x=\frac{-5}{2}\)

\(C=5x-x^2\)

=> \(C=-\left(x^2-5x\right)\)

=> \(C=-\left(x^2-2.x.\frac{5}{2}+\frac{25}{4}\right)+\frac{25}{4}\)

=> \(C=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)

Có \(\left(x-\frac{5}{2}\right)^2\ge0\)với mọi x

=> \(-\left(x-\frac{5}{2}\right)^2\le0\)

=> \(C=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu "=" xảy ra <=> \(\left(x-\frac{5}{2}\right)^2=0\)<=> \(x-\frac{5}{2}=0\)<=> \(x=\frac{5}{2}\)

KL: C_max = \(\frac{25}{4}\)<=> \(x=\frac{5}{2}\)

B=2x²+10x-1=2(x²+5x-1/2)=2(x²+2*5/2*x+25/4-27/4)=2[x²+2*5/2*x+(5/2)²]-27/2=2(x+5/2)²-27/2

Ta có: (x+5/2)^2>=0(với mọi x)

=> 2(x+5/2)^2>=0(với mọi x)

=> 2(x+5/2)^2-27/2>=-27/2(với mọi x)

hay B>=-27/2( với mọi x)

Do đó, GTNN của B là -27/2 khi:

x+5/2=0

x=-5/2

Vậy GTNN của B là -27/2 khi x=-5/2

C=5x-x^2=-x^2+5x=-x^2+2*5/2*x-25/4+25/4=-[x^2-2*5/2*x+(5/2)^2]+25/4=-(x-5/2)^2+25/4

Ta có: (x-5/2)^2>=0(với mọi x)

=>-(x-5/2)^2<=0(với mọi x)

=> -(x-5/2)^2+25/4<=25/4(với mọi x) hay C<=25/4(với mọi x)

Do đó, GTLN của C là 25/4 khi: x-5/2=0

                                              x=5/2

Vậy GTLN của C là 25/4 tại x=5/2

sin a=3/5

=>cos a=4/5

tan a=3/5:4/5=3/4; cot a=1:3/4=4/3

M=(4/3+3/4):(4/3-3/4)=25/7

\(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)=15^4-\left(15^4-1\right)=15^4-15^4+1=1\)

\(B=2x^2+10x-1\)

\(\Rightarrow B=2\left(x^2+5x\right)-1\)

\(\Rightarrow B=2\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{27}{2}\)

\(\Rightarrow B=\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)

Ta có : \(2\left(x+\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\)

Dấu "=" xảy rak hi và chỉ khi \(\left(x+\frac{5}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{5}{2}=0\)

\(\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(Min_B=\frac{-27}{2}\Leftrightarrow x=\frac{-5}{2}\)

\(A=\frac{-5x}{21}+\frac{-5y}{21}+\frac{-5z}{21}\)

\(=\frac{-5x-5y-5z}{21}\)

\(=\frac{-5\left(x+y+z\right)}{21}\)

Do  \(x+y=-z\) =>    \(x+y+z=0\)

Như vậy  \(A=0\)