Bài 1: 

b) Ta có: \(D=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)

\(=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot0\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)

=0

0,2-0,375+5/11/-0,3+9/16-15/22

Ta có: \(x^2\ge0;3\left|y-2\right|\ge0\)

\(\Rightarrow x^2+3\left|y-2\right|\ge0\)

\(\Rightarrow x^2+3\left|y-2\right|-1\ge-1\)

\(\Rightarrow A\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x^2=0\\3\left|y-2\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}}\)

Vậy GTNN của A = -1 khi x = 0 và y = 2

\(A=x^2+3\left|y-2\right|-1\)

Có \(x^2\ge0;3\left|y-2\right|\ge0\)

\(\Rightarrow A\ge0+0-1=-1\)

Dấu '=" xảy ra khi MinA=-1\(\Leftrightarrow x=0;y=2\)

bai 1.

giai chi tiet cho ban mot bai

\(x\ge\)0  (vi neu x<0 thi ve trai luon >0 VP <0 vo ly)

=>x+3>0=>Ix+3I=x+3

x+4>0=> Ix+4I=x+4

Ix+3I+Ix+4I=(x+3)+(x+4)=2x+7

2x+7=3x

7=3x-2x=x

x=7 

1)5^x+1 + 6.5^x+1 = 875

   5^x+1 ( 1+6 ) = 875

   5^x+1 . 7 = 875

5^x+1 = 875 : 7

5^x+1 = 125

5^x+1 = 5³

x+1 = 3

x = 3 - 1

x = 2

2)3^x+1 + 3^x+3 = 810

  3^x . 3 + 3² . 3^x+1 = 810

  3^x . 3 + 9 . 3^x . 3 = 810

  3^x .3 ( 1 + 9 ) = 810

  3^x+1 . 10 = 810

  3^x+1 = 810 : 10

  3^x+1 = 81

  3^x+1 = 3⁴ 

x+1 = 4

x = 4-1

x = 3

\(2^{x+3}+2^x=144\)

\(\Rightarrow2^x.2^3+2^x.1=144\)

\(\Rightarrow2^x.\left(2^3+1\right)=144\)

\(\Rightarrow2^x.9=144\)

\(\Rightarrow2^x=144:9\)

\(\Rightarrow2^x=16\)

\(\Rightarrow2^x=2^4\)

\(\Rightarrow x=4\)

x+ ( x + 1) + (x+2) + ...+ (x+30) = 1240

x.31 + ( 1 + 2 + ...+ 30) = 1240

x.31 + [(1+30).30:2] = 1240

x.31 + 465 = 1240

x.31 = 775

x = 25

=31x+(1+2+3+..+30)=1240

=31x+465=1240

=31x=1240-465

31x=775

x=25

vậy x = 25

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)

\(\left(x+x+x+...+x\right)+\left(1+2+...+30=1240\right)\)  ( 31 chư số x )

\(31x+\frac{30\cdot31}{2}=1240\)

\(31x+465=1240\)

\(31x=1240-465=775\)

\(x=\frac{775}{31}=25\)

Vậy ....