a) 3(2a - 1) + 5(3 - a)

= 6a - 3 + 15 -5a

= a + 12

Thay a=\(-\frac{3}{2}\) vào biểu thức a) ta có:

⇒ \(-\frac{3}{2}+12=\frac{21}{2}\)

b) 25x - 4(3x - 1) + 7(5 - 2x)

= 25x -12x + 4 + 35 - 14x

= -x + 39

Thay x= 2,1 vào biểu thức b) ta có:

⇒ -2,1 + 39 = 36,9

c) 4a - 2(10a - 1) + 8a - 2

= 4a -20a + 2 + 8a - 2

= -8a

Thay a= -0,2 vào biểu thức c) ta có:

⇒ -8.(-0,2)= 1,6

d) 12(2 - 3b) + 35b - 9(b + 1)

= 24 - 36b + 35b - 9b -9

= 15 - 10b

Thay b=\(\frac{1}{2}\) vào biểu thức d) ta có:

⇒ 15 - 10. \(\frac{1}{2}=\) 10

\(a,a=-\dfrac{3}{2}\)

\(\Rightarrow3\left[2\left(-\dfrac{3}{2}\right)-1\right]+5\left(3+\dfrac{3}{2}\right)=3.\left(-3-1\right)+5.\dfrac{9}{2}=-12+\dfrac{45}{2}=\dfrac{21}{2}\)

\(b,x=2,1\)

\(\Rightarrow25.2,1-4\left(3.2,1-1\right)+7\left(5-2.2,1\right)=52,5-4.5,3+7.0,8=36,9\)

\(c,b=\dfrac{1}{2}\)

\(\Rightarrow12\left(2-3.\dfrac{1}{2}\right)+35.\dfrac{1}{2}-9\left(\dfrac{1}{2}+1\right)=12.\dfrac{1}{2}+\dfrac{35}{2}-9.\dfrac{3}{2}=6+\dfrac{35}{2}-\dfrac{27}{2}=10\)

\(d,a=-0,2\)

\(\Rightarrow4.\left(-0,2\right)^2-2\left(10.\left(-0,2\right)-1\right)+4.\left(-0,2\right)\left(2-\left(-0,2\right)^2\right)\)

\(=4.0,04-2.\left(-3\right)-0,8.1,96\)

\(=0,16+6-1,568\)

\(=4,592\)

a: A=6a-3+15-5a=a+12

Khi a=-3/2 thì A=-3/2+12=10,5

b: B=25x-12x+4+35-8x=5x+39

Khi x=2,1 thì B=10,5+39=49,5

c: C=24-6b+35b-9b-9=20b+15

Khi b=0,5 thì C=10+15=25

d: D=4a^2-20a+2+8a-4a^3=-4a^3+4a^2-12a+2

Khi a=-0,2 thì 

D=-4*(-1/5)^3+4*(-1/5)^2-12*(-1/5)+2=4,592

\(A=1+2-3-4+5+6-7-8+.........+2013\)

\(=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+........+\left(2010-2011-2012+2013\right)\)

\(=1+0+0+..........+0\)

\(=1\)

26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

mình thi toán volympic lớp 4 , mình hok lớp 4 

k nha 

Mình cũng thế nè

Mình cũng thế nè

Câu  1: Gọi a là số sách cần tìm

\(a\in BC\left(10;12;15;18\right)\)và \(200< a< 500\)

Ta có: \(10=2.5\)

          \(12=2^2.3\)

         \(15=3.5\)

          \(18=2.3^2\)

\(\Rightarrow BCNN\left(10;12;15;18\right)=2^2.3^3.5\)\(=180\)

\(\Rightarrow BC\left(10;12;15;18\right)=BC\left(180\right)=\left(0;180;360;540;720;...\right)\)

Mà \(200< a< 500\)

\(\Rightarrow a=360\)

Câu 2

a)Ta có \(A=x+16\)

       \(\Rightarrow A=-4+16\)

       \(\Rightarrow A=12\)

b)Ta có:\(B=\left(-103\right)+y\)

         \(\Rightarrow B=-103+3\)

          \(\Rightarrow B=-100\)

Câu 3:Tương tự bn nhé-câu 1

1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)

\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)

2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)

\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)

4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) 

\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)

\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)

e: Ta có: \(E=\left(x-8\right)^2+\left(x+7\right)^2\)

\(=x^2-16x+64+x^2+14x+49\)

\(=2x^2-2x+113\)

\(=2\left(x^2-x+\dfrac{113}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{225}{4}\right)\)

\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(15\left(2a^2-1\right)+5\left(3-\frac{1}{5a}-6a^2\right)\)

\(=30a^2-15+15-\frac{1}{a}-30a^2\)

\(=-\frac{1}{a}\)

tại \(a=2017\)=> M= \(\frac{-1}{a}=\frac{-1}{2017}\)

\(\left(x-y\right)\left(x^2+xy+y^2\right)+y^3\)

\(=x^3-y^3+y^3\)

\(=x^3\)

ại \(x=2\)=> N= \(x^3=2^3=8\)