bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

a)tổng :\(=-\dfrac{1}{3}x^3y^4+\dfrac{4}{3}x^3y^4=\dfrac{3}{3}x^3y^4=x^3y^4\)

hiệu :\(=-\dfrac{1}{3}x^3y^4-\dfrac{4}{3}x^3y^4=-\dfrac{5}{3}x^3y^4\)

b)tôngr : \(=\left(-2+1\right)xyz=-xyz\)

hiệu :\(=\left(-2-1\right)xyz=-3xyz\)

a) \(\left( {\dfrac{1}{2}{x^3}} \right).\left( -{4{x^2}} \right) = \left( {\dfrac{1}{2}.(-4)} \right).\left( {{x^3}.{x^2}} \right) = (-2).{x^5}\).

Hệ số: -2

Bậc: 5

b) \(\dfrac{1}{2}{x^3} - \dfrac{5}{2}{x^3} = \left( {\dfrac{1}{2} - \dfrac{5}{2}} \right){x^3} = \dfrac{{ - 4}}{2}.{x^3} =  - 2{x^3}\)

Hệ số: -2

Bậc: 3

\(a,\left(\dfrac{1}{2}x^3\right).\left(-4x^2\right)=\left(-4.\dfrac{1}{2}\right).\left(x^3.x^2\right)=-2x^5\\ Hệ.số:-2;bậc:5\\ b,\dfrac{1}{2}x^3-\dfrac{5}{2}x^3=-2x^3\\ Hệ.số:-2;bậc:3\)

a) Ta thấy: Hình 4 được chia thành 10 phần bằng nhau, tô màu 6 phần.

Vậy đã tô màu $\frac{6}{{10}} = \frac{3}{5}$ hình 4.

b) So sánh hai phân số $\frac{7}{8}$ và $\frac{3}{5}$

Ta có: $\frac{7}{8} = \frac{{7 \times 5}}{{8 \times 5}} = \frac{{35}}{{40}}$   ;    $\frac{3}{5} = \frac{{3 \times 8}}{{5 \times 8}} = \frac{{24}}{{40}}$

Vì $\frac{{35}}{{40}} > \frac{{24}}{{40}}$ nên $\frac{7}{8}$> $\frac{3}{5}$

Thực hiện tính:

$\frac{7}{8} + \frac{3}{5} = \frac{{35}}{{40}} + \frac{{24}}{{40}} = \frac{{59}}{{40}}$

$\frac{7}{8} - \frac{3}{5} = \frac{{35}}{{40}} - \frac{{24}}{{40}} = \frac{{11}}{{40}}$

$\frac{7}{8} \times \frac{3}{5} = \frac{{7 \times 3}}{{8 \times 5}} = \frac{{21}}{{40}}$

$\frac{7}{8}:\frac{3}{5} = \frac{7}{8} \times \frac{5}{3} = \frac{{35}}{{24}}$

1 chiều dài hcn là:\(5\dfrac{1}{4}:\dfrac{3}{2}=\dfrac{42}{12}=\dfrac{7}{2}\)(m)

chu vi hcn là:\(\left(\dfrac{3}{2}+\dfrac{7}{2}\right).2=10\)(m)

cứ coi đó là cd ik

a. x²=16.4

    x²=64

vì x<0 => x= -8

b: Tỉ số giữa 48 giờ và 3,2 giờ là:

48:3,2=15:1

c: \(\dfrac{-36}{7}\cdot\dfrac{2}{3}=\dfrac{-72}{21}=\dfrac{-24}{7}\)

B1:
a.(-14/39)
b.45 độ
B2:
a.x=(-13/10)
b.
c.0

Bài 1:

a/\(\dfrac{17}{3}-\dfrac{5}{3}=\dfrac{17-5}{3}=\dfrac{12}{3}\)=4

b/Tam giác ABC có:

góc A+góc B+góc C=180 độ

=>70 độ+65 độ+góc C=180 độ

=>góc C =180 độ-70 độ-65 độ=45 độ

Bài 2:

a/\(x+3\dfrac{1}{2}=\dfrac{11}{5}=>x+\dfrac{7}{2}=\dfrac{11}{5}=>x=\dfrac{11}{5}-\dfrac{7}{2}=\dfrac{-13}{10}\)

c/\(\left(\dfrac{-3}{7}+\dfrac{5}{11}\right):\dfrac{4}{31}+\left(\dfrac{-4}{7}+\dfrac{6}{11}\right):\dfrac{4}{31}\)

=>\(\left(\dfrac{-3}{7}+\dfrac{5}{11}\right).\dfrac{31}{4}+\left(\dfrac{-4}{7}+\dfrac{6}{11}\right).\dfrac{31}{4}\)

=>\(\dfrac{31}{4}.\left(\dfrac{-3}{7}+\dfrac{5}{11}+\dfrac{\left(-4\right)}{7}+\dfrac{6}{11}\right)=\dfrac{31}{4}.0=0\)

25cm = \(\dfrac{1}{4}m\)
\(\dfrac{3}{4}:\dfrac{1}{4}=3\)
b.30p=\(\dfrac{1}{2}\)giờ
\(\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{3}{4}\)
c.0,4kg=400g 
\(\dfrac{400}{340}=\dfrac{20}{17}\)
d.\(\dfrac{2}{5}:\dfrac{3}{4}=\dfrac{8}{15}\)