Thầy giáo dạy phải biết yêu nước thương nòi.

Ai cũng phải nhớ về quê cha đắt tổ.

?????????????????????????????/

-Đi đâu xa chũng ta vẫn nhớ về quê cha đất tổ

-Ông bà đã dạy phải bt yêu thương giống nòi

\(a,x\ge2\\ \Leftrightarrow P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left(x-2\right)\\ P=\dfrac{1}{2}-3x+\dfrac{1}{8}-2x+4\\ P=-5x+\dfrac{37}{8}\)

\(b,x< 2\\ \Leftrightarrow P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left(2-x\right)\\ P=\dfrac{1}{2}-3x+\dfrac{1}{8}-4+2x\\ P=-x-\dfrac{27}{8}\)

1: \(\dfrac{15x}{7y^3}\cdot\dfrac{2y^2}{x^2}\)

\(=\dfrac{15x\cdot2y^2}{7y^3\cdot x^2}=\dfrac{30xy^2}{7x^2y^3}=\dfrac{30}{7xy}\)

2: \(\dfrac{6x^3}{7y^4}\cdot\dfrac{35y^2}{24x}\)

\(=\dfrac{6x^3}{24x}\cdot\dfrac{35y^2}{7y^4}\)

\(=\dfrac{x^2}{4}\cdot\dfrac{5}{y^2}=\dfrac{5x^2}{4y^2}\)

3: \(\dfrac{4y^2}{x^4}\cdot\dfrac{-3x^2}{8y}\)

\(=\dfrac{4y^2}{8y}\cdot\dfrac{-3x^2}{x^4}=\dfrac{y}{2}\cdot\dfrac{-3}{x^2}=\dfrac{-3y}{2x^2}\)

4: \(\dfrac{-18y^3}{25x^4}:\dfrac{9y^3}{-15x^2}\)

\(=\dfrac{18y^3}{25x^4}\cdot\dfrac{15x^2}{9y^3}\)

\(=\dfrac{18y^3}{9y^3}\cdot\dfrac{15x^2}{25x^4}=2\cdot\dfrac{3}{5x^2}=\dfrac{6}{5x^2}\)

5: \(\dfrac{8y^2}{9x^2}:\dfrac{4y}{3x^2}\)

\(=\dfrac{8y^2}{9x^2}\cdot\dfrac{3x^2}{4y}=\dfrac{8y^2}{4y}\cdot\dfrac{3x^2}{9x^2}=\dfrac{1}{3}\cdot2y=\dfrac{2y}{3}\)

6: \(\dfrac{-20x}{3y^2}:\dfrac{-4x^3}{5y}\)

\(=\dfrac{20x}{3y^2}:\dfrac{4x^3}{5y}\)

\(=\dfrac{20x}{3y^2}\cdot\dfrac{5y}{4x^3}=\dfrac{20x}{4x^3}\cdot\dfrac{5y}{3y^2}=\dfrac{5}{3y}\cdot\dfrac{5}{x^2}=\dfrac{25}{3x^2y}\)

7: \(\dfrac{\left(x+4\right)^2}{4x+12}:\dfrac{x+4}{3x+9}\)

\(=\dfrac{\left(x+4\right)^2}{4\left(x+3\right)}:\dfrac{x+4}{3\left(x+3\right)}\)

\(=\dfrac{\left(x+4\right)^2}{4\left(x+3\right)}\cdot\dfrac{3\left(x+3\right)}{x+4}=\dfrac{3\left(x+4\right)}{4}\)

8: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}\)

\(=\dfrac{5\cdot\left(-2\right)}{4}=-\dfrac{10}{4}=-\dfrac{5}{2}\)

9: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)

\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)

\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)

10: \(\dfrac{x^2-4}{x^2-x}:\dfrac{x^2+2x}{x-1}\)

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

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

DĐây là TA lớp 4

?

???

❓

Câu1:a) \(\left(\dfrac{-2}{3}\right)^2\)+\(\dfrac{-7}{6}\)\(\div\dfrac{3}{8}\)=\(\dfrac{4}{9}\)+\(\dfrac{-7}{6}\times\dfrac{8}{3}\)=\(\dfrac{4}{9}\)+\(\dfrac{-28}{9}\)=\(\dfrac{-24}{9}\)=\(\dfrac{-8}{3}\)

b)=\(\dfrac{-1}{10}\)\(\times\dfrac{9}{2}\)\(-\)\(\dfrac{1}{4}\)=\(\dfrac{-9}{20}-\dfrac{5}{20}\)=\(\dfrac{-14}{20}\)=\(\dfrac{-7}{10}\)

Câu 3: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{4+5+7}=\dfrac{-32}{16}=-2\)

Do đó: x=-8; y=-10; z=-14

help me