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

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

Bài 2:

1: ĐKXĐ: x<>1

\(\dfrac{x}{x-1}+\dfrac{1}{1-x}\)

\(=\dfrac{x}{x-1}-\dfrac{1}{x-1}\)

\(=\dfrac{x-1}{x-1}=1\)

2: ĐKXĐ: x<>3/2

\(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)

\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)

\(=\dfrac{11x+x-18}{2x-3}=\dfrac{12x-18}{2x-3}\)

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

=6

3: ĐKXĐ: x<>1/2

\(\dfrac{4x+5}{2x-1}+\dfrac{5-9x}{1-2x}\)

\(=\dfrac{4x+5}{2x-1}+\dfrac{9x-5}{2x-1}\)

\(=\dfrac{4x+5+9x-5}{2x-1}=\dfrac{13x}{2x-1}\)

4: ĐKXĐ: x<>2/5

\(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)

\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)

\(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)

5: ĐKXĐ: \(x\ne\pm y\)

\(\dfrac{xy}{x^2-y^2}-\dfrac{x^2}{y^2-x^2}\)

\(=\dfrac{xy}{x^2-y^2}+\dfrac{x^2}{x^2-y^2}\)

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

6: ĐKXĐ: \(x\notin\left\{0;7\right\}\)

\(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)

\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)

\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)

\(=\dfrac{5x-35}{x\left(5x-35\right)}=\dfrac{1}{x}\)

7: ĐKXĐ: \(x\ne1\)

\(\dfrac{x+2}{x-1}-\dfrac{x-9}{1-x}-\dfrac{x-9}{1-x}\)

\(=\dfrac{x+2}{x-1}+\dfrac{x-9}{x-1}+\dfrac{x-9}{x-1}\)

\(=\dfrac{x+2+x-9+x-9}{x-1}=\dfrac{3x-16}{x-1}\)

8: ĐKXĐ:x<>1

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

\(=\dfrac{2x^2-x}{x-1}-\dfrac{x+1}{x-1}+\dfrac{2-x^2}{x-1}\)

\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}\)

=x-1

9: ĐKXĐ: x<>3

\(\dfrac{4-x^2}{x-3}+\dfrac{2x-x^2}{3-x}+\dfrac{5-4x}{x-3}\)

\(=\dfrac{4-x^2}{x-3}+\dfrac{x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)

\(=\dfrac{4-x^2+x^2-2x+5-4x}{x-3}=\dfrac{-6x+9}{x-3}\)

10: ĐKXĐ: x<>5

\(\dfrac{x+1}{x-5}+\dfrac{x-18}{5-x}+\dfrac{x+2}{x-5}\)

\(=\dfrac{x+1}{x-5}-\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)

\(=\dfrac{x+1-x+18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)

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

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???

❓

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