Các bạn giúp mình giải bài này nha!!!
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NN
1
Những câu hỏi liên quan
PH
Các bạn giúp mình giải bài này bằng bài giải nha !
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 6
Em đăng câu hỏi lên cộng đồng Olm thì cần đăng rõ yêu cầu câu hỏi ra. Có như vậy em mới nhận được sự trợ giúp tốt nhất từ cộng đồng Olm.
HL
3
S
20 tháng 11 2021
Thầy giáo dạy phải biết yêu nước thương nòi.
Ai cũng phải nhớ về quê cha đắt tổ.
20 tháng 11 2023
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
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{5-x}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1}{x-5}-\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1-x+18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)
GS
1
20 tháng 11 2023
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: \(\dfrac{x^2-4}{x^2-x}:\dfrac{x^2+2x}{x-1}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-1\right)}\cdot\dfrac{x-1}{x\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)}{x^2\left(x+2\right)}\)
TA
1
NH
2
27 tháng 10 2021
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)=\(\dfrac{-1}{10}\)\(\times\dfrac{9}{2}\)\(-\)\(\dfrac{1}{4}\)=\(\dfrac{-9}{20}-\dfrac{5}{20}\)=\(\dfrac{-14}{20}\)=\(\dfrac{-7}{10}\)
27 tháng 10 2021
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
JW
1
\(a,x\ge2\\ \Leftrightarrow P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left(x-2\right)\\ P=\dfrac{1}{2}-3x+\dfrac{1}{8}-2x+4\\ P=-5x+\dfrac{37}{8}\)
\(b,x< 2\\ \Leftrightarrow P=\dfrac{1}{2}-\dfrac{1}{2}\left(6x-\dfrac{1}{4}\right)-2\left(2-x\right)\\ P=\dfrac{1}{2}-3x+\dfrac{1}{8}-4+2x\\ P=-x-\dfrac{27}{8}\)