Bài 2:

a: \(\dfrac{7}{8}+x=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-35}{40}=\dfrac{-11}{40}\)

b: \(\dfrac{17}{2}:x=5\)

=>\(x=\dfrac{17}{2}:5\)

=>\(x=\dfrac{17}{2\cdot5}=\dfrac{17}{10}\)

c: \(x-\dfrac{3}{8}=2+\dfrac{1}{4}\)

=>\(x-\dfrac{3}{8}=\dfrac{9}{4}\)

=>\(x=\dfrac{9}{4}+\dfrac{3}{8}=\dfrac{18}{8}+\dfrac{3}{8}=\dfrac{21}{8}\)

d: \(\dfrac{1}{2}+\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}\)

=>\(\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}-\dfrac{1}{2}=\dfrac{-3}{10}\)

=>\(x-2=-\dfrac{1}{2}\)

=>\(x=2-\dfrac{1}{2}=\dfrac{3}{2}\)

Bài 1:

a: \(\dfrac{-3}{4}+\dfrac{1}{5}=\dfrac{-15}{20}+\dfrac{4}{20}=\dfrac{-15+4}{20}=\dfrac{-11}{20}\)

b: \(\dfrac{-2}{5}-\dfrac{1}{3}=\dfrac{-6}{15}-\dfrac{5}{15}=\dfrac{-6-5}{15}=\dfrac{-11}{15}\)

c: \(\dfrac{3}{7}\cdot\dfrac{2}{5}-\dfrac{2}{5}=\dfrac{2}{5}\left(\dfrac{3}{7}-1\right)=\dfrac{2}{5}\cdot\dfrac{-4}{7}=\dfrac{-8}{35}\)

d: \(\dfrac{1}{4}+\dfrac{3}{4}\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{4-3}{6}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{1}{6}=\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{3}{8}\)

e: \(\dfrac{7}{2}\cdot\dfrac{8}{13}+\dfrac{8}{13}\cdot\dfrac{-5}{12}+\dfrac{8}{13}\)

\(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)\)

\(=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)

f: \(1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)

\(=1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}\)

\(=1+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{10\cdot12}\right)\)

\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)

\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)

Đổi: 4h 48 phút = 288 phút

       6h 45 phút = 405 phút

Tổng thời gian chị Hà làm hết 9 sản phẩm là:

288+405 = 693 (phút)

Tổng số sản phẩm chị Hà làm là:

4+5=9(sản phẩm)

Trung bình mỗi sản phẩm chị Hà làm hết số thời gian là:

693:9 = 77(phút)=1h 17 phút

Đáp số:...

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}=x+y+z\)

\(\Rightarrow\left\{{}\begin{matrix}x=a\left(x+y+z\right)\\y=b\left(x+y+z\right)\\z=c\left(x+y+z\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=a^2\left(x+y+z\right)^2\\y^2=b^2\left(x+y+z\right)^2\\z^2=c^2\left(x+y+z\right)^2\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2=a^2\left(x+y+z\right)^2+b^2\left(x+y+z\right)^2+c^2\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=\left(a^2+b^2+c^2\right)\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=1.\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=\left(x+y+z\right)^2\)

Tổng số phần bằng nhau là 2+3=5(phần)

Số gạo tẻ còn lại là \(135\cdot\dfrac{3}{5}=81\left(kg\right)\)

Số gạo nếp còn lại là 81:1,5=54(kg)

Số gạo tẻ ban đầu là \(81:\left(1-\dfrac{2}{5}\right)=81:\dfrac{3}{5}=135\left(kg\right)\)

Số gạo nếp ban đầu là:

\(54:\left(1-\dfrac{2}{3}\right)=54:\dfrac{1}{3}=54\cdot3=162\left(kg\right)\)

Vận tốc trung bình của bác Bình:

      \(20:0,5=40\left(km/giờ\right)\)

Mình bổ sung đề nhé: Trên cả quãng đường bác Bình chuyển động với vận tốc đều (không đổi)

Do 40km/giờ < 50km/giờ

Nên bác Bình đi như vậy không vi phạm nha bạn

cacmr ơn nha

a) \(A=\dfrac{2x^2-4x+8}{x^3+8}\left(x\ne-2\right)\\ =\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\dfrac{2}{x+2}\)

b) Thay x=2 (TMDK) vào bt A:

\(A=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2}\)

c) \(A=\dfrac{2}{x+2}\inℤ\Rightarrow2⋮\left(x+2\right)\\ \Rightarrow x+2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\\ \Rightarrow x\in\left\{-1;-3;0;-4\right\}\) (TMDK)

d) \(A=-\dfrac{3}{2}\\ \Rightarrow\dfrac{2}{x+2}=-\dfrac{3}{2}\\ \Rightarrow-3\left(x+2\right)=4\\ \Rightarrow-3x-6=4\\ \Rightarrow3x=-10\\ \Rightarrow x=-\dfrac{10}{3}\left(TMDK\right)\)

Vậy x=-10/3 thì A=-3/2

a: Sửa đề: x<>-2

\(A=\dfrac{2x^2-4x+8}{x^3+8}\)

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

\(=\dfrac{2}{x+2}\)

b: Thay x=2 vào A, ta được: \(A=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2}\)

c: Để A là số nguyên thì \(x+2\inƯ\left(2\right)\)

=>\(x+2\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{-1;-3;0;-4\right\}\)

d: \(A=-\dfrac{3}{2}\)

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

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

=>\(x=-\dfrac{4}{3}-2=-\dfrac{10}{3}\)

3 ngày 8 giờ = 80 giờ

80 giờ

Tick mik

a) 

\(P\left(x\right)=6x^4+2x+4x^3-3x^2-10+x^3+3x\)

\(=6x^4+\left(4x^3+x^3\right)-3x^2+\left(2x+3x\right)-10\)

\(=6x^4+5x^3-3x^2+5x-10\)

\(Q\left(x\right)=4-5x^3+2x^2-x^3+5x^4+11x^3-4x\)

\(=5x^4+\left(-5x^3-x^3+11x^3\right)+2x^2-4x+4\)

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

b) 

\(P\left(x\right)+Q\left(x\right)\)

\(=\left(6x^4+5x^3-3x^2+5x-10\right)+\left(5x^4+5x^3+2x^2-4x+4\right)\)

\(=6x^4+5x^3-3x^2+5x-10+5x^4+5x^3+2x^2-4x+4\)

\(=\left(6x^4+5x^4\right)+\left(5x^3+5x^3\right)+\left(-3x^2+2x^2\right)+\left(5x-4x\right)+\left(-10+4\right)\)

\(=11x^4+10x^3-x^2+x-6\)

\(P\left(x\right)-Q\left(x\right)\)

\(=\left(6x^4+5x^3-3x^2+5x-10\right)-\left(5x^4+5x^3+2x^2-4x+4\right)\)

\(=6x^4+5x^3-3x^2+5x-10-5x^4-5x^3-2x^2+4x-4\)

\(=\left(6x^4-5x^4\right)+\left(5x^3-5x^3\right)+\left(-3x^2-2x^2\right)+\left(5x+4x\right)+\left(-10-4\right)\)

\(=x^4-5x^2+9x-14\)

-1/4 + 2/5 . x = 4/15

2/5 . x = 4/15 + 1/4

2/5 . x = 31/60

x = 31/60 : 2/5

x = 31/24

-1/4+2/5.x=4/15

        2/5.x=4/15-1/4

        2/5.x=1/60

              X=1/60:2/5

              X=1/60.5/2

             X= 1/24

-1/4+2/5.x=4/15

153 phút