Toán

1:

Trong 1 ngày, đội 1 làm được: \(\dfrac{1}{15}\)(công việc)

Trong 1 ngày, đội 2 làm được \(\dfrac{1}{12}\left(côngviệc\right)\)

Trong 1 ngày, 2 đội làm được:

\(\dfrac{1}{12}+\dfrac{1}{15}=\dfrac{5}{60}+\dfrac{4}{60}=\dfrac{9}{60}=\dfrac{3}{20}\left(côngviệc\right)\)

=>Hai đội cần \(\dfrac{20}{3}\left(ngày\right)\) để hoàn thành công việc khi làm chung

\(\left(x^3-2x^2+x+4\right):\left(x+1\right)\)

\(=\left(x^3-3x^2+4x+x^2-3x+4\right):\left(x+1\right)\)

\(=\left[x\left(x^2-3x+4\right)+\left(x^2-3x+4\right)\right]:\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-3x+4\right):\left(x+1\right)\)

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

`x^2 -4x+4-y^2`

`=(x^2 -4x+4)-y^2`

`=(x-2)^2 -y^2`

`=(x-2-y)(x-2+y)`

`x^2+2xy+y^2-x-y`

`=(x^2+2xy+y^2) -(x+y)`

`=(x+y)^2 -(x+y)`

`=(x+y)(x+y-1)`

`x^2-2xy+y^2-9`

`=(x^2-2xy+y^2)-3^2`

`=(x-y)^2-3^3`

`=(x-y-3)(x-y+3)`

Tách ra đi cậu.

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\\x< >4\end{matrix}\right.\)

b: \(A=\dfrac{\left(\sqrt{x}-3\right)}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

c: A nguyên khi \(\sqrt{x}-2\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{3;1\right\}\)

Kết hợp ĐKXĐ, ta được: \(\sqrt{x}=3\)

=>x=9

a: \(\widehat{AIK}=180^0-\widehat{HAC}-\widehat{AKB}\)

\(=90^0-\widehat{HAC}+90^0-\widehat{AKB}\)

\(=\widehat{ABK}+\widehat{C}=\widehat{KBC}+\widehat{BAH}\)

b: \(\widehat{AKI}=90^0-\widehat{ABK}\)

\(\widehat{AIK}=\widehat{BIH}=90^0-\widehat{KBC}\)

mà \(\widehat{ABK}=\widehat{KBC}\)

nên \(\widehat{AKI}=\widehat{AIK}\)

\(x^2-2>-\sqrt{3}\)

=>\(x^2>2-\sqrt{3}\)

=>\(\left[{}\begin{matrix}x>\sqrt{2-\sqrt{3}}=\dfrac{1}{\sqrt{2}}\cdot\sqrt{4-2\sqrt{3}}=\dfrac{\sqrt{2}}{2}\left(\sqrt{3}-1\right)\\x< -\sqrt{2-\sqrt{3}}=-\dfrac{\sqrt{2}}{2}\left(\sqrt{3}-1\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>\dfrac{\sqrt{6}-\sqrt{2}}{2}\\x< \dfrac{-\sqrt{6}+\sqrt{2}}{2}\end{matrix}\right.\)

x² - 2 > -√3

⇔ x² > 2 - √3

⇔ x > √(2 - √3) hoặc x < -√(2-√3)

1/2 - (1/3 x - 2) = 2/3

1/3 x - 2 = 1/2 - 2/3

1/3 x - 2 = -1/6

1/3 x = -1/6 + 2

1/3 x = 11/6

x = 11/6 : 1/3

x = 11/2

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

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

\(\Rightarrow\dfrac{1}{3}x-2=-\dfrac{1}{6}\)

\(\Rightarrow\dfrac{1}{3}x=-\dfrac{1}{6}+2\)

\(\Rightarrow\dfrac{1}{3}x=\dfrac{11}{6}\)

\(\Rightarrow x=\dfrac{11}{6}:\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{11}{6}\cdot3=\dfrac{11}{2}\)