BÀI 1

1.a

2.c

3.d

4.d

9)Ta có: \(\sqrt{x^2-6x+9}=5\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=5\)

\(\Leftrightarrow\left|x-3\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy: S={8;-2}

19) Ta có: \(\sqrt[3]{x^3+9x^2}=x+3\)

\(\Leftrightarrow x^3+9x^2=\left(x+3\right)^3\)

\(\Leftrightarrow x^3+9x^2=x^3+9x^2+27x+27\)

\(\Leftrightarrow27x=-27\)

hay x=-1

Vậy: S={-1}

miêu tả

chúc bạn hok tốt

miêu tả+ biểu cảm

C - B - D

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Yes, they can

To help consumers know what they are buying in a paper product

No, they aren’t

It means the paper that you and I return to recycling centers

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A - D - C - A - A - B - C - A - B - C

erase - spend - was invented - has done - belives - have been given - think - represent - are given - saw/putting - are advised

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D - ? - C - C - D - D - D - C - D - D

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D - A - A

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B - D - A - A - D - B - C

3m 27cm=3,27m

3m27cm=3,27m hok toot

3m 27cm = 3,27 m

\(\text{#TNam}\) 

`5,A`

Gọi các cạnh của Tam giác `ABC` lần lượt là `x,y,z (x,y,z \ne 0)`

`3` góc của tam giác lần lượt tỉ lệ với `2:3:4`

Nghĩa là: `x/2=y/3=z/4`

Tổng số đo `3` góc trong `1` tam giác là `180^0`

`-> x+y+z=180`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`x/2=y/3=z/4=`\(\dfrac{x+y+z}{2+3+4}=\dfrac{180}{9}=20\) 

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=20\\\dfrac{y}{3}=20\\\dfrac{z}{4}=20\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\cdot2=40\\y=20\cdot3=60\\z=20\cdot4=80\end{matrix}\right.\) 

Vậy, độ dài các cạnh của Tam giác `ABC` lần lượt là `40^0, 60^0, 80^0`.

`6,B`

Gọi số người thợ của `3` nhóm lần lượt là `x,y,z (x,y,z \ne 0)`

Vì năng suất làm việc của các người thợ như nhau `->` số thợ và số ngày là `2` đại lượng tỉ lệ nghịch

`-> 40x=60y=50z` hay \(\dfrac{x}{\dfrac{1}{40}}=\dfrac{y}{\dfrac{1}{60}}=\dfrac{z}{\dfrac{1}{50}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{\dfrac{1}{40}}=\dfrac{y}{\dfrac{1}{60}}=\dfrac{z}{\dfrac{1}{50}}=\dfrac{x-z}{\dfrac{1}{40}-\dfrac{1}{50}}=\dfrac{3}{\dfrac{1}{200}}=600\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{\dfrac{1}{40}}=600\\\dfrac{y}{\dfrac{1}{60}}=600\\\dfrac{z}{\dfrac{1}{50}}=600\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=600\cdot\dfrac{1}{40}=15\\y=600\cdot\dfrac{1}{60}=10\\z=600\cdot\dfrac{1}{50}=12\end{matrix}\right.\) 

Vậy, số thợ của nhóm `1,2,3` lần lượt là `15,10,12`.

Giải cho mình bài 5A và bài 6B thôi nhé

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