Giúp mình với ạ, mình đang cần vội
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DT
Đỗ Thanh Hải
CTVVIP
1 tháng 8 2021
1 Where are the picture which were on this wall
2 My father is the person who has great influence on my behavior
3 Do you like the flowers which I bought this morning?
4 All of us are interested in the man who is teaching us PE
5 What's the name of the girl who lent you this book?
6 I don't know the man who wrote lots of good books
7 The girl who has worked here for 5 days will leave me the message
8 All the man who are businessman attended the meeting
9 Do you know the lady who is presenting the topic
10 I haven't read the novel which was written by Nam cao
9 tháng 11 2021
vì đưởng thẳng c đi qua a và b tạo thành 2 cặp góc vuông bằng nhau
⇒ a//b
9 tháng 11 2021
b) Do B1 và A1 ở vị trí trong cùng phía :
=>a1+b1=180 độ.
=>B1=180 độ - 120 độ
B1=60 độ
M
2
25 tháng 6 2021
19) Ta có: \(\sqrt[3]{x^3+9x^2}=x+3\)
\(\Leftrightarrow x^3+9x^2=\left(x+3\right)^3\)
\(\Leftrightarrow x^3+9x^2=x^3+9x^2+27x+27\)
\(\Leftrightarrow27x+27=0\)
\(\Leftrightarrow27x=-27\)
hay x=-1
Vậy: S={-1}
6) Ta có: \(\sqrt{9x^2-6x+1}-x=4\)
\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}=x+4\)
\(\Leftrightarrow\left|3x-1\right|=x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x+4\left(x\ge\dfrac{1}{3}\right)\\1-3x=x+4\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-x=4+1\\-3x-x=4-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\-4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(nhận\right)\\x=\dfrac{-3}{4}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{-3}{4}\right\}\)
25 tháng 6 2021
8)
ĐKXĐ: \(x>2\)
Ta có: \(\sqrt{x^2+2x+4}=x-2\)
\(\Leftrightarrow x^2+2x+4=\left(x-2\right)^2\)
\(\Leftrightarrow x^2+2x+4-x^2+4x-4=0\)
\(\Leftrightarrow6x=0\)
hay x=0(loại)
Vậy: \(S=\varnothing\)
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}
DD
0
BT
7
SS
15 tháng 2 2022
\(1-\frac{1}{3}-\frac{1}{9}=\frac{9}{9}-\frac{3}{9}-\frac{1}{9}=\frac{9-3-1}{9}=\frac{5}{9}\)
\(a,2\left|3x-1\right|+1=5\\ \Rightarrow2\left|3x-1\right|=5-1\\ \Rightarrow2\left|3x-1\right|=4\\ \Rightarrow\left|3x-1\right|=4:2\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(b,\left|\dfrac{x}{2}-1\right|=3\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
\(c,\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3\\ \Rightarrow\left[{}\begin{matrix}-x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-x=\dfrac{13}{5}\\-x=-\dfrac{17}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{13}{5}\\x=\dfrac{17}{5}\end{matrix}\right.\)
\(d,\left|x-\dfrac{1}{3}\right|=2\dfrac{3}{5}\\ \Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{13}{5}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{13}{5}\\x-\dfrac{1}{3}=-\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}+\dfrac{1}{3}\\x=-\dfrac{13}{5}+\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{44}{15}\\x=-\dfrac{34}{15}\end{matrix}\right.\)