Giúp e vs ạ
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1. ĐKXĐ: \(x\ne5\)
\(\dfrac{3x+1}{x-5}+\dfrac{-2x-6}{x-5}\)
\(=\dfrac{3x+1-2x-6}{x-5}\)
\(=\dfrac{x-5}{x-5}\)
\(=1\)
2. ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x+1}{2x-2}-\dfrac{x^2+3}{2x^2-2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x^2-1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-2}{\left(2x-2\right)\left(x+1\right)}\)
\(=\dfrac{1}{x+1}\)
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Trắc nghiệm
Câu 1:B
Câu 2:C
Câu 3:A
Câu 4:C
Câu 5:A
Câu 6:B
Câu 7:D
Câu 8:D
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a: BC=căn 6^2+8^2=10cm
AM=BC/2=5cm
b:
ΔAEH vuông tại A có AI là trung tuyến
nên IH=IA
=>góc IHA=góc IAH
góc IAH+góc MAB
=góc MBA+góc IHA=90 độ
=>góc IAM=90 độ
=>AI vuông góc AM
![](https://rs.olm.vn/images/avt/0.png?1311)
1 telling
2 to come
3 having
4 talking
5 to speak
6 giving
7 carry
8 to study
9 waiting
10 to start
11 to help
12 going
13 to bring
14to visit
15 going
--------------------------------
1 preparing
2 working - finishing
3 to give - smoking
4 talking - eating
5 arguing - working
6 to think - making
7 to come - standing
8 solving
9to lock - going
10 to persuade - change
-------------------------------------
1 watching - reading
2 playing - doing
3 to go
4 Did you see
5 to dream - were
6 showing - to send
7 going
8 doing
9 reading
10 to seeing
#\(Vion.Serity\)
#\(yGLinh\)
1 telling
2 to come
3 having
4 talking
5 to speak
6 giving
7 carry
8 to study
9 waiting
10 to start
11 to help
12 going
13 to bring
14to visit
15 going
--------------------------------
1 preparing
2 working - finishing
3 to give - smoking
4 talking - eating
5 arguing - working
6 to think - making
7 to come - standing
8 solving
9to lock - going
10 to persuade - change
-------------------------------------
1 watching - reading
2 playing - doing
3 to go
4 Did you see
5 to dream - were
6 showing - to send
7 going
8 doing
9 reading
10 to seeing
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\Rightarrow3:\dfrac{9}{4}=\dfrac{3}{4}:x\\ \Rightarrow\dfrac{4}{3}=\dfrac{3}{4}:x\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{4}{3}=\dfrac{9}{16}\\ 2,\)
Nửa chu vi là \(50:2=25\left(cm\right)\)
Gọi cd là a, cr là b (cm)(a,b>0)
Ta có \(a:b=3:2\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\) và \(a+b=25\left(cm\right)\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{a+b}{3+2}=\dfrac{25}{5}=5\\ \Rightarrow\left\{{}\begin{matrix}a=15\\b=10\end{matrix}\right.\)
Vậy ...
25. Hàm \(y=ax^3+bx^2+cx+d\) có pt đường thẳng qua 2 cực trị dạng:
\(y=\left(\dfrac{2c}{3}-\dfrac{2b^2}{9a}\right)x+d-\dfrac{bc}{9a}\)
Ở bài này a=1;b=0, c=-3, d=1 thay vào công thức trên ta được:
\(y=-2x+1\) hay \(y=1-2x\)
30.
\(\left\{{}\begin{matrix}y'=3x^2-2mx+2m-3\\y''=6x-2m\end{matrix}\right.\)
Hàm đạt cực đại tại x=1 khi: \(\left\{{}\begin{matrix}y'\left(1\right)=0\\y''\left(1\right)< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3-2m+2m-3=0\\6-2m< 0\end{matrix}\right.\) \(\Rightarrow m>3\)
Bạn cần câu nào trong 3 câu này nhỉ?