giúp e phần viet vs ạ
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Bài 3:
a: Xét ΔADC có
\(\dfrac{AM}{MD}=\dfrac{AP}{PC}\)
Do đó: MP//DC
Xét ΔCAB có
\(\dfrac{CQ}{QB}=\dfrac{CP}{PA}\)
Do đó: PQ//AB
hay PQ//CD
Xét ΔBCD có
\(\dfrac{BQ}{QC}=\dfrac{BN}{ND}\)
Do đó: NQ//DC
Ta có: PQ//CD
NQ//DC
mà PQ và NQ có điểm chung là Q
nên Q,P,N thẳng hàng(1)
Ta có: PQ//CD
PM//CD
mà PQ và PM có điểm chung là P
nên M,P,Q thẳng hàng(2)
Từ (1) và (2) suy ra M,N,P,Q thẳng hàng
1: Khi x=3-2 căn 2 thì \(A=\dfrac{\sqrt{2}-1+2}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
2: \(B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
3: \(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x-4}{x}\)
\(x\cdot P< =10\sqrt{x}-29-\sqrt{x-25}\)
=>\(x-4< =10\sqrt{x}-29-\sqrt{x-25}\)
\(\Leftrightarrow x-4-10\sqrt{x}+29< =-\sqrt{x-25}\)
=>\(x-10\sqrt{x}+25< =-\sqrt{x-25}\)
=>(căn x-5)^2<=-căn x-25
=>x-25=0
=>x=25
1 traditional
2 Traditionally
3 enjoyable
4 villagers
5 enjoyably
6 entertaining
7 entertainment
8 peace
9 peaceful
10 nature
11 naturally
12 noisily
13 Noise
14 peacefully
15 natural
16 quietly
17 refreshment
18 entertainments
19 entertainment
20 peaceful
1. We decorate the house before Tet
2. In the New Year's Eve, they watch fireworks display
3. What do you do at Christmast
4. Children get a lot of present on Children's Day
5. Teachers's Day is on the twentieth of November
1. decorate / the / we / house / Tet / before
We decorate the house before Tet.
2. in / the / Eve / they / New Year’s / watch / displays / firework
They watch firework displays in the New Year's Eve.
3. you / Christmas / do / what / at / do
What do you do at Christmas?
4. a / presents / of / get / Children’s / lot / on / Children / Day
Children get a lot of presents on Children's Day.
5. twentieth / Teachers’ / is / the / Day / on / of / November
Teacher's Day is on the twentieth of November.
e: \(\left(-4156+2021\right)-\left(119+2021-4156\right)\)
\(=-4156+2021-119-2021+4156\)
\(=\left(-4156+4156\right)+\left(2021-2021\right)-119\)
=0+0-119
=-119
g: \(315\cdot75-\left(15\cdot100-315\cdot25\right)\)
\(=315\cdot75-15\cdot100+315\cdot25\)
\(=315\left(75+25\right)-15\cdot100\)
\(=315\cdot100-15\cdot100=300\cdot100=30000\)
h: \(\left(-489\right)\cdot125-\left(125\cdot11-500\cdot25\right)\)
\(=-489\cdot125-125\cdot11+500\cdot25\)
\(=125\left(-489-11\right)+500\cdot25\)
\(=125\cdot\left(-500\right)+500\cdot25\)
\(=500\left(-125+25\right)\)
\(=500\cdot\left(-100\right)=-50000\)
Bài 2:
a: \(-415-3\left(2x-1\right)^2=-490\)
=>\(3\left(2x-1\right)^2+415=490\)
=>\(3\left(2x-1\right)^2=75\)
=>\(\left(2x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Pt hoành độ giao điểm:
\(x^2=mx+2\Leftrightarrow x^2-mx-2=0\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=-2\end{matrix}\right.\)
\(\Rightarrow\) \(x_1;x_2\) trái dấu
Mà \(\left|x_2\right|+1>0;\forall x_2\Rightarrow\dfrac{4}{x_1}>0\Rightarrow x_1>0\)
\(\Rightarrow x_2< 0\)
\(\Rightarrow\left|x_2\right|=-x_2\)
Đồng thời: \(x_1x_2=-2\Rightarrow x_2=-\dfrac{2}{x_1}\Rightarrow-2x_2=\dfrac{4}{x_1}\)
Do đó ta có:
\(\dfrac{4}{x_1}=\left|x_2\right|+1\)
\(\Rightarrow-2x_2=-x_2+1\)
\(\Leftrightarrow x_2=-1\)
Thế vào \(x_1x_2=-2\Rightarrow x_1=2\)
Thế vào \(x_1+x_2=m\)
\(\Rightarrow m=2+\left(-1\right)=1\)