(2x+15)⋮(x-2) tìm x e N
mng giải giúp mik ạ mik cảm ơn 🥰
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\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)
\(\Leftrightarrow3x^2-6x+3+9x-21=0\)
\(\Leftrightarrow3x^2+3x-18=0\)
\(\Leftrightarrow3\left(x^2+x-6\right)=0\)
\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy \(S=\left\{2;-3\right\}\)
\(1,\sqrt{5x^2-2x+2}=x+1\)
\(\Leftrightarrow\left(\sqrt{5x^2-2x+2}\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow5x^2-2x+2=x^2+2x+1\)
\(\Leftrightarrow5x^2-x^2-2x-2x=1-2\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
\(2,\sqrt{4x^2-x+1}-2x=3\)
\(\Leftrightarrow\left(\sqrt{4x^2-x+1}\right)^2=\left(3+2x\right)^2\)
\(\Leftrightarrow4x^2-x+1=9+12x+4x^2\)
\(\Leftrightarrow4x^2-4x^2-x-12x=9-1\)
\(\Leftrightarrow-13x=8\)
\(\Leftrightarrow x=-\dfrac{8}{13}\)
Vậy \(S=\left\{-\dfrac{8}{13}\right\}\)
1: =>x>=-1 và 5x^2-2x+2=x^2+2x+1
=>x>=-1 và 4x^2-4x+1=0
=>x=1/2
2: =>\(\sqrt{4x^2-x+1}=2x+3\)
=>x>=-3/2 và 4x^2-x+1=4x^2+12x+9
=>x>=-3/2 và -11x=8
=>x=-8/11(nhận)
Lời giải:
Theo đề ta có:
\(\text{sđc(AD)}=\frac{1}{3}\text{sđc(AB)}=\frac{1}{9}[\text{sđc(AB)+sđc(BC)+sđc(CD)}]\)
\(=\frac{1}{9}(360^0-\text{sđc(AD)})\)
\(\Rightarrow \text{sđc(AD)}=36^0\)
\(\widehat{BEC}=\frac{\text{sđc(BC)-sđc(AD)}}{2}=\frac{3\text{sđc(AD)}-\text{sđc(AD)}}{2}=\text{sđc(AD)}=36^0\)
3 . ( 2x - 1 ) - 2 = 13
3 . ( 2x - 1 ) = 12 + 3
3 . ( 2x - 1 ) = 15
2x - 1 = 15 : 3
2x - 1 = 5
2x = 5 + 1 = 6
x = 6 : 2 = 3
Vậy x = 3
\(3\left(2x-1\right)-2=13\)
\(3\left(2x-1\right)=15\)
\(2x-1=5\)
\(2x=6\)
\(x=3\)
3:
#include <bits/stdc++.h>
using namespace std;
double x,y;
int main()
{
cin>>x>>y;
cout<<fixed<<setprecision(2)<<sqrt(x*x+y*y);
return 0;
}
Bn ơi bn viết r chụp lên đc k ạ ? Mik k định dạng đc ý
N(x) = 2x + x3 + x2 - 4x - x3
= x2 - 2x
N(x) = 0 <=> x2 - 2x = 0
<=> x(x - 2) = 0
<=> x = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = 2
Vậy nghiệm của N(x) là 0 và 2
\(N\left(x\right)=2x+x^3+x^2-4x-x^3=x^2-2x=x\left(x-2\right)\)
Để N(x) có nghiệm => x(x-2)=0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
Vậy x=0; x=2
59 saving
60 to reward
61 haven't seen
62 to buy
63 spoke
64 stays
65 did-do
66 row
67 go
68 sets
69 will build
70 am working
71 were-doing
72 to do
74 will go
75 don't eat
76 gets
78 was visiting
79 is shining
80 Was
83 went
84 reads
91 will buy
92 takes
93 have taught
94 has lived
95 was
96 came, were playing
97 has just won
98 has already begun
99 goes
100 drinking
101 has worked
102 has
103 hasn't washed
104 drink
105 had run
106 is giving
\(\dfrac{2\left(x-2\right)+19}{x-2}=2+\dfrac{19}{x-2}\Rightarrow x-2\inƯ\left(19\right)=\left\{\pm1;\pm19\right\}\)
ai giúp mik với ạ:(