Giúp t vs mn
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Câu 2:
\(a,\Leftrightarrow\Delta'=\left(1-m\right)^2-\left(m^2-m\right)>0\\ \Leftrightarrow m^2-2m+1-m^2+m>0\\ \Leftrightarrow1-m>0\Leftrightarrow m< 1\\ b,\text{Áp dụng Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(1-m\right)\\x_1x_2=m^2-m\end{matrix}\right.\\ \left(2x_1-1\right)\left(2x_2-1\right)-x_1x_2=1\\ \Leftrightarrow2x_1x_2-2\left(x_1+x_2\right)+1-x_1x_2=1\\ \Leftrightarrow x_1x_2-2\left(x_1+x_2\right)=0\\ \Leftrightarrow m^2-m-4\left(1-m\right)=0\\ \Leftrightarrow m^2+3m-4=0\\ \Leftrightarrow\left(m-1\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-4\left(tm\right)\end{matrix}\right.\)
Vậy m=-4
Câu 1:
\(1,\Leftrightarrow2x-2=3\Leftrightarrow x=\dfrac{5}{2}\\ 2,ĐK:x\ne\pm1\\ PT\Leftrightarrow\dfrac{2x^2+2x-1}{x^2-1}=2\\ \Leftrightarrow2x^2+2x-1=2x^2-2\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\\ 3,\Leftrightarrow\left[{}\begin{matrix}3x-2=2x-1\\3x-2=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\)
\(4,\Leftrightarrow\left[{}\begin{matrix}3x-1=2-x\left(x\ge\dfrac{1}{3}\right)\\3x-1=x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(tm\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\\ 5,\Leftrightarrow4x^2-2x+10=9x^2-6x+1\left(x\le\dfrac{1}{3}\right)\\ \Leftrightarrow5x^2-4x-9=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(6,\Leftrightarrow3x^2-9x+1=x^2-4x+4\left(x\ge2\right)\\ \Leftrightarrow2x^2-5x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ 7,\Leftrightarrow2x^2+3x-4=7x+2\left(x\ge-\dfrac{2}{7}\right)\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
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1. likes
2. enjoy
3. welcomed
4. doesn't like
5. works
6. eat - sing
7. don't have
8. make
9. eat - light
10. decorate - get
11. clean
12. does
13. don't like
14. doesn't like
15. to go
16. watch
17. plays - gets
18. starts - ends
19. is - need
20. to eat
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Cách 1: Cho lai phân tích
Nếu đời con cho cả quả đỏ và quả vàng thì kiểu gen của cây cà chua quả đó là Aa.
Nếu đời con chỉ có quả vàng thì kiểu gen của cây quả đó là AA
- Cách 2. Cho cây tự thụ phấn
Nếu đời con cho cả quả đỏ và quả vàng thì kiểu gen của cây cà chua quả đó là Aa.
Nếu đời con chỉ có quả vàng thì kiểu gen của cây quả đó là AA
![](https://rs.olm.vn/images/avt/0.png?1311)
-472 + (235 - 28) - (35 - 350)
= -472 + 235 - 28 - 35 + 350
= (-472 - 28) + (235 - 35) + 350
= -500 + 200 + 350
= -500 + 550
= 50
![](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)
Uses crt;
var i,t: integer;
begin clrscr;
i:=0;
while(i<=120) do begin
i:=i+1;
t:=t+i;
end;
write(t);
readln;
end.
a: \(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{x-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1-x+\sqrt{x}+2}\)
\(=\dfrac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}=\dfrac{x}{\sqrt{x}-1}\)
b; Để E>1 thì E-1>0
=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
=>\(\sqrt{x}-1>0\)
=>x>1
c: \(E=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)
=>\(E>=2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\dfrac{1}{\sqrt{x}-1}}+2=4\)
Dấu = xảy ra khi \(\left(\sqrt{x}-1\right)^2=1\)
=>\(\left[{}\begin{matrix}\sqrt{x}-1=-1\\\sqrt{x}-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=0\left(loại\right)\end{matrix}\right.\)
d: Để E là số nguyên thì \(x⋮\sqrt{x}-1\)
=>\(x-1+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0\right\}\)
=>\(x\in\left\{4;0\right\}\)
Kết hợp ĐKXĐ, ta được: x=4
e: E=9/2
=>\(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)
=>\(2x=9\sqrt{x}-9\)
=>\(2x-3\sqrt{x}-6\sqrt{x}+9=0\)
=>\(\left(2\sqrt{x}-3\right)\left(\sqrt{x}-3\right)=0\)
=>x=9 hoặc x=9/4
a) \(E=\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left[\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\dfrac{x}{\sqrt{x}-1}\)
b) \(E>1\) khi:
\(\dfrac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
Mà:
\(x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
\(\Rightarrow x>1\)
c) Ta có:
\(E=\dfrac{x}{\sqrt{x}-1}\) với \(x>1\)
\(E=\dfrac{x-1+1}{\sqrt{x}-1}\)
\(E=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\)
\(E=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\)
\(E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)
\(\Rightarrow E\ge2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\dfrac{1}{\sqrt{x}-1}}+2=2\cdot1+2=4\)
Dấu "=" xảy ra:
\(\left(\sqrt{x}-1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\left(ĐK:x>1\right)\)
Vậy: ...
d) \(E\in Z\) khi:
\(\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\in Z\)
\(\Rightarrow1\) ⋮ \(\sqrt{x}-1\)
\(\Rightarrow\sqrt{x}-1\) \(\in\) Ư(1)
Mà: \(Ư\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2;0\right\}\)
\(\Rightarrow x\in\left\{4;0\right\}\)
Vậy: ...
e) \(E=\dfrac{9}{2}\) khi:
\(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)
\(\Leftrightarrow2x=9\sqrt{x}-9\)
\(\Leftrightarrow2x-9\sqrt{x}+9=0\)
\(\Leftrightarrow2x-6\sqrt{x}-3\sqrt{x}+9\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-3\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\2\sqrt{x}-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3^2\\x=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=\dfrac{9}{4}\left(tm\right)\end{matrix}\right.\)