Giải hệ pt:
\(\left\{{}\begin{matrix}\dfrac{60}{x_1+x_2}=\dfrac{4}{3}\\\dfrac{60}{x_2}_{_{ }}-\dfrac{60}{x_1}=2\end{matrix}\right.\)
K
Khách
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30 tháng 12 2022
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1-10\right)=\left(x_2-10\right)=\left(x_3-10\right)=...=\left(x_9-10\right)\\x_1+x_2+x_3+...+x_9=90\end{matrix}\right.\)
=>x1=x2=x3=...=x9=10
10 tháng 8 2021
,có \(ac< 0\)=>pt đã cho luôn có 2 nghiệm phân biệt
vi ét \(=>\left\{{}\begin{matrix}x1+x2=2\\x1x2=-1\end{matrix}\right.\)
a,\(A=\left(x1+x2\right)^2-2x1x2=.....\) thay số tính
b,\(B=\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)=.......\)
c,\(C=x1^{2^2}+x2^{2^2}=\left(x1^2+x2^2\right)^2-2\left(x1x2\right)^2=\left[\left(x1+x2\right)^2-2x1x2\right]^2-2\left(x1x2\right)^2=....\)
\(D=x1x2\left(x1+x2\right)=.....\)
\(x1,x2\ne0=>E=\dfrac{\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)}{x1x2}=...\)
\(F=\sqrt{\left(x1-x2\right)^2}=\sqrt{\left(x1+x2\right)^2-4x1x2}=....\)
\(x1,x2\ne-1=>G=\dfrac{\left(x1+x2\right)^2-2x1x2+x1x2}{x1x2+x1+X2+1}=...\)
\(x1,x2\ne0=>H=\left(\dfrac{x1x2+2}{x2}\right)\left(\dfrac{x1x2+2}{x1}\right)=\dfrac{\left(x1x2+2\right)^2}{x1x2}\)
\(=\dfrac{\left(x1x2\right)^2+4x1x2+4}{x1x2}=..\)
N
11 tháng 1 2019
\(\left\{{}\begin{matrix}x_1+x_2+...+x_{2000}=a\left(1\right)\\x_1^2+x_2^2+...+x_{2000}^2=a^2\left(2\right)\\x_1^{2000}+x_2^{2000}+...+x_{2000}^{2000}=a^{2000}\left(3\right)\end{matrix}\right.\)
Từ (2)(3)\(\Rightarrow2\left(x_1x_2+x_2x_3+...+x_{2000}x_1\right)=0\)
\(\Rightarrow x_1=x_2=...=x_{2000}=0\)
Vậy hpt có nghiệm là x=0.
Đúng không ạ?
7 tháng 10 2021
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
7 tháng 10 2021
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
16 tháng 9 2017
coi như đoạn trên bạn đúng nhé (làm tiếp)
\(S=\dfrac{m^2+2m}{m^2+2m+2017}=\dfrac{m^2+2m+2017-2017}{m^2+2m+2017}=1-\dfrac{2017}{\left(m+1\right)^2+2016}\)
có \(s_1=\left(m+1\right)^2+2016\ge2016\Rightarrow\dfrac{1}{\left(m+1\right)^2+2016}\le\dfrac{1}{2016}\)\(\Rightarrow-\dfrac{1}{\left(m+1\right)^2+2016}\ge\dfrac{1}{2016}\)
\(\Rightarrow\Rightarrow-\dfrac{2017}{\left(m+1\right)^2+2016}\ge\dfrac{-2017}{2016}\)
\(\Rightarrow\Rightarrow\Rightarrow1-\dfrac{2017}{\left(m+1\right)^2+2016}\ge1-\dfrac{2017}{2016}=\dfrac{-1}{2016}\)
\(S\ge-\dfrac{1}{2016}\)
đẳng thức khi m =-1
AH
Akai Haruma
Giáo viên
20 tháng 5 2022
$|x_1-x_2|=2$ sẵn rồi thì việc gì phải tính nữa bạn?
20 tháng 5 2022
ý em là rút gọn biểu thức đó rồi áp dụng hệ thức vi ét để tìm n ( n là ẩn khi thay vào )
Lời giải:
PT (1)\(\rightarrow x_1+x_2=\frac{60.3}{4}=45\)
\(\Rightarrow x_2=45-x_1\)
Thay vào pt (2)
\(\frac{60}{x_2}-\frac{60}{x_1}=2\)
\(\Leftrightarrow \frac{60}{45-x_1}-\frac{60}{x_1}=2\)
\(\Leftrightarrow \frac{1}{45-x_1}-\frac{1}{x_1}=\frac{1}{30}\Leftrightarrow \frac{x_1-(45-x_1)}{x_1(45-x_1)}=\frac{1}{30}\)
\(\Leftrightarrow 30(2x_1-45)=x_1(45-x_1)\)
\(\Leftrightarrow x_1^2+15x_1-1350=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=30\rightarrow x_2=15\\x_1=-45\rightarrow x_2=90\end{matrix}\right.\)
(đều thỏa mãn)
Vậy \((x_1,x_2)=(30;15);(-45;90)\)