Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
a, \(\left\{{}\begin{matrix}2x-3y=3\\-4x=3x-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\\-4x-3x=13\end{matrix}\right.\)\(\left\{{}\begin{matrix}-4x+6y=-6\\-4x-3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9y=-19\\-4x+6y=-6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{19}{9}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=3\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=9\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=2\\\dfrac{3}{x}+\dfrac{3}{y}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\left(TM\right)\\y=\dfrac{1}{2}\left(TM\right)\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=3\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{x}=16\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{16}\left(TM\right)\\y=\dfrac{13}{7}\left(TM\right)\end{matrix}\right.\)
d, \(\left\{{}\begin{matrix}\sqrt{x+1}-3\sqrt{y-1}=-4\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\left(x\ge-1,y\ge1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+1}-6\sqrt{y-1}=-8\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{y-1}=-10\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-1}=2\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\y=5\left(TM\right)\end{matrix}\right.\)
bạn có thể dùng phương pháp thế biến đổi pt (2) thành
x=2 căn y cộng 1 trên 4 y trừ 1
sau đó thế vào pt (1) tính ra y=1
sau đó thế y vào pt (1) hoặc pt (2) tính ra x=1
vậy pt có nghiệm duy nhất (1;1)
\(\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{1}{x+y}=\dfrac{3}{2\sqrt{x}}\\1-\dfrac{1}{x+y}=\dfrac{1}{2\sqrt{y}}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=4\\\dfrac{3}{\sqrt{x}}-\dfrac{1}{\sqrt{y}}=\dfrac{4}{x+y}\end{matrix}\right.\)Nhân vế theo vế, ta được:\(\dfrac{9}{x}-\dfrac{1}{y}=\dfrac{16}{x+y}\)
Qui đồng giải phương trình bậc nhât 2 ẩn...
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
a) \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\) (ĐKXĐ: \(x\ne-1;y\ne-4\))
Đặt \(\dfrac{x}{x+1}=a;\dfrac{1}{y+4}=b\left(a\ne0;b\ne0\right)\)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}3a-2b=4\left(1\right)\\2a-5b=9\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow2a=9+5b\Leftrightarrow a=\dfrac{9+5b}{2}\)
Thay \(a=\dfrac{9+5b}{2}\) vào \(\left(1\right)\), ta có:
\(\dfrac{3\left(9+5b\right)}{2}-2b=4\)
\(\Leftrightarrow27+15b-4b=8\)
\(\Leftrightarrow11b=-19\Leftrightarrow b=\dfrac{-19}{11}\)
Thay \(b=\dfrac{-19}{11}\) vào \(\left(2\right)\), ta có:
\(2a-5\cdot\dfrac{-19}{11}=9\)
\(\Leftrightarrow a=\dfrac{2}{11}\)
Với \(a=\dfrac{2}{11}\Rightarrow\dfrac{x}{x+1}=\dfrac{2}{11}\)
\(\Leftrightarrow11x=2x+2\Leftrightarrow x=\dfrac{2}{9}\)
Với \(b=\dfrac{-19}{11}\Rightarrow\dfrac{1}{y+4}=\dfrac{-19}{11}\)
\(\Leftrightarrow-19y-76=11\)
\(\Leftrightarrow y=\dfrac{-90}{19}\)
b,Ta có:
\(PT\Leftrightarrow7+3.\sqrt[3]{2+x}.\sqrt[3]{5-x}\left(\sqrt[3]{2+x}+\sqrt[3]{5-x}\right)=1\)
Thay \(\sqrt[3]{2+x}+\sqrt[3]{5-x}=1\) vào PT
\(\Rightarrow\) \(3.\sqrt[3]{2+x}.\sqrt[3]{5-x}=-6\)
\(\Leftrightarrow\sqrt[3]{2+x}.\sqrt[3]{5-x}=-2\)
\(\Leftrightarrow\left(2+x\right)\left(5-x\right)=-8\)
\(\Leftrightarrow x^2-3x-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
Thử lại thấy x= - 3, x=6 thỏa mãn
Vậy x= -3, x = 6
a: Đặt 1/x=a; 1/y=b
Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}3a+5b=-\dfrac{3}{2}\\5a-2b=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
b: Đặt \(\dfrac{1}{x+y-1}=a;\dfrac{1}{x-y+1}=b\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a-4b=\dfrac{-14}{5}\\3a+2b=-\dfrac{13}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-1=-1\\x-y+1=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
a) ĐKXĐ: \(x\ne0,\text{ }y\ne0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\), hệ phương trình đã cho trở thành:
\(\left\{{}\begin{matrix}3a+5b=\dfrac{-3}{2}\\5a-2b=\dfrac{8}{3}\end{matrix}\right.\)
Giải hệ này bằng phương pháp cộng đại số hoặc thế tìm được 1 nghiệm duy nhất: \(\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{-1}{2}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=\dfrac{-1}{2}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)(thỏa mãn ĐKXĐ)
Vậy hệ đã cho có 1 nghiệm duy nhất \(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\).
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5