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H9
HT.Phong (9A5)
CTVHS
20 tháng 11 2023
\(\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\end{matrix}\right.\left(x\ne1;y\ne-3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x-2+8}{x-1}+\dfrac{3y+9+5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2\left(x-1\right)}{x-1}+\dfrac{8}{x-1}+\dfrac{3\left(y+3\right)}{y+3}+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\2+\dfrac{8}{x-1}+3+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{8}{x-1}+\dfrac{5}{y+3}=13\end{matrix}\right.\) (I)
Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{x-1}\\v=\dfrac{1}{y+3}\end{matrix}\right.\)
Hệ (I) trở thành:
\(\Leftrightarrow\left\{{}\begin{matrix}12u+7v=19\\8u+5v=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}24u+14v=38\\24u+15v=39\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12u+7=19\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12u=12\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)
Trả ẩn phụ:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=1\\\dfrac{1}{y+3}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y+3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\left(tm\right)\)
Vậy hệ pt có 1 cặp nghiệm duy nhất là: (2;-2)
MK
20 tháng 11 2023
⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192x+6x−1+3y+14y+3=18(x≠1;y≠−3){12�−1+7�+3=192�+6�−1+3�+14�+3=18(�≠1;�≠−3)
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192x−2+8x−1+3y+9+5y+3=18⇔{12�−1+7�+3=192�−2+8�−1+3�+9+5�+3=18
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192(x−1)x−1+8x−1+3(y+3)y+3+5y+3=18⇔{12�−1+7�+3=192(�−1)�−1+8�−1+3(�+3)�+3+5�+3=18
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192+8x−1+3+5y+3=18⇔{12�−1+7�+3=192+8�−1+3+5�+3=18
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=198x−1+5y+3=13⇔{12�−1+7�+3=198�−1+5�+3=13 (I)
Đặt: ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩u=1x−1v=1y+3{�=1�−1�=1�+3
Hệ (I) trở thành:
⇔{12u+7v=198u+5v=13⇔{12�+7�=198�+5�=13
⇔{24u+14v=3824u+15v=39⇔{24�+14�=3824�+15�=39
⇔{12u+7=19v=1⇔{12�+7=19�=1
⇔{12u=12v=1⇔{12�=12�=1
⇔{u=1v=1⇔{�=1�=1
Trả ẩn phụ:
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1x−1=11y+3=1⇔{1�−1=11�+3=1
⇔{x−1=1y+3=1⇔{�−1=1�+3=1
⇔{x=2y=−2(tm)⇔{�=2�=−2(��)
Vậy hệ pt có 1 cặp nghiệm duy nhất là: (2;-2)
27 tháng 7 2021
4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)
28 tháng 9 2021
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
5 tháng 5 2017
Câu 2/
Điều kiện xác định b tự làm nhé:
\(\frac{6}{x^2-9}+\frac{4}{x^2-11}-\frac{7}{x^2-8}-\frac{3}{x^2-12}=0\)
\(\Leftrightarrow x^4-25x^2+150=0\)
\(\Leftrightarrow\left(x^2-10\right)\left(x^2-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=10\\x^2=15\end{cases}}\)
Tới đây b làm tiếp nhé.
CH
Cô Hoàng Huyền
Admin
VIP
6 tháng 5 2017
a. ĐK: \(\frac{2x-1}{y+2}\ge0\)
Áp dụng bđt Cô-si ta có: \(\sqrt{\frac{y+2}{2x-1}}+\sqrt{\frac{2x-1}{y+2}}\ge2\)
\(\)Dấu bằng xảy ra khi \(\frac{y+2}{2x-1}=1\Rightarrow y+2=2x-1\Rightarrow y=2x-3\)
Kết hợp với pt (1) ta tìm được x = -1, y = -5 (tmđk)
b. \(pt\Leftrightarrow\left(\frac{6}{x^2-9}-1\right)+\left(\frac{4}{x^2-11}-1\right)-\left(\frac{7}{x^2-8}-1\right)-\left(\frac{3}{x^2-12}-1\right)=0\)
\(\Leftrightarrow\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}+\frac{1}{x^2-8}+\frac{1}{x^2-12}\right)=0\)
\(\Leftrightarrow x^2-15=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)
5 tháng 7 2021
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm