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PT 2
\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))
\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Rightarrow2x^2-3x+6=0\)
=> PT vô nghiệm.
c: \(\Leftrightarrow2x+2x-6=12-2x\)
=>4x-6=12-2x
=>6x=18
hay x=3
b: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)+x=2x-1\)
\(\Leftrightarrow x^2-1+x=2x-1\)
=>x2-x=0
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
lớp 9 gì như lớp 6 thế
a) đề sai
c) <=>x/3 +x/3 -1 =2-x/3
<=>3.x/3 =3 => x=3
b) x<> 0; -2 <=>
x^2 -1 +x =2x-1
<=>x^2 -x =0 => x =0 (l) x =1 nhận
d ; <=> (x+1)/65+1 +(x+3)/63 +1 =(x+5)/61+1 +(x+7)/59+1
<=>(x+66) [1/65+1/63-1/61-1/59] =0
[...] khác 0
x=-66
\(\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\end{matrix}\right.\left(x\ne1;y\ne-3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x-2+8}{x-1}+\dfrac{3y+9+5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2\left(x-1\right)}{x-1}+\dfrac{8}{x-1}+\dfrac{3\left(y+3\right)}{y+3}+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\2+\dfrac{8}{x-1}+3+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{8}{x-1}+\dfrac{5}{y+3}=13\end{matrix}\right.\) (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)
⎪ ⎪⎨⎪ ⎪ ⎪⎩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)
\(ĐK:x\ne3\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(-x-1+x^2-2x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\x^2-3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=\dfrac{3+\sqrt{41}}{2}\\x=\dfrac{3-\sqrt{41}}{2}\end{matrix}\right.\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290