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Đk : x >= -70
Đặt : \(\sqrt{x+70}=a\); \(\sqrt{2x^2+4x+16}=b\)
=> 6x^2+10x-92 = 3b^2 - 2a^2
pt trở thành :
3b^2 - 2a^2 + ab = 0
<=> (3b^2+3ab)-(2ab+2a^2) = 0
<=> (a+b).(3b-2a) = 0
<=> a+b=0 hoặc 3b-2a = 0
<=> a=-b hoặc 2a=3b
Đến đó bạn tự thay vào mà làm nha
Tk mk nha
a)Đk:\(x\ge\frac{1}{2}\)
\(pt\Leftrightarrow4x^2-12x+4+4\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(2x-1\right)^2-4\left(2x-1\right)-1+4\sqrt{2x-1}=0\)
Đặt \(t=\sqrt{2x-1}>0\Rightarrow\hept{\begin{cases}t^2=2x-1\\t^4=\left(2x-1\right)^2\end{cases}}\)
\(t^4-4t^2+4t-1=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}t-1=0\\t^2+2t-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}t=1\\t=\sqrt{2}-1\end{cases}\left(t>0\right)}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2-\sqrt{2}\end{cases}}\) là nghiệm thỏa pt
ĐKXĐ:...
a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành:
\(3a^2-2b^2+ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow3a=2b\)
\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
Phương trình trở thành:
\(a^2+2+ab=3a+b\)
\(\Leftrightarrow a^2-3a+2+ab-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) \(3x^3+6x^2-4x=0\) \(\Leftrightarrow\) \(x\left(3x^2+6x-4\right)=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{3}\\x=\dfrac{-3-\sqrt{21}}{3}\end{matrix}\right.\end{matrix}\right.\)
vậy phương trình có 2 nghiệm \(x=0;x=\dfrac{-3+\sqrt{21}}{3};x=\dfrac{-3-\sqrt{21}}{3}\)
a)
DK: x\(\ge\)-2,x\(\ge\)\(\dfrac{1}{2}\)
=>\(\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0\)
\(\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0\)
\(\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0\)
\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)
<=>25x+50=2x-1
=>23x=-51
=>x=\(-\dfrac{51}{23}\)(ko thỏa mãn dk)
=> phương trình vô nghiệm..
b)
ĐKXĐ:\(x\ge1,x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)(nhận)
Vậy S={1;8}
c) ĐKXĐ:
\(x\ge0\)
\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)
\(\Leftrightarrow-11\sqrt{2x}=-11\)
\(\Leftrightarrow\sqrt{2x}=1\)
\(\Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)
Câu a :\(\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0\) ( ĐK : \(x\ge\dfrac{1}{2}\) )
\(\Leftrightarrow\sqrt{4x+8}+\sqrt{9x+18}=\sqrt{2x-1}\)
\(\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=\sqrt{2x-1}\)
\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)
\(\Leftrightarrow25\left(x+2\right)=2x-1\)
\(\Leftrightarrow25x+50=2x-1\)
\(\Leftrightarrow23x=-51\)
\(\Leftrightarrow x=-\dfrac{51}{23}< -\dfrac{1}{2}\)
Vậy phương trình vô nghiệm .
Câu b :
\(\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0\) ( ĐK : \(x\ge1\) )
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-3\sqrt{\left(x-1\right)}=0\)
\(\Leftrightarrow\sqrt{\left(x-1\right)}\left(\sqrt{x+1}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
Vậy \(S=\left\{1;8\right\}\)
Câu c : \(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\) ( ĐK : \(x\ge\dfrac{5}{6}\) )
\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)
\(\Leftrightarrow-11\sqrt{2x}+11=0\)
\(\Leftrightarrow-11\left(\sqrt{2x}-1\right)=0\)
\(\Leftrightarrow\sqrt{2x}-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
Chúc bạn học tốt
Điều kiện tự xử nhé!
\(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)(*)
Đặt \(a=\sqrt{x+70};\sqrt{2x^2+4x+16}=b\), (*) trở thành:
\(6x^2+10x-92+ab=0\)
\(\Leftrightarrow6x^2+12x+48-2x-140+ab=0\)
\(\Leftrightarrow3b^2-2a^2+ab=0\)
\(\Leftrightarrow3b^2+3ab-2ab-2a^2=0\)
\(\Leftrightarrow3b\left(a+b\right)-2a\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(3b-2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\3b=2a\end{matrix}\right.\)
Tới đây dễ rồi UwU
Chú ý rằng \(3\left(2x^2+4x+16\right)-2\left(x+70\right)=6x^2+10x-92\)
ĐKXĐ: \(x\ge-70\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+70}=a\ge0\\\sqrt{2x^2+4x+16}=b>0\end{matrix}\right.\) \(\Rightarrow a+b>0\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành: \(3a^2+ab-2b^2=0\Leftrightarrow\left(3a-2b\right)\left(a+b\right)=0\)
\(\Leftrightarrow3a-2b=0\Rightarrow3a=2b\)
\(\Rightarrow3\sqrt{x+70}=2\sqrt{2x^2+4x+16}\Leftrightarrow9\left(x+70\right)=4\left(2x^2+4x+16\right)\)
\(\Leftrightarrow8x^2+7x-566=0\Rightarrow\left[{}\begin{matrix}x=...\\x=...\end{matrix}\right.\)