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3 x - 2 6 - 5 = 3 - 2 x + 7 4
⇔ 2(3x – 2) – 5.12 = 3[3 – 2(x + 7)]
⇔ 6x – 4 – 60 = 9 – 6(x + 7)
⇔ 6x – 64 = 9 – 6x – 42
⇔ 6x + 6x = 9 – 42 + 64
⇔ 12x = 31 ⇔ x = 31/12
Phương trình có nghiệm x = 31/12
\((3x+\frac{1}{4})-\frac{1}{3}\times(6x+\frac{9}{5})=1\)1
<=> \(3x+\frac{1}{4}-2x-\frac{3}{5}=1\)
<=>\(x-\frac{7}{20}=1\)
<=>\(20x-7=20\)
<=>\(20x=20+7\)
<=>\(20x=27\)
<=> \(x=\frac{27}{20}\)
\(\left(3x+\frac{1}{4}\right)-\frac{1}{3}.\left(6x+\frac{9}{5}\right)=1\)
\(\Leftrightarrow3x+\frac{1}{4}-2x-\frac{3}{5}=1\)
\(\Leftrightarrow x=1+\frac{3}{5}-\frac{1}{4}\)
\(\Leftrightarrow x=\frac{27}{20}\)
Thay thế t cho x\(^2\)
Ta có: \(3t^2-12t+9=0\)
Áp dụng phương trình bặc 2:
t=\(\dfrac{-\left(-12\right)\pm\sqrt{\left(-12\right)^2-4.3.9}}{2.3}\)
t=\(\dfrac{12\pm6}{6}\)
=>\(\left\{{}\begin{matrix}t=3\\t=1\end{matrix}\right.\)
Vì thay x\(^2\)=t nên giá trị x=\(\sqrt{t}\)
=>\(\left\{{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\\x=1\\x=-1\end{matrix}\right.\)
Mình mới lớp 8 nên chưa quen giải PT bậc 2, nếu có sai sót mong bạn thông cảm
\(\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)
⇒ \(3x=-5\)
⇒ \(x=-\dfrac{5}{3}\)
\(a,\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y+x+2y=\left(-4\right)+\left(-1\right)\\x+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=-5\\x+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\-\dfrac{5}{3}+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\2y=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}3x+5y=11\\2x+5y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=11\\3x+5y-2x-5y=11-9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3.2+5y=11\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6+5y=11\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=5\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
\(\sqrt{x^2-6x+9}-4=3x\left(đkxđ:x\ge-\dfrac{4}{3}\right)\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=3x+4\\ \Leftrightarrow\left|x-3\right|=3x+4\\ \Leftrightarrow\left[{}\begin{matrix}x-3=3x+4\\x-3=-3x-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-3x=4+3\\x+3x=-4+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=7\\4x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
(3x+1).9-74=16
(3x+1).9=16+74
(3x+1).9=90
(3x+1)=90/9
3x+1=10
3x=9
x=3