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\(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
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VM
14 tháng 2 2020
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 10 2019
a/ \(x\ge-3\)
\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow3x^2-10x-8=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)
b/ \(x\ge-\frac{5}{2}\)
\(\Leftrightarrow\left(4x+7\right)^2=\left(2x+5\right)^2\)
\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
c/ \(x\ge1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=5x-5\\2x^2-3x-5=5-5x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x=0\\2x^2+2x-10=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=4\\x=\frac{-1+\sqrt{21}}{2}\\x=\frac{-1-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 10 2019
d/ \(x\ge\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\left(l\right)\\x=\sqrt{22}\\x=-\sqrt{22}\left(l\right)\end{matrix}\right.\)
e/ \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-x-2=x-2\\3x^2-x-2=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=0\\3x^2=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{2}{3}\left(l\right)\\x=\frac{2\sqrt{3}}{3}\\x=\frac{-2\sqrt{3}}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 11 2019
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+2x\right)\left(2x+y\right)=9\\x^2+2x+2x+y=6\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+2x=a\\2x+y=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}ab=9\\a+b=6\end{matrix}\right.\) theo Viet đảo, a và b là nghiệm của:
\(t^2-6t+9=0\Rightarrow t=3\Rightarrow a=b=3\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+2x=3\\2x+y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+2x-3=0\\y=3-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\end{matrix}\right.\)
9 tháng 2 2017
Làm cho bạn 1 con thôi dài quá trôi hết màn hình:
c) có vẻ khó nhất (con khác tương tự)
đặt 2x+2=t=> x+1=t/2
\(\left(t-1\right).\left(\frac{t}{2}\right)^{^2}.\left(t+1\right)=18\Leftrightarrow\left(t^2-1\right)t^2=4.18\)
\(t^4-t^2=4.18\Leftrightarrow y^2-2.\frac{1}{2}y+\frac{1}{4}=4.18+\frac{1}{4}=\frac{16.18+1}{4}=\left(\frac{17}{2}\right)^2\)
<=> \(\left(y-\frac{1}{2}\right)^{^2}=\left(\frac{17}{2}\right)^2\Rightarrow\left[\begin{matrix}y=\frac{1}{2}-\frac{17}{2}=-8\\y=\frac{1}{2}+\frac{17}{2}=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x+2=-8\Rightarrow x=-5\\2x+2=9\Rightarrow x=\frac{7}{2}\end{matrix}\right.\)
\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow\left(4x-4\right)\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy ...
a) \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x+7\right)^2-\left[3\left(x+2\right)\right]^2=0\)
\(\Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(2x+7-3x-6\right)\left(2x+7+3x+6\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\5x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b)\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+6\right)\left(x+2+3x-6\right)=0\)
\(\Leftrightarrow\left(8-2x\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8-2x=0\\4x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(4-x\right)=0\\4\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)