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\(\Leftrightarrow\left(x-1\right)\left(2x^2-3x+2013\right)=0\Leftrightarrow\left(x-1\right)\left(2\left(x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}\right)+2013-\frac{9}{8}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2\left(x-\frac{3}{4}\right)^2+2011\frac{7}{8}\right)=0\)(1)
Do \(\left(2\left(x-\frac{3}{4}\right)^2+2011\frac{7}{8}\right)>0\forall x\)nên
(1) <=> x - 1 = 0 <=> x = 1.
đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)
\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)
TH1: \(x-1=0\Leftrightarrow x=1\)
TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
\(\left(3x-5\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)
\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)
\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)
\(\Leftrightarrow3x^2+2+7x=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)
b. (x2-0,5):2x-(3x-1)2:(3x-1)=0
<=> \(\frac{1}{2}\)x-0,25-3x+1=0
<=>\(-\frac{5}{2}\)x+0,75=0
<=> \(-\frac{5}{2}\)x=-0,75
<=> x=0,3
chúc bạn học tốt
\(a.\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=4\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=4\)
\(\Leftrightarrow\left(x^2+x+5x+5\right)\left(x^2+4x+2x+8\right)=4\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=4\)
\(\text{Đặt a = }x^2+6x+5\text{ }\Rightarrow\text{ }a+3=x^2+6x+8\)
\(\Leftrightarrow a\left(a+3\right)=4\)
\(\Leftrightarrow a^2+3a-4=0\)
\(\Leftrightarrow a^2+4a-a-4=0\)
\(\Leftrightarrow a\left(a+4\right)-\left(a+4\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left(x^2+6x+9\right)\left(x^2+6x+4\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2\left[\left(x^2+6x+9\right)-5\right]=0\)
\(\Leftrightarrow\left(x+3\right)^2\left[\left(x+3\right)^2-5\right]=0\)
\(\text{Hoặc }\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(\text{Hoặc }\left(x+3\right)^2-5=0\Leftrightarrow\left(x+3\right)^2=5\Leftrightarrow\hept{\begin{cases}x+3=\sqrt{5}\\x+3=-\sqrt{5}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{5}-3\\x=-\sqrt{5}-3\end{cases}}}\)
\(\text{Vậy }x\in\left\{-3;\sqrt{5}-3;-\sqrt{5}-3\right\}\)
a. \(9\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow9x+18-3x-6=0\)
\(\Leftrightarrow6x+12=0\)
\(\Leftrightarrow x=-2\)
e. \(\left(2x-1\right)^2-45=0\)
\(\Leftrightarrow4x^2-2x+1-45=0\)
\(\Leftrightarrow4x^2-2x-44=0\)
Đến đó tự giải tiếp nha!
c. \(2\left(2x-5\right)-3x=0\)
\(\Leftrightarrow4x-10-3x=0\)
\(\Leftrightarrow x-10=0\)
\(\Leftrightarrow x=10\)
g. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt
(x-1)(2x2-3x+2016) =0
x-1 =0 => x = 1
còn lại 2x2 -3x +2016 =0
giải theo denta cho nhanh