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\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
Bạn chú ý cách viết phương trình.
Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.
\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=4\left(x+\frac{1}{x}\right)^2\left(2-x^2-\frac{1}{x^2}\right)+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4\left(x+\frac{1}{x}\right)^2\left(x-\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4\left(x^2-\frac{1}{x^2}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4x^4+8-\frac{4}{x^4}+4x^4+8+\frac{4}{x^4}\)
\(=16\)
Phương trình đã cho trở thành
\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
\(x\left(x+1\right)^4+x\left(x+1\right)^3+x\left(x+1\right)^2+\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[x\left(x+1\right)^2+x\left(x+1\right)+x+1\right]\)
\(=\left(x+1\right)^2\left[x\left(x+1\right)\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x\left(x+1\right)+x+1\right]\right\}\)
\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x^2+x+x+1\right]\right\}\)
\(=\left(x+1\right)^2\left[\left(x+1\right)\left(x^2+2x+1\right)\right]\)
\(=\left(x+1\right)^2\cdot\left(x+1\right)^3\)
\(=\left(x+1\right)^5\left(đpcm\right)\)
\(9x^2-1+\left(3x-1\right).\left(x+2\right)=0\)
\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)
\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)
\(\Leftrightarrow12x^2+5x=3\)
\(\Leftrightarrow12x^2+5x-3=0\)
\(\Leftrightarrow12x^2-4x+9x-3=0\)
\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)
\(\Leftrightarrow\left(4x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)
Đặt \(u=x^2-x\)
Phương trình trở thành \(u^2-4u+4=0\)
\(\Leftrightarrow\left(u-2\right)^2=0\)
\(\Leftrightarrow u-2=0\)
\(\Rightarrow x^2-x=2\)
\(\Rightarrow x^2-x-2=0\)
Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+3}{2}=2\\x=\frac{1-3}{2}=-1\end{cases}}\)
Đặt \(2x+1=w\)
Phương trình trở thành \(w^2-w=2\)
\(\Rightarrow\orbr{\begin{cases}w=2\\w=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=2\\2x+1=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
ko ai giải đc à, giúp mk đi mà mau lên đang cần gấp, please
RẤT nhieu bn giai dc vi các pt này dễ nhung k ai giai vi nó dài ,làm mệt mà kè nhờ vả k biet ơn, k coi trọng chât xám
toàn là h tảo lao nên ng tài k dc trọng dụng , kẻ bât tai thi k giai dc, bởi z ng tài chỉ xem bài nào khó, k dài thi giai, dc kdc h cũng k cần
Ta có : \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2-x+2x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-1=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)
<=> 2 trường hợp sảy ra là bằng 5 hoặc -5 nhé
a) Sửa đề
\(\left(x+1\right)^4-\left(x-3\right)^4=82\)
Đặt x - 1 = a
\(\left(a+2\right)^4-\left(a-2\right)^4=82\)
\(\Rightarrow\left[\left(a+2\right)^2\right]^2-\left[\left(a-2\right)^2\right]^2=82\)
\(\Rightarrow\left(a^2+4a+4\right)^2-\left(a^2-4a+4\right)^2=82\)
\(\Rightarrow\left(a^2+4\right)^2+8a\left(a^2+4\right)+16a^2+\left(a^2+4\right)^2-8a\left(a^2+4\right)+16a^2=82\)
\(\Rightarrow\left(a^2+4\right)^2+16a^2=41\)
\(\Rightarrow a^4+8a^2+16+16a^2=41\)
\(\Rightarrow a^4+24a^2=25\)
\(\Rightarrow a^4+24a^2-25=0\)
\(\Rightarrow a^4-a^2+25a^2-25=0\)
\(\Rightarrow a^2\left(a^2-1\right)+25\left(a^2-1\right)=0\)
\(\Rightarrow\left(a^2-1\right)\left(a^2+25\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+25=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-1\\a^2=-25\end{matrix}\right.\)
Do a2= -25 không tồn tại
Vậy a = 1 ; a = -1
b) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-2\right)=24\)
\(\Rightarrow\left[\left(x-1\right)\left(x-2\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]=24\)
\(\Rightarrow\left(x^2-3x+2\right)\left(x^2+3x+2\right)=24\)
\(\Rightarrow\left(x^2+2\right)^2-\left(3x\right)^2=24\)
\(\Rightarrow x^4+4x^2+4-9x^2-24=0\)
\(\Rightarrow x^4-5x^2-20=0\)
\(\Rightarrow\left(x^2\right)^2-2.x^2\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}-20=0\)
\(\Rightarrow\left(x^2-\dfrac{5}{2}\right)^2-\dfrac{105}{4}=0\)
\(\Rightarrow\left(x^2-\dfrac{5}{2}\right)^2=\dfrac{105}{4}\)
\(\Rightarrow\left(x^2-\dfrac{5}{2}\right)=\left(\dfrac{\sqrt{105}}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x^2-\dfrac{5}{2}=\dfrac{\sqrt{105}}{2}\\x^2-\dfrac{5}{2}=-\dfrac{\sqrt{105}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=\dfrac{5+\sqrt{105}}{2}\\x^2=\dfrac{5-\sqrt{105}}{2}\end{matrix}\right.\)
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