Giải các phương trình :
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(b,x\left(2x-7\right)-4x+14=0\)
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a, \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\Rightarrow2x=-5\Rightarrow x=\frac{-5}{2}\\x-3=0\Rightarrow x=3\end{cases}}\)
Vậy \(S=\left\{-\frac{5}{2};3\right\}\)
T**k mik nhé!
Hok tốt!
b,\(x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow x\left(2x-7\right)-2\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-7=0\Rightarrow2x=7\Rightarrow x=\frac{7}{2}\\x-2=0\Rightarrow x=2\end{cases}}\)
Vậy \(S=\left\{\frac{7}{2};2\right\}\)
T**k mik nhé!
Hok tốt!
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-5\\x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
Vậy .........
\(b,\left(x^2-4\right)+\left(x-2\right)\left(3-2x=0\right)\)
\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)
\(\Leftrightarrow-x^2+7x-10=0\)
\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
Vậy ..................
\(c,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
\(d,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-7x-4x+14=0\)
\(\Leftrightarrow2x^2-11x+14=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
Vậy ............
\(e,\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)
\(\Leftrightarrow3x^2-24x+21=0\)
\(\Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)
Vậy .....................
\(f,x^2-x-\left(3x-3\right)=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy ..............
a) Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2+\left(x^2-2x\right)-3\left(x^2-2x\right)-3=0\)
\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-2x+1\right)-3\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x^2-2x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Vậy: S={1;-1;3}
\(\left(x-2\right)^2+\left|x-5\right|-x^2-14=0.\)
\(\left(x^2-4x+4\right)+\left|x-5\right|-x^2-14=0.\)
\(x\text{}\text{}\text{}^2-4x+4+\left|x-5\right|-x^2-14=0.\)
\(x\text{}\text{}\text{}^2-x^2-4x+4-14+\left|x-5\right|=0.\)
\(-4x-10+\left|x-5\right|=0\)
.. đến đây xét tiếp để ra kq ạ -,-
\(a,\) Đặt \(x^2+2x=a\), pt trở thành:
\(a^2-3a+2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=0\left(1\right)\\x^2+2x-2=0\left(2\right)\end{matrix}\right.\)
\(\left[{}\begin{matrix}\Delta\left(1\right)=4+4=8\\\Delta\left(2\right)=4+8=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{8}}{2}\\x=\dfrac{-2+\sqrt{8}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{12}}{2}\\x=\dfrac{-2+\sqrt{12}}{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{2}\\x=-1+\sqrt{2}\\x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)
\(b,\) Đặt \(x^2+x=b\), pt trở thành:
\(b\left(b+1\right)-6=0\\ \Leftrightarrow b^2+b-6=0\\ \Leftrightarrow\left[{}\begin{matrix}b=2\\b=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\x\in\varnothing\left[x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(d,x^4-2x^3+x=2\\ \Leftrightarrow x^4-2x^3+x-2=0\\\Leftrightarrow\left(x^3+1\right)\left(x-2\right)=0 \\ \Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x\in\varnothing\left[x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Lời giải:
a.
PT $\Leftrightarrow (x^2+2x)^2-(x^2+2x)-2[(x^2+2x)-1]=0$
$\Leftrightarrow (x^2+2x)(x^2+2x-1)-2(x^2+2x-1)=0$
$\Leftrightarrow (x^2+2x-1)(x^2+2x-2)=0$
$\Leftrightarrow x^2+2x-1=0$ hoặc $x^2+2x-2=0$
$\Leftrightarrow x=-1\pm \sqrt{2}$ hoặc $x=-1\pm \sqrt{3}$
b.
PT $\Leftrightarrow (x^2+x)^2+(x^2+x)-6=0$
$\Leftrightarrow (x^2+x)^2-2(x^2+x)+3(x^2+x)-6=0$
$\Leftrightarrow (x^2+x)(x^2+x-2)+3(x^2+x-2)=0$
$\Leftrightarrow (x^2+x-2)(x^2+x+3)=0$
$\Leftrightarrow x^2+x-2=0$ (chọn) hoặc $x^2+x+3=0$ (loại do $x^2+x+3=(x+0,5)^2+2,75>0$)
$\Leftrightarrow x=-1\pm \sqrt{3}$
c. Nghiệm khá xấu. Bạn coi lại đề.
d.
PT $\Leftrightarrow x^3(x-2)+(x-2)=0$
$\Leftrightarrow (x^3+1)(x-2)=0$
$\Leftrightarrow x^3+1=0$ hoặc $x-2=0$
$\Leftrightarrow x=-1$ hoặc $x=2$
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!
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)