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\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
\(a,\left(2x-3\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-3+x+1\right)\left(2x-3-x-1\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-4\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\x=4\end{matrix}\right. \\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{2}{3};4\right\}\)
\(b,x^2-6x+9=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2-9\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-3^2\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left[3\left(x-1\right)\right]^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left(3x-3\right)^2=0\\ \Leftrightarrow\left(x-3+3x-3\right)\left(x-3-3x+3\right)=0\\ \Leftrightarrow-2x\left(4x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x=0\\4x-6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\dfrac{3}{2}\right\}\)
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)
\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)
\(\Leftrightarrow6x=2x^2+4\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)
\(x^4+2x^3-2x^2+2x-3=0\\ \Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\\ \Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^3-x^2+x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\left(\text{vì }x^2+1\ge1>0\right)\)
Vậy ...
\(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^2+5x-2\right)-\left(x^2+x+1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
\(x^2+\left(x+2\right)\left(11x-7\right)=4\\ \Leftrightarrow x^2-4+\left(x+2\right)\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
nghiệm đâu bạn ưi...nó là phương trình vô nghiệm hay vô số nghiệm vậy m :))
`a,4x^2+(x-1)^2-(2x+1)^2=0`
`<=>4x^2+3x(-x-2)=0`
`<=>x(4x-3x-6)=0`
`<=>x(x-6)=0`
`<=>` $\left[ \begin{array}{l}x=0\\x=6\end{array} \right.$
`b)(x^2-3x)^2+5(x^2-3x)+6=0`
Đặt `x^2-3x=a(a>=-9/4)`
`pt<=>a^2+5a+6=0`
`<=>(a+2)(a+3)=0`
`<=>` $\left[ \begin{array}{l}a=-2\\a=-3(l)\end{array} \right.$
`<=>x^2-3x=-2`
`<=>x^2-3x+2=0`
`<=>(x-1)(x-2)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=1\end{array} \right.$