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\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
a. \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\)
\(\Leftrightarrow-x^3+5x^2-5x=0\)
\(\Leftrightarrow-x\left(x^2-5x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)
a) \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-2-x^2+4x-3\right)=0\)
\(\Leftrightarrow x\left(-x^2+5x-5\right)=0\)
\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)
=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)
b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)
\(\Leftrightarrow-12=0\left(vn\right)\)
c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)
\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)
\(\Leftrightarrow2x=15\)
\(\Rightarrow x=\frac{15}{2}\)
a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)
\(\Leftrightarrow2x^2+4x-4=0\)
\(\Leftrightarrow x^2+2x-2=0\)
\(\Leftrightarrow x^2+2x+1-3=0\)
\(\Leftrightarrow\left(x+1\right)^2=3\)
hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)
b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)
\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)
=>6=0(vô lý)
c: \(\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)
=>x=-2 hoặc x=2
đ: \(\Rightarrow2x^2-2x-5x+5=0\)
=>(x-1)(2x-5)=0
=>x=1 hoặc x=5/2
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) x(x-1) - (x+1)(x+2) = 0
x\(^2\)- x -x\(^{^2}\)-2x +x+2=0
-2x+2=0
-2x=0+2
-2x=2
x=-1
Vậy x bằng -1
a. 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0
<=> 3(6x2-5x+1)-(18x2-29x+3)=0
<=> 14x=0
<=> x=0
b. (x - 3)(x - 5) + 3 (x - 1) = (x - 1)(x - 3)
<=> (x-3)(x-5-x+1)+3(x-1)=0
<=> -4(x-3)+3(x-1)=0
<=> -x+9=0
<=> x=9
c. (x - 1)(x - 2) - (x + 2)(x + 1) = 8
<=> x2-3x+2-(x2+3x+2)=8
<=> -6x=8
<=> \(x=\frac{-4}{3}\)
a)2x.(x+3)-3.(x^2+1)=x+1-x.(x-2)
<=> 2x2 + 6x - 3x2 - 3 = x - 1 - x2 + 2x
<=> 2x2 + 6x - 3x2 - x + x2 - 2x = -1 +3
<=> 3x = 2
<=> x = 2/3
b)(x+2).(x-2)-(x-3).(x+5)=0
<=> x2 - 4 - x2 - 5x - 3x - 15 = 0
<=> -5x - 3x = 4 + 15
<=> -8x = 19
<=> x = -19/8
Phần c tương tự ạ