Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-5\\x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
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) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)
\(\Rightarrow x^3+27-x^3+4x=27\)
\(\Rightarrow27+4x=27\)
\(\Rightarrow4x=0\)
\(\Rightarrow x=0\)
b) \(2x^2+7x+3=0\)
\(\Rightarrow2x^2+x+6x+3=0\)
\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
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. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)
\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{5}\)
m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)
\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)
\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=9\)
e) \(\left|5x\right|=3x-2\)
\(\Rightarrow5\cdot\left|x\right|=3x-2\)
\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
g) \(\left|-2,5x\right|=x-12\)
\(\Rightarrow2,5\cdot\left|x\right|=x-12\)
\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
\(a.\left(2-3x\right)\left(x^2+2x+3\right)=0.\)
\(\left(2-3x\right)=0\)
\(\left(x^2+2x+3\right)=0\)
\(TH1:2-3x=0\Leftrightarrow x=\frac{-2}{-3}\)
\(TH2:x^2+2x+3=0\Leftrightarrow\left(x^2+2x+1\right)+3\Leftrightarrow\left(x+1\right)^2+3>0\)
b) \(3x-3x=5+2\) ( vô nghiệm)
c) vô nghiệm
d-\(x^2-5x-6=0\Leftrightarrow\left(x^2-x\right)+\left(6x-6\right)\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
vậy ...
x=1
x=-6
E) \(\frac{2\left(x-3\right)^2}{3}=\frac{3x^2}{2}\) quy đồng khử mẫu ta được
\(4\left(x-3\right)^2-9x^2=0\Leftrightarrow4\left(x-3\right)^2-\frac{4.1.9x^2}{4}\) rút 4 ta được
\(4\left\{\left(x-3\right)^2-\frac{9x^2}{4}\right\}=0\Leftrightarrow4\left\{\left(x-3\right)^2-\left(\frac{3}{2}x\right)^2\right\}\Leftrightarrow4\left(x-3+\frac{3}{2}x\right)\left(x-3-\frac{3}{2}x\right)=0\) ( hằng đẳng thức số 3 )
tích = 0
vậy ....
F) trị tuyệt đối + bình phương của 1 số thực luôn lớn hơn hoặc = 0( định lí Pain)
phá trị tuyệt đối ta được
\(\left(x+5\right)^2-\left(3x-2\right)^2=0\)
\(\left(x+5-3x-2\right)\left(x+5+3x-2\right)=0\) ( hẳng đẳng thức số 3 )
tích = 0 suy ra 2 TH vậy .....
g) câu G bạn lên coccoc math bạn ghi là nó ra kết quả phân tích thành nhân tử chứ làm = tay vừa dài vừa hại não :)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24=0\)
\(x\left(x-5\right)x\left(x^2-5x+10\right)=0\) ( coccoc math)
\(\left(x^2-5x+10\right)=0\Leftrightarrow\left(x^2-\frac{2x.5}{2}+\left(\frac{5}{2}\right)^2\right)+10-\frac{25}{4}=0\) ( 10-25/4) = 15/4
\(\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\) ( vô nghiệm)
vậy....
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow4x^2-20x+25-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)
\(\Leftrightarrow3x^2-24x+21=0\)
\(\Leftrightarrow3x^2-21x-3x+21=0\)
\(\Leftrightarrow3x\left(x-7\right)-3\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
\(x^2-x-\left(3x-3\right)=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)