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a, \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=4\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{-\frac{2}{3};4\right\}\)
b, \(5x\left(x-3\right)+3-x=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\frac{1}{5}\right\}\)
d, \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x^3+3^3\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3-;0;2\right\}\)
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
a, \(x^4-5x^3+2x^2+10x+2=0\)
\(\Rightarrow x^4+x^3-6x^3-6x^2+8x^2+8x+2x+2=0\)
\(\Rightarrow x^3\left(x+1\right)-6x^2\left(x+1\right)+8x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)
Vì \(x^3-6x^2+8x+2>0\) nên \(x+1=0\Rightarrow x=-1\)
Các câu còn lại tương tự!
Chúc bạn học tốt!!!
B1:
a) \(x^3-2x^2+x-2\)
= \(x^2\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+1\right)\)
b) \(2x^3+3x^2-3x-2\)
= \(2x^3-2x^2+5x^2-5x+2x-2\)
= \(2x^2\left(x-1\right)+5x\left(x-1\right)+2\left(x-1\right)\)
= \(\left(x-1\right)\left(2x^2+5x+2\right)\)
= \(\left(x-1\right)\left(2x^2+4x+x+2\right)\)
= \(\left(x-1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
c) \(5x^2+5y^2-x^2z+2xyz-y^2z-10xy\)
= \(5\left(x^2+2xy+y^2\right)+z\left(x^2+2xy+y^2\right)\)
= \(5\left(x+y\right)^2+z\left(x+y\right)^2\)
= \(\left(x+y\right)^2\left(5+z\right)\)
d) \(x^3-3x^2y+3xy^2-x+y-y^3\)
= \(\left(x-y\right)^3-\left(x-y\right)\)
= \(\left(x-y\right)\left[\left(x-y\right)^2-1\right]\)
= \(\left(x-y\right)\left(x-y-1\right)\left(x-y+1\right)\)
B2:
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\left(2x-5\right).\left(-2\right)=0\)
\(\Rightarrow2x-5=0\Rightarrow x=\dfrac{5}{2}\)
b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\left(x+3\right)\left(x^2-2x\right)=0\)
\(\left(x+3\right).x.\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
c) \(2x^3+3x^2+2x+3=0\)
\(x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\left(2x+3\right)\left(x^2+1\right)=0\)
Ta thấy \(x^2+1>0\) với mọi x
\(\Rightarrow2x+3=0\Rightarrow x=\dfrac{-3}{2}\)
a: ta có: \(x^2+3x-\left(2x+6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
b: Ta có: \(5x+20-x^2-4x=0\)
\(\Leftrightarrow\left(x+4\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)