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a, x^3 - x^2 - x + 1 = 0
x^2 (x-1) - (x-1) =0
(x^2 -1) (x-1) =0
\(\Rightarrow\orbr{\begin{cases}x^2-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=1\\x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=+-1\\x=1\end{cases}}}\)
Vậy x= +- 1
b, 3x^2 - 3xy + 5y - 5x = 0
3x (x-y) - 5(x-y) =0
(3x -5)(x-y) =0
(làm tương tự như bài trên)
A.
\(\Leftrightarrow\) 9x - 2x - 6 = 3x + 1
\(\Leftrightarrow\) 4x = 7
\(\Leftrightarrow\) x = \(\dfrac{7}{4}\)
B.
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-13}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\) 5x + 15 - 4x +12 = x - 13
\(\Leftrightarrow\) 0x = -40 ( phương trình vô nghiệm)
C.
\(\Leftrightarrow\) 7x + 8 \(\ge\) 3x -3
\(\Leftrightarrow\) 4x \(\ge\) - 11
\(\Leftrightarrow\)\(x\ge\dfrac{-11}{4}\)
a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)
c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)
a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)
b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương
c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)
a: \(A=-4x^2+4x-1\)
\(=-\left(4x^2-4x+1\right)\)
\(=-\left(2x-1\right)^2\le0\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
b: \(B=-x^2+5x\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
a) x3 - x2 - x + 1
= x2.(x - 1) - (x - 1)
= (x - 1).(x2 - 1)
= (x - 1).(x - 1).(x + 1)
= (x - 1)2.(x + 1)
b) 3x2 - 3xy + 5y - 5x
= 3x.(x - y) - 5.(x - y)
= (x - y).(3x - 5)
c) x3 - 7x - 6
= x3 - 4x - 3x - 6
= x.(x2 - 4) - 3.(x + 2)
= x.(x - 2).(x + 2) - 3.(x + 2)
= (x + 2).[x.(x - 2) - 3]
= (x + 2).(x2 - 2x - 3)
= (x + 2).(x2 - 3x + x - 3)
= (x + 2).[x.(x - 3) + (x - 3)]
= (x + 2).(x - 3).(x + 1)
a, x^3 - x^2 - x + 1
= x^2 (x-1) - (x-1)
= (x^2 -1) (x-1)
b, 3x^2 - 3xy + 5y - 5x
= 3x(x-y) - 5(x-y)
= (3x - 5) (x-y)