Tim nghiem
a) 2x-3
b)x^3+1
c) x^2-2x+1
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a.
⇔ \(\left\{{}\begin{matrix}x-2y=4.3-5\\2x+y=3.3\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x-2y=7\\2x+y=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}-2x+4y=-14\\2x+y=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}5y=-5\\2x+y=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}y=-1\\2x-1=9\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}y=-1\\x=5\end{matrix}\right.\)
Vậy nghiệm của hpt là: (5;1)
Đặt C(x)=0
\(\Leftrightarrow-2x\left(2x-3\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow-4x^2+6x-2x+2=0\)
\(\Leftrightarrow-4x^2+4x+2=0\)
\(\Leftrightarrow4x^2-4x-2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}\\2x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3}+1\\2x=-\sqrt{3}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{2}\\x=\dfrac{-\sqrt{3}+1}{2}\end{matrix}\right.\)
Đặt Q(x)=0
\(\Leftrightarrow2\left(x-3\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x-6-x+1=0\)
\(\Leftrightarrow x=5\)
a. Bạn tự giải
b. \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\4x+2y=6m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=4m-5\\5x=10m-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2m-1\\y=-m+2\end{matrix}\right.\)
\(\dfrac{2}{x}-\dfrac{1}{y}=-1\Rightarrow\dfrac{2}{2m-1}-\dfrac{1}{-m+2}=-1\) (\(m\ne\left\{\dfrac{1}{2};2\right\}\))
\(\Leftrightarrow2\left(-m+2\right)-\left(2m-1\right)=\left(m-2\right)\left(2m-1\right)\)
\(\Leftrightarrow2m^2-m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=\dfrac{3}{2}\end{matrix}\right.\)
a/ 3(1 - x) - 5(2x - 2) = 0
=> 3 - 3x - 10x + 10 = 0
=> -13x = -13
=> x = 1
Vậy x = 1
b/ |3x - 2| - 4 = 0 => |3x - 2| = 4
Suy ra 2 trường hợp:
Vậy x = 2 , x = -2/3
c/ 2x - x3 = 0 => x.(2 - x2) = 0
=> x = 0
hoặc 2 - x2 = 0 => x2 = 2 => x = \(\sqrt{2}\) hoặc x = \(-\sqrt{2}\)
Vậy \(x=\left\{0;\sqrt{2};-\sqrt{2}\right\}\)
d/ x(1 - 2x) + (2x2 - x + 4) = 0
=> x - 2x2 + 2x2 - x + 4 = 0
=> 4 = 0 (vô lí)
Vậy vô nghiệm
Lời giải:
a.
a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$
$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$
$\Leftrightarrow 3x+1=5x-3$
$\Leftrightarrow 4=2x$
$\Leftrightarrow x=2$
b.
$(2x-1)(x+3)-(x-2)(x+3)=3x+1$
$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$
$\Leftrightarrow x^2+5x+1=3x+1$
$\Leftrightarrow x^2+2x=0$
$\Leftrightarrow x(x+2)=0$
$\Leftrightarrow x=0$ hoặc $x=-2$
c.
$x^2(x-1)-x(x-1)(x+1)=0$
$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$
$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$
$\Leftrightarrow (x-1)(-x)=0$
$\Leftrightarrow x-1=0$ hoặc $-x=0$
$\Leftrightarrow x=1$ hoặc $x=0$
d.
$4x(x-5)-(2x-3)(2x+3)=9$
$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$
$\Leftrightarrow -20x=0$
$\Leftrightarrow x=0$
a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow2x=4\)
hay x=2
b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)
\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)
\(\Leftrightarrow4x^2-20x-4x^2+9=9\)
hay x=0
a) \(2x-3=0\Leftrightarrow x=\frac{3}{2}\)
b) \(x^3+1=0\)
\(\Leftrightarrow x^3=-1\)
\(\Leftrightarrow x^3=\left(-1\right)^3\Leftrightarrow x=-1\)
c) \(x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
a) \(2x-3=0\)
\(\Leftrightarrow2x=0+3\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=3\div2\)
\(\Leftrightarrow x=1,5\)
Vậy nghiệm của đa thức 2x - 3 là 1,5.
b. \(x^3+1=0\)
\(\Leftrightarrow x^3=0-1\)
\(\Leftrightarrow x^3=-1\)
\(\Leftrightarrow x^3=-1^3\)
\(\Leftrightarrow x=-1\)
Vậy nghiệm của đa thức x3 + 1 là - 1.
c) \(x^2-2x+1=0\)
\(\Leftrightarrow x^2-2x=0-1\)
\(\Leftrightarrow x\left(x-2\right)=-1=-1.1=1.\left(-1\right)\)
Vậy đa thức \(x^2-2x+1\) vô nghiệm