Bài 2 Tìm x
1) 5(x+3)-2x(3+x)=0
2)4x(x-2004)-x+2004=0
3)(x-5)² =16
4)25=(3-x)²
5) x(x-5)-4x+20=0
6) x²-5x²+x-5+0
7)$x^{4}$ – 2x ³+10x ²-20x = 0
Bài 2 Tìm x
1) 5(x+3)-2x(3+x)=0
2)4x(x-2004)-x+2004=0
3)(x-5)² =16
4)25=(3-x)²
5) x(x-5)-4x+20=0
6) x²-5x²+x-5+0
7)$x^{4}$ – 2x ³+10x ²-20x = 0
1) 5(x+3)-2x(3+x)=0
=>(x+3)(5-2x)=0 => \(\left[ \begin{array}{l}x+3=0\\5-2x=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=-3\\x=5/2\end{array} \right.\)
2) 4x(x-2004)-x+2004=0 =>4x(x-2004)-(x-2004)=0
=>(4x-1)(x-2004)=0 =>\(\left[ \begin{array}{l}4x-1=0\\x-2004=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=1/4\\x=2004\end{array} \right.\)
3) (x-5)²=16 =>(x-5)²=(±4)² => x-5=±4
Nếu x-5=4 => x=9
Nếu x-5=-4 => x=1
4) 25=(3-x)² => (±5)²=(3-x)² => 3-x = ±5
Nếu 3-x=5 => x=-2
Nếu 3-x=-5 => x=8
5) x(x-5) -4x+20=0 => x(x-5) -4(x-5)=0
=> (x-4)(x-5)=0 => \(\left[ \begin{array}{l}x-4=0\\x-5=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=4\\x=5\end{array} \right.\)
6) x²-5x²+x-5=0 => (x-5)x²+(x-5)=0
=> (x-5)(x²-1)=0 => (x-5)(x-1)(x+1)=0
+) x-5=0 => x=5
+) x-1=0 => x=1
+) x+1=0 => x=-1
7) $x^{4}$ -2x³+10x²=20x=0 => x³(x-2)+x²(x-2)=0
=> x²(x+1)(x-2)=0
+) x²=0 => x=0
+) x+1=0 => x=-1
+) x-2=0 => x=2
Đáp án:
Giải thích các bước giải:
1) $5(x+3)-2x(3+x)=0$
⇔$\left \{ {{5.(x+3)=0} \atop {2x.(3+x)=0}} \right.$
⇔$\left \{ {{x=-3} \atop {x=-3}} \right.$
Vậy…
2) $4x(x-2004)-x+2004=0$
$4x(x-2004)-(x-2004)=0$
$(4x-1)(x-2004)=0$
⇔\(\left[ \begin{array}{l}4x-1=0\\x-2004=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{1}{4}\\x=2004\end{array} \right.\)
Vậy ….
3) $(x-5)² = 16$
⇔$(x-5)² = 4²$
⇒$x-5=4$
⇒$x=9$
Vậy ….
4) $25=(3-x)²$
⇔$5² = (3-x)²$
⇒$5=3-x$
⇒$x=-2$
Vậy …
5) $x(x-5)-4x+20=0$
⇔$x(x-5)-4(x-5)=0$
⇔$(x-5)(x-4)=0$
⇒\(\left[ \begin{array}{l}x-5=0\\x-4=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=5\\x=4\end{array} \right.\)
Vậy….
6) $x²-5x+x-5=0$
⇔$x(x-5)+(x-5)=0$
⇔$(x+1)(x-5)=0$
⇒\(\left[ \begin{array}{l}x+1=0\\x-5=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)
Vậy…
7) $x^{4} – 2x³ + 10x² – 20x=0$
⇔$x^{3}.(x-2) + 10x.(x-2)=0$
⇔$(x^{3}+10x).(x-2)=0$
⇔$x.(x^{2}+10).(x-2)=0$
⇒\(\left[ \begin{array}{l}x=0\\x-2=0\\x²+10=0 (KTM)\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy…