1a, Tự giải
b,$2x-3|-x=|2-x|$
-Nếu: $|2x-3|=0\Rightarrow 2x-3=0\Rightarrow x=\frac{3}{2}$
$|2-x|=0\Rightarrow 2-x=0\Rightarrow x=2$
-Ta có bảng sau:
| x | $\frac{3}{2}$ | 2 | |
| 2x-3 | -0 | +| | + |
| 2-x | -| | -0 | + |
[TBODY]
[/TBODY]
-Với: $x<\frac{3}{2}\Rightarrow \left\{\begin{matrix} |2x-3|=3-2x & \\ |2-x|=2-x & \end{matrix}\right.$
-Suy ra:$(3-2x)-x=2-x$
$=>3-3x=2-x$
$\Rightarrow x=\frac{1}{2}(t/mđk-> nhận)$
-Với:$\frac{3}{2}\leq x< 2\Rightarrow \left\{\begin{matrix} |2x-3|=2x-3 & \\ |2-x|=2-x & \end{matrix}\right.$
$=>(2x-3)-x=2-x$
$=>x-3=2-x$
$=>x=\frac{5}{2}(ko t/m đk->loại)$
-Với:$x\geq 2\Rightarrow \left\{\begin{matrix} |2x-3|=2x-3 & \\ |2-x|=x-2 & \end{matrix}\right.$
$\Rightarrow (2x-3)-x=x-2$
$\Rightarrow x-3=x-2$
$\Rightarrow 0=1(vô lí->vô nghiệm)$
-Vậy.....
b, d, e tương tự
c,$\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}$
$\Leftrightarrow \frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1$
$\Leftrightarrow \frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0$
$\Leftrightarrow (x-2005).(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001})$
...
f, $\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}$
$\Leftrightarrow \frac{3}{8}x=\frac{3}{8}.\frac{y}{8}=\frac{3}{8}.\frac{z}{27}$
....
2a, $3^{n+2}-2^{n+2}+3^{n}-2^n \vdots 10$
$\Leftrightarrow (3^{n+2}+3^{n})-(2^{n+2}+2^{n})\vdots 10$
$\Leftrightarrow 3^n(3^2+1)-2^{n-1}(2^3+2)\vdots 10$
$\Leftrightarrow 3^{n}.10-2^{n-1}.10\vdots 10(đpcm)$
.....
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
