Đặt t=x-a

\(\left|2x+4-2a\right|+\left|x-2+a\right|\le3\)

=>\(\left|2\left(x-a\right)+4\right|+\left|x+a-2\right|\le3\)

=>|2t+4|+|x+a-2|<=3

mà |x+a-2|>=0∀x,a

nên |2t+4|<=3

=>-3<=2t+4<=3

=>-7<=2t<=-1

=>-7/2<=t<=-1/2

=>Tập giá trị của x-a là [-7/2;-1/2]

Đặt t=x-a

Đặt u=x-a+2=t+2; v=x+a-2

=>u+v=2x; u-v=-2a+4

Theo đề, ta có: \(\left|2u\right|+\left|v\right|\le3\)

=>2|u|<=3-|v|<=3

=>\(\left|u\right|\le\frac32\)

=>\(\left|t+2\right|\le\frac32\)

=>\(-\frac32\le t+2\le\frac32\)

=>\(-\frac72\le t\le-\frac12\)

=>-7/2<=x-a<=-1/2

cau biet toan lop 10 ha

​

cos đó bạn

Lời giải:

a)

\(\cos 2a=\frac{2}{5}\Rightarrow \sin ^22a=1-(\cos 2a)^2=1-(\frac{2}{5})^2=\frac{21}{25}\)

Vì $a\in (0; \frac{\pi}{4})\Rightarrow 2a\in (0; \frac{\pi}{2})$

$\Rightarrow \sin 2a>0\Rightarrow \sin 2a=\frac{\sqrt{21}}{5}$

$\tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\sqrt{21}}{5.\frac{2}{5}}=\frac{\sqrt{21}}{2}$

$\cot 2a=\frac{1}{\tan 2a}=\frac{2}{\sqrt{21}}$

-------------------------

$\sin 2a=\frac{24}{25}\Rightarrow \cos ^22a=1-(\sin 2a)^2=\frac{49}{625}$

$a\in [\frac{-3}{4}\pi; \frac{-\pi}{2}]\Rightarrow 2a\in [\frac{-3}{2}\pi ; -\pi]\Rightarrow \cos 2a< 0$

$\Rightarrow \cos 2a=\frac{-7}{25}$

$\Rightarrow \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{24}{25.\frac{-7}{25}}=\frac{-24}{7}$

$\Rightarrow \cot 2a=\frac{-7}{24}$