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\(\dfrac{1}{1-x}\)+\(\dfrac{1}{1+x}\)+\(\dfrac{2}{1+x^2}\)+\(\dfrac{4}{1+x^4}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)
=
=\(\dfrac{4}{1-x^4}\)+\(\dfrac{4}{1+x^4}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)
=\(\dfrac{8}{1-x^8}\)+\(\dfrac{8}{1+x^8}\)+\(\dfrac{16}{1+x^{16}}\)
=\(\dfrac{16}{1-x^{16}}\)+\(\dfrac{16}{1+x^{16}}\)
=\(\dfrac{32}{1-x^{32}}\)
a) Ta có I 2x - 1 I = 2015
=> 2x-1=2015 hoặc 2x-1=2015
+,Th1: 2x-1=2015
2x=2015+1
2x=2016
x=2016:2
x=1008
+,Th2: 2x-1=-2015
2x=-2015+1
2x=-2014
x=-2014:2
x=-1007
Vậy x=1008, x=-1007
|x-1|<5
th1: x-1<5=> x<6
th2: x-1<-5=> x<-4
vậy x <6 hoặc<-4
|x-1|>5 cũng tương tự như thế
còn mấy câu khác Nguyễn Diệu Thảo làm thế chắc bạn cũng biết cách làm rồi
L_I_K_E CHO MÌNH NHA!!!
a) Chương trình bị lỗi
b) Chương trình bị lỗi
c) Chương trình bị lỗi luôn
|7 + 5x| = 1 - 4x
=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)
=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)
|4x2 - 2x| + 1 = 2x
=> |4x2 - 2x| = 2x - 1
=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)
Vậy ...
#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
a) |2x +1| = 7
Th1: 2x + 1 = 7
<=> x = 3
Th2: 2x + 1 = -7
<=> x = -4
\(1)\) Ta có :
\(\left|2x-1\right|\ge0\)
\(\Leftrightarrow\)\(A=\left|2x-1\right|+8\ge8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left|2x-1\right|=0\)
\(\Leftrightarrow\)\(2x-1=0\)
\(\Leftrightarrow\)\(2x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy GTNN của \(A\) là \(8\) khi \(x=\frac{1}{2}\)
Chúc bạn học tốt ~
\(2)\) Ta có :
\(B=\left|x-3\right|+\left|x-9\right|-1\)
\(B=\left|x-3\right|+\left|9-x\right|-1\ge\left|x-3+9-x\right|-1=\left|6\right|-1=6-1=5\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x-3\right)\left(9-x\right)\ge0\)
Trường hợp 1 :
\(\hept{\begin{cases}x-3\ge0\\9-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le9\end{cases}\Leftrightarrow}3\le x\le9}\)
Trường hợp 2 :
\(\hept{\begin{cases}x-3\le0\\9-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge9\end{cases}}}\) ( loại )
Vậy GTNN của \(B\) là \(5\) khi \(3\le x\le9\)
Chúc bạn học tốt ~
1. I'm very glad (meet)to meet you again
2.I am learning how (play)to play chess
3.Don't let him (eat)eat ice cream too much
4.Would you please (call)calling him?
5.Are you keen on (go)(ski)going skiing?
6Don't forget (turn)to turn of the TV before (go)going to bed
7.After (have)having diner, we often watch TV
8.Remember (check)to check your answer again
9How about (give)giving him some money?
10. I advise you (not get)not to get up too late.
1. I'm very glad (meet) to meet you again
2.I am learning how (play) to play chess
3.Don't let him (eat) to eat ice cream too much
4.Would you please (call) to call him?
5.Are you keen on (go)(ski) going skiing ?
6Don't forget (turn) of the TV before (go) going to bed
7.After (have) having diner, we often watch TV
8.Remember (check) to check your answer again
9How about (give) giving him some money?
10. I advise you (not get) not to get up too late.