\(\left(1999X1998+1998+1997\right)X\left(1:1\frac{1}{2}-1\frac{1}{3}\right)\)
X là nhân chứ không phải x đâu nhá
GIÚP MK NHEN!
Tính \(A=1997+P\left(\frac{1}{1998}\right)+P\left(\frac{2}{1998}\right)+...+P\left(\frac{1997}{1998}\right)\)
với \(P\left(x\right)=\frac{2^{2x+1}}{2^{2x}-2}\)và \(x\ne\frac{1}{2}\)
Ta có
\(D=\frac{2^{2x+1}}{2^{2x}-2}+\frac{2^{2\left(1-x\right)+1}}{2^{2\left(1-x\right)}-2}=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{2^{1-2x}-1}\)
Mà \(2^{1-2x}=\frac{1}{2^{2x-1}}\)(do 1-2x+2x-1=0)
=>\(D=\frac{2^{2x}}{2^{2x-1}-1}+\frac{2^{2\left(1-x\right)}}{\frac{1}{2^{2x-1}}-1}=\frac{2^{2x}-2^{2\left(1-x\right)}.2^{2x-1}}{2^{2x-1}-1}=\frac{2^{2x}-2^1}{2^{2x-1}-1}=\frac{2\left(2^{2x-1}-1\right)}{2^{2x-1}-1}=2\)
Áp dụng D ta được
\(P\left(\frac{1}{1998}\right)+P\left(\frac{1997}{1998}\right)=2\)
\(P\left(\frac{2}{1998}\right)+P\left(\frac{1996}{1998}\right)=2\)
..............................................................
Do \(x\ne\frac{1}{2}\)nên không có \(P\left(\frac{999}{1998}\right)\)
\(P\left(\frac{998}{1998}\right)+P\left(\frac{1000}{1998}\right)=2\)
=> \(A=1997+2+2+....+2\)(998 số 2)
=> \(A=1997+2.998=3993\)
Vậy A=3993
a, \(\left(\frac{x+2}{98}+18\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
b, \(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)
\(\Rightarrow x=-100\)
b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)
\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)
\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)
b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0
\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0
\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))
\(\Leftrightarrow\)x=-1999
Vậy x=-1999
Tính nhanh hộ mình nhé:
\(\left(1999x1998+1998x1997\right)x\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
\((1999x1998+1998x1997)x(1+\frac{1}{2}\)\(:1\frac{1}{2}\)\(-1\frac{1}{3}\)\()\)
= \((1999x1998+1998x1997)x\)\((1+\frac{1}{2}\)\(:\frac{3}{2}\)\(-\frac{4}{3}\)\()\)
= \((1999x1998+1998x1997)x\)\((1+\frac{1}{3}\)\(-\frac{4}{3}\)\()\)
= \((1999x1998+1998x1997)x\)\((\frac{4}{3}-\frac{4}{3}\)\()\)
=\((1999x1998+1998x1997)x\)0
= 0
Chúc bạn học tốt!
Ta có:
\((1999x1998+1998x1997)x(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3})\)
\(=(1999x1998+1998x1997)x\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
\(=\left(1999x1998+1998x1997\right)x\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(=\left(1999x1998+1998x1997\right)x\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(1999x1998+1998x1997\right)x0=0\)
Tìm x, biết:
\(\left(\frac{1999}{2}+\frac{1998}{3}+\frac{1997}{4}+.......+\frac{1}{2000}+4000\right)x=1+\frac{1}{2}+\frac{1}{3}\)\(\frac{1}{3}\)
Ta có:(1+1999/2)+(1+1998/3)+...(2/1999)(có 1998 tổng<=>1998 số 1)+(2000 - 1998)+400
= 2001/2+2001/3+...+2001/1999+402
=2001.(1/2+1/3+...+1/1999)+402(1)
Thay (1) vào biểu thức trên và tính(tự tính nha!,tk cho mk!!!)
Bài 1: Tính
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)
b)\(\frac{1}{2000.1999}-\frac{1}{1999-1998}-\frac{1}{1998-1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
c)\(-3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{3}}}}\)
a)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n}{n+1}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot n}{2\cdot3\cdot4\cdot...\cdot\left(n+1\right)}\)
\(=\frac{1}{n+1}\)
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)
\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{n+1}{n+1}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n}{n+1}=\frac{1}{n+1}\)
Tính giá trị biểu thức :b,[1999x1998+1998 x 1997 ] x [1+\(\frac{1}{2}\): 1\(\frac{1}{2}\)- 1\(\frac{2}{3}\)]
Tìm GTNN, GTLN:
\(C=\frac{1996x+1}{1997x-1997}\left(\cdot x+1\right)\)
\(F=\frac{2\left|x\right|+3}{3\left|x\right|-1}\)
\(K=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
\(M=\left|x-2\right|+\left|x+28\right|+\left|x-60\right|\)
\(K=|x-1|+|x-2|+|x-3|\)
\(=\left(|x-1|+|x-3|\right)+|x-2|\)
\(=\left(|x-1|+|3-x|\right)+|x-2|\)
Đặt \(A=|x-1|+|3-x|\ge|x-1+3-x|\)
Hay \(A\ge2\left(1\right)\)
Dấu "= " xảy ra \(\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1< 0\\3-x< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le3\end{cases}}\)hoặc \(\hept{\begin{cases}x< 1\\x>3\end{cases}\left(loai\right)}\)
\(\Leftrightarrow1\le x\le3\)
Đặt \(B=|x-2|\)
Ta có: \(|x-2|\ge0;\forall x\)
Hay \(B\ge0;\forall x\left(2\right)\)
Dấu "=" xảy ra \(\Leftrightarrow|x-2|=0\)
\(\Leftrightarrow x=2\)
Từ \(\left(1\right);\left(2\right)\Rightarrow A+B\ge2+0\)
Hay \(K\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}1\le x\le3\\x=2\end{cases}\Leftrightarrow}x=2\)
Vậy MIN K=2 \(\Leftrightarrow x=2\)
Kiệt ơi phần M là x+28 hay là x-28 đấy
GPT:\(\frac{\left(x+1\right)\left(x+28\right)\left(x+4\right)\left(x-10\right)\left(-5\right)}{\sqrt{x}\left(x-6\right)^{\frac{1}{2}}}\ln\left(x^2-10\right)=0\)
Nhân tài đâu giúp mình với mình tick cho
1, \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
2, \(2.\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
3,\(-\frac{3}{2}.\left(5-\frac{1}{6}\right)+4.\left(x-\frac{1}{2}\right)=\frac{1}{2}+x\)
4,\(-\frac{5}{7}.\left(\frac{2}{5}-x\right)-\frac{1}{3}=\frac{1}{5}-\frac{3}{10}\)
5,\(4-\frac{2}{3}.\left(x-3\right)=2-\frac{1}{2}+\frac{2}{3}\)
6,\(\frac{2}{3}-\frac{5}{3}.x=\frac{7}{10}.x+\frac{5}{6}\)
7,\(3.\left(x-\frac{5}{3}\right)+\frac{1}{2}=2\left(x-\frac{1}{4}\right)+\frac{5}{2}\)
Phần nào có bn giải rầu các men đừng giải lại nha mk sẽ ko tk đâu chỉ tik những phần chưa lm
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
Bước cưối 58/21 minh man viết nhầm nên sai
\(\left(x-\frac{9}{4}\right)=\frac{58}{21}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{87}{7}\)
\(x=\frac{87}{7}+\frac{9}{4}\)
\(x=\frac{411}{28}\)