(1-\(\frac{1}{2}\))x (1-\(\frac{1}{3}\))x(1-\(\frac{1}{4}\))x(1-\(\frac{1}{5}\))x......x(1-\(\frac{1}{2016}\))
Help me
Help me:
Tìm x:
a, \(\frac{x+7}{7}+\frac{x+14}{6}=\frac{x-2008}{2016}+\frac{x+2023}{2015}\)
b, \(\frac{x+1}{8}+\frac{x+2}{7}=\frac{x}{3}-\frac{x+3}{6}\)
c, 1+\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\cdot\left(x+1\right)}=\frac{99}{50}\)(x thuộc N*)
Tìm x:
\(x.\frac{1}{2}-x.\frac{2}{3}+x.\frac{3}{4}-x.\frac{5}{6}=\frac{5}{6}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
help me pleas
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\)= \(\frac{1}{4}\)=> x. (\(\frac{1}{2}\)- \(\frac{2}{3}\) + \(\frac{3}{4}\)- \(\frac{5}{6}\)) = \(\frac{1}{4}\)=> x.( \(\frac{6}{12}\)- \(\frac{8}{12}\)+\(\frac{9}{12}\)-\(\frac{10}{12}\))= \(\frac{1}{4}\)=> x. \(\frac{-1}{4}\)=\(\frac{1}{4}\)=> x = \(\frac{1}{4}\): \(\frac{-1}{4}\)=> x = -1=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
\(GPT\)
\(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
HELP ME
đặt x-2016=a
y-2017=b
z-2018=c
ta có\(\frac{1}{\sqrt{a}}-\frac{1}{a}+\frac{1}{\sqrt{b}}-\frac{1}{b}+\frac{1}{\sqrt{c}}-\frac{1}{c}=\frac{3}{4}\)
=>\(\left(\frac{1}{\sqrt{a}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{2}\right)^2=0\)
=>\(a=b=c=4\)
còn lại tự lm nốt
Đặt \(\hept{\begin{cases}a=\sqrt{x-2009}\\b=\sqrt{y-2010}\\c=\sqrt{z-2011}\end{cases}}\)(với a,b,c>0). Khi đó phương trình đã cho trở thành
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2\)
\(\Leftrightarrow a=b=c=2\)\(\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
Giải phương trình:
1,\(\left(x^2-x+1\right)^4+5x^4=6\left(x^2-x+1\right)^4\)
2,\(\frac{x+4}{x-1}+\frac{x-4}{x+1}=\frac{x-8}{x+2}+\frac{x+8}{x-2}+\frac{8}{3}\)
3,\(\left|x-2015\right|^{2015}+\left|x-2016\right|^{2016}=1\)
4,\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
5,\(\left(x+2008\right)^4+\left(x+2009\right)^4=\frac{1}{8}\)
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
Bài 1: Tìm các cặp số nguyên x,y có tích bằng 1 biết: \(\frac{x+y-2}{4}\)= \(\frac{1}{x-1}\)+ \(\frac{1}{y-1}\).
Bài 2: Tính giá trị của biểu thức:
A= \(\frac{1}{2016^{2016}+1}\)+ \(\frac{1}{2016^{-2015}+1}+...+\frac{1}{2016^{-1}+1}+\frac{1}{2016^0+1}+\frac{1}{2016^1+1}+...+\frac{1}{2016^{2016}+1}\)
Help me!!! Mình cần gấp lắm bạn nào giỏi toán giúp mình với !!!! Cảm ơn nhiều!!!
Ví dụ : Tìm tập hợp các ước của 24
Ư(24) = {1 ; 2 ; 3 ; 4 ; 6 ; 8 ; 12 ; 24 }
Ta có thể tìm các ước của a bằng cách lần lượt chia a cho
các số tự nhiên từ 1 đến a để xét xem a chia hết cho những
số nào ,khi đó các số ấy là ước của a
Câu 1Tính giá trị biểu thức A biết
A=\(\frac{4+\frac{5}{6}-\frac{1}{9}}{10-\frac{7}{12}+\frac{1}{16}}-\frac{3-\frac{1}{5}+\frac{1}{3}-\frac{1}{9}}{9-\frac{3}{5}+1-\frac{1}{3}}\)
Câu 3 : Tìm x biết : 2016.x+x.\(\frac{1}{2016}\)-2016=\(\frac{1}{2016}\)
Câu 4 : Tìm tất cả các cặp số nguyên x,y biết rằng : (x-y).(y+3)2=9
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
Cảm ơn bạn rất nhiều mình đã hiểu rồi
Chúc bạn học tốt nhé
help me
1) \(\frac{3-7x}{1+x}=\frac{1}{2}\)
2)\(\frac{x+1}{x-2}=\frac{1}{x^2-4}\)
3) \(\frac{y-1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)
4) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2+1}\)
5) \(1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
6) \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
7) \(\frac{2x}{x+2}-\frac{x}{x-2}=\frac{-4x}{x^2-4}\)
8) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
9) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)
10) \(\frac{2x-3}{x+2}-\frac{x+2}{x-2}=\frac{2}{x^2-4}\)
11)\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)
12)\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
bài 1
\(ĐKXĐ:1+x\ne0\Rightarrow x\ne-1\)
\(\frac{3-7x}{1+x}=\frac{1}{2}\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\\
\Leftrightarrow-14x-x=1-6\\
\Leftrightarrow-15x=-5\\
\Leftrightarrow x=\frac{1}{3}\left(N\right)\)
Tìm x, biết:
a) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)\div\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
b) \(-\frac{3x}{4}.\left(\frac{1}{x}+\frac{2}{7}\right)=0\)
c) \(3-\frac{1-\frac{1}{2}}{1+\frac{1}{x}}=2\frac{2}{3}\)
Help me, cần gấp lắm. Nhanh lên ạ!
a) ĐKXĐ : \(x\ne0\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=\frac{-5}{4}\)
\(\left(\frac{-9x}{3x}+\frac{9}{3x}-\frac{x}{3x}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=\frac{-5}{4}\)
\(\frac{-9x+9-x}{3x}:\frac{15+6+10}{15}=\frac{-5}{4}\)
\(\frac{-10x+9}{3x}:\frac{31}{15}=\frac{-5}{4}\)
\(\frac{-10x+9}{3x}=\frac{-31}{12}\)
\(\Leftrightarrow12\left(-10x+9\right)=-31\cdot3x\)
\(\Leftrightarrow-120x+108=-93x\)
\(\Leftrightarrow-120x+93x=-108\)
\(\Leftrightarrow-27x=-108\)
\(\Leftrightarrow x=4\)
b) ĐKXĐ : \(x\ne0\)
\(\frac{-3x}{4}\cdot\left(\frac{1}{x}+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{-3x}{4}=0\\\frac{1}{x}+\frac{2}{7}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\\frac{-2}{-2x}=\frac{-2}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\x=\frac{-7}{2}\end{cases}}\)
Vậy.....
c) phân tích ra rồi làm thôi e :)) a bận rồi
Trần Thanh Phương làm đúng rồi đó !
k bạn ý đi !