Gỉai pt sau giúp mk với:
a) \(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+.....+\frac{8}{49.57}\right)+2.\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
\(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+.....+\frac{8}{49.57}\right)+2.\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
Ta có: \(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+...+\frac{8}{49.57}\right)+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
\(\Leftrightarrow1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+\frac{1}{17}-\frac{1}{25}+....+\frac{1}{49}-\frac{1}{57}+2x-2=\frac{8x+28+15x-24}{12}\)
\(\Leftrightarrow1-\frac{1}{57}+2x-2=\frac{23x+4}{12}\)
\(\Leftrightarrow2x-\frac{58}{57}=\frac{23x+4}{12}\)
\(\Leftrightarrow24x-\frac{232}{19}=23x+4\)
\(\Leftrightarrow x=\frac{308}{19}\)
1.Tìm x biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
2. Tìm x biết:
\(\left(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}\right)+\frac{58}{57}+2.\left(x-1\right)=2x+\frac{7}{3}+5x-\frac{8}{4}\)
3. Chứng minh với mọi n>1 thì:
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{n.\left(n+2\right)}\right)<2\)
1/
\(1+\frac{2014}{2}+...+\frac{4024}{2012}=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{2012}{2012}\right)\)
\(=2012+2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
Phương trình đã cho tương đương:
\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
\(\Leftrightarrow503x=2012\)
\(\Leftrightarrow x=4\)
2/
\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}+\left(1+\frac{1}{57}\right)-2-\frac{7}{3}+\frac{8}{4}=5x\)
\(\Leftrightarrow\)\(5x=\frac{17}{3}\Leftrightarrow x=\frac{17}{15}\)
3/
Ta có: \(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{n\left(n+2\right)}\right)\)\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.......\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(=2.\frac{n+1}{n+2}
Tìm x:
\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
Ta có: \(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}=1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}=1-\frac{1}{57}=\frac{56}{57}\)
Vậy: 56/57 + 58/57 + 2x - 2 = 2x + 7/3 + 5x - 8/4
2 + 2x - 2 = 2x + 7/3 + 5x - 8/4
2x = 2x + 7/3 + 5x - 8/4
=> 7/3 + 5x - 8/4 = 0
1/3 + 5x = 0
=> 5x = -1/3
=> x = -1/3 : 5=-1/15
Giải pt sau:
a) \(\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+....+\dfrac{8}{49.57}\right)+2.\left(x-1\right)=\dfrac{2x+7}{3}+\dfrac{5x-8}{4}\)
b) \(\left(x+2\right).\left(x-2\right).\left(x^2-10\right)=72\)
c) (x+3)4+ (x+5)4 = 2
a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)
\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)
=>1/12x=77/57
=>x=308/19
b: =>(x^2-4)(x^2-10)=72
=>x^4-14x^2+40-72=0
=>x^4-14x^2-32=0
=>(x^2-16)(x^2+2)=0
=>x^2-16=0
=>x^2=16
=>x=4 hoặc x=-4
Giải phương trình:
h) \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
i) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
k) \(\left(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}\right)+\frac{58}{57}+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
l) \(\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-660}{46}=0\)
các bạn không giải thì làm ơn đừng trả lời
\(h.\) \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow\) \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}+4=0\)
\(\Leftrightarrow\) \(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}\right)=0\)
\(\Leftrightarrow\) \(\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow\) \(\left(200-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow\) \(200-x=0\) \(\Leftrightarrow\) \(x=200\)
Tìm x :
\(\frac{8}{1.9}+\frac{8}{1.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
giải các pt
\(a,\frac{2x-13}{2x-16}+\frac{2\left(x-6\right)}{x-8}=\frac{7}{8}+\frac{2\left(5x-39\right)}{3x-24}\)
\(b,x\left(x-2\right)\left(x-1\right)\left(x+1\right)=24\)
\(c,x^4+2x^3+5x^2+4x-12=0\)
câu a tự quy đồng cùng mẫu rồi làm thôi :"))
b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
gải pt
\(\frac{2x-13}{2x-16}+\frac{2\left(x-6\right)}{x-8}=\frac{7}{8}+\frac{2\left(5x-39\right)}{3x-24}\)
máy tính mik khó viết nhưng bài này có mẫu chung nên dễ làm mà
bn cứ đưa mẫu ra có x-8 chung đó
sau đó tính tiếp theo bt là ra mà
bạn ơi bạn làm chi tiết ra ik mk thư rôi nhưng không đc