ai on làm nè:
tìm x biết: \(\frac{2030-x}{15}+\frac{2041-x}{13}+\frac{2048-x}{11}+\frac{1961-x}{9}=0\)
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Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
(x+2)/17+(x+4)/15+(x+6)/13=(x+8)/11+(x+10)/9+(x+12)/7
=>(x+2+17)/17+(x+4+15)/15+(x+6+13)/13=(x+8+11)/11+(x+10+9)/9+(x+12+7)/7
=>(x+19)/17+(x+19)/15+(x+19)/13=(x+19)/11+(x+19)/9+(x+19)/7
=>(x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0
=>(x+19)*(1/17+1/15+1/13-1/11-1/9-1/7)=0
=>x+19=0
=>x=19
áp dụng tc tỉ lệ thức ta có :
\(\Leftrightarrow\frac{671x+2804}{3315}=\frac{239x+2462}{693}\Rightarrow\left(671x+2804\right)693=3315\left(239x+2462\right)\)
=>(671x+2804)693=693(671x+2804) (VT)
<=>693(671x+2804)=3315(239x+2462)
=>465003x+1943172=792285x+8161530
=>-327282x=621835
=>x=621835:(-327282)
=>x=-19
Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.
Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.
Chuyển 2 phân số ở vế kia sang rồi đặt (x+2) ra ngoài....Từ đó có (x+2)=0=>x=-2
\(3x.\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)
\(3x=0\Rightarrow x=0\)
\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy..
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\\ \Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=10-10\\ \Leftrightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\\ \Leftrightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)\Leftrightarrow x-100=0\left(Vi\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\right)\Leftrightarrow x=100\)
\(\frac{2030-x}{15}+\frac{2041-x}{13}+\frac{2048-x}{11}+\frac{1961-x}{9}=0\)
\(\Leftrightarrow\frac{2030-x}{15}-1+\frac{2041-x}{13}-2+\frac{2048-x}{11}-3+\frac{1961-x}{9}+6=0\)
\(\Leftrightarrow\frac{2015-x}{15}+\frac{2015-x}{13}+\frac{2015-x}{11}+\frac{2015-x}{9}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Mà \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow2015-x=0\Leftrightarrow x=2015\)