Tim x,biet : (3/20 + 1/20 - x) : 32/9 = 21/128
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Nếu đúng thì ấnĐúng 2 không những thees sau khi ấn sẽ may mắn cả năm
\(\left(\frac{3}{20}+\frac{1}{2}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\left(\frac{13}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\frac{13}{20}-x=\frac{21}{128}\times\frac{32}{9}\)
\(\frac{13}{20}-x=\frac{7}{12}\)
\(x=\frac{13}{20}-\frac{7}{12}\)
\(x=\frac{1}{15}\)
\(\left(\frac{3}{20}+\frac{1}{2}-\chi\right):\frac{32}{9}=\frac{21}{128}\)
\(\Rightarrow\left(\frac{3}{20}+\frac{10}{20}-\chi\right)=\frac{21}{128}\times\frac{32}{9}\)
\(\Rightarrow\frac{13}{20}-\chi=\frac{7}{12}\)
\(\Rightarrow\chi=\frac{13}{20}-\frac{7}{12}\)
\(\Rightarrow\chi=\frac{1}{15}\)
c,32+48:x=40 =>48:x=8=>x=48:8=>x=6
x-128=-128(cái này tự ấn máy tính nha viết hết ra lâu lắm)
x=-128+128
x=0
ko biết đúng ko nữa nếu đúng thì tick cho mình nha
a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
x = 1
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
x = 17
\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Rightarrow x=1\)
\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)
\(\Rightarrow x+1=18\Leftrightarrow x=17\)
Ta có : A=20/11×13 + 20/13×15 +20/15×17+...+20/53×55
A = 10 ×( 2/11×13+2/13×15+...12/53×55)
A = 10 ×(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)
A = 10 × (1/11-1/55)
A =10 × 4/55
A = 8/11
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\Leftrightarrow\frac{4}{20}-x=\frac{21}{128}.\frac{32}{9}\)
\(\Leftrightarrow\frac{1}{5}-x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{1}{5}-\frac{7}{12}\)
\(\Leftrightarrow x=-\frac{23}{60}\)
Vậy \(x=\frac{-23}{60}\)
Bài làm
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\left(\frac{3}{20}+\frac{1}{20}-x\right)=\frac{21}{128}.\frac{32}{9}\)
\(\frac{4}{20}-x=\frac{7}{4}.\frac{1}{3}\)
\(\frac{4}{20}-x=\frac{7}{12}\)
\(x=\frac{4}{20}-\frac{7}{12}\)
\(x=\frac{12}{60}-\frac{35}{60}\)
\(x=-\frac{23}{60}\)
Vậy \(x=-\frac{23}{60}\)
\(x=-\frac{37}{15}\)