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Ta có : \(\frac{1}{4}+\frac{1}{3}:\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow\frac{1}{3}:\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}:\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}:\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\times\frac{3}{2}\)
\(\frac{1}{x}=\frac{1}{2}\)
=> x = 2
a) \(\frac{x\div3-16}{2}+21=38\)
\(\frac{x\div3-16}{2}=38+21\)
\(\frac{x\div3-16}{2}=59\)
\(x\div3-16=59.2\)
\(x\div3-16=118\)
\(x\div3=118+16\)
\(x\div3=134\)
\(x=134.3\)
\(x=402\)
b) \(\frac{1}{4}+\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\div\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{2}\)
Vậy x = ....
a,
\(x+\frac{7}{15}=1\frac{1}{20}\)
\(x+\frac{7}{15}=\frac{21}{20}\)
\(x=\frac{21}{20}-\frac{7}{15}=\frac{63}{60}-\frac{28}{60}\)
\(x=\frac{35}{60}=\frac{7}{12}\)
b,
\(\left[3\frac{1}{2}-x\right]\cdot1\frac{1}{4}=\frac{15}{16}\)
\(\left[\frac{7}{2}-x\right]\cdot\frac{5}{4}=\frac{15}{16}\)
\(\frac{7}{2}-x=\frac{15}{16}:\frac{5}{4}=\frac{3}{4}\)
\(\frac{7}{2}-x=\frac{3}{4}\Rightarrow x=\frac{7}{2}-\frac{3}{4}\)
\(x=\frac{11}{4}\)
c,
\(1\frac{1}{5}x+\frac{2}{3}x=\frac{56}{125}\Leftrightarrow\frac{6}{5}x+\frac{2}{3}x=\frac{56}{125}\)
\(\frac{28}{15}x=\frac{56}{125}\Rightarrow x=\frac{6}{25}\)
d,
\(60\%x+0,4x+x:3=2\)
\(\frac{3}{5}x+\frac{2}{5}x+\frac{1}{3}x=2\)
\(\frac{4}{3}x=2\Rightarrow x=\frac{3}{2}\)
Nguyễn Anh Thiện
a)
x + \(\frac{7}{15}\) = \(1\frac{1}{20}\)
X + \(\frac{7}{15}=\frac{21}{20}\)
X \(=\frac{21}{20}-\frac{7}{15}\)
X \(=\frac{63}{60}-\frac{28}{60}=\frac{35}{60}=\frac{7}{12}\)
^^ Học tốt !
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(x\cdot4+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(x\cdot4+\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=1\)
\(x\cdot4+\frac{13}{16}=1\)
\(x\cdot4=1-\frac{13}{16}\)
\(x\cdot4=\frac{3}{16}\)
\(x=\frac{3}{16}:4\)
\(x=\frac{3}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(=\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(=4x+\frac{15}{16}=1\)
\(\Rightarrow4x=1-\frac{15}{16}=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{16}:4=\frac{1}{64}\)
\(a,\)\(71+65\times4=\frac{x+140}{x}+260\)
\(\Rightarrow71+260=\frac{x-140}{x}+260\)
\(\Rightarrow71=\frac{x-140}{x}\)
\(\Rightarrow71x=x-140\)
\(\Rightarrow71x-x=-140\)
\(\Rightarrow70x=-140\)
\(\Rightarrow x=-2\)
\(b,\)\(y\times\frac{15}{2}-\frac{1}{3}\times\left(\frac{1}{4}+y\right)=90\frac{2}{3}\)
\(\Rightarrow\frac{15y}{2}-\frac{1}{12}-\frac{y}{3}=\frac{272}{3}\)
\(\Rightarrow\frac{90y}{12}-\frac{1}{12}-\frac{4y}{12}=\frac{1088}{12}\)
\(\Rightarrow90y-1-4y=1088\)
\(\Rightarrow86y=1089\)
\(\Rightarrow y=\frac{1089}{86}\)
a ) 1 + 2 + 3 + 4 + ... + x = 1275 ( có x số tự nhiên )
( x + 1 ) . x : 2 = 1275
( x + 1 ) . x = 1275 x 2
( x + 1 ) . x = 2550
( x + 1 ) . x = 50 . 51
Mà x , x + 1 là hai số tự nhiên liên tiếp => x = 50
Vậy x = 50
1+2+3+4+...+x=1275
\(\frac{x.\left(x+1\right)}{2}=1275\)
x(x+1)=1275x2=2550
x(x+1)=50.51
x=50
\(\Rightarrow\frac{11}{4}-X=\frac{1}{3}.\frac{6}{2}\)
\(\Rightarrow\frac{11}{4}-X=1\)
\(\Rightarrow\) \(X\) \(=\frac{11}{4}-1\)
\(\Rightarrow\) \(X=\frac{7}{4}\)
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
\(\frac{70}{3}-\left(x+4\frac{1}{5}\right)=16\)
\(\Rightarrow x+4\frac{1}{5}=\frac{22}{3}\)
\(\Rightarrow x+\frac{21}{5}=\frac{22}{3}\)
\(\Rightarrow x=\frac{22}{3}-\frac{21}{5}\)
\(\Rightarrow x=\frac{47}{15}\)
1 / 1 x 2 = 1-1/2 và 1/2x3 = (1/2)-(1/3) tương tự đến 1/(x-1).x=(1/x-1)-(1/x).
Cuối cùng ta có phép tính
1+(1/x-1)-(1/x)=15/16.