a, [(4.x+28):3+55]:5=35
b, (2/3.5+2/5.7+2/7.9+.....+2/19.21).x=9/7
tìm x
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1/3.5+1/5.7+1/7.9+...+1/(2x+1)(2x+3)=5/31
1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=5/31
1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3=5/31:1/2
1/3-1/2x+3=10/31
1/2x+3=1/3-10/31
1/2x+3=1/63
suy ra : 2x+3=63
2x=63-3
2x=60
x=60:2
x=30
vay x=30
nhớ **** cho mình nha
Đặt \(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{101}\)
\(\Rightarrow\frac{1}{2}A=\frac{101}{303}-\frac{3}{303}\)
\(\Rightarrow\frac{1}{2}A=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}\div\frac{1}{2}\)
\(\Rightarrow A=\frac{199}{303}\)
\(x+\frac{3}{22}=\frac{27}{121}.\frac{11}{9}\)
\(\Leftrightarrow x+\frac{3}{22}=\frac{27.11}{121.9}\)
\(\Leftrightarrow x+\frac{3}{22}=\frac{3.1}{11.1}\)
\(\Leftrightarrow x+\frac{3}{22}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}-\frac{3}{22}\)
\(\Leftrightarrow x=\frac{6}{22}-\frac{3}{22}\)
\(\Leftrightarrow x=\frac{3}{22}\)
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)
Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)
=> A = 98B
<=> 2/1.3 + 2/3.5 + 2/5.7 +....+ 2/(x+2)(x+4) = 100/101
<=> 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.....+ 1/x+2 - 1/x+4 = 100/101
<=> 1 - 1/x+4 = 100/101
<=> 1/x+4 = 1 - 100/101 <=> 1/x+4 = 1/101 <=> x+4 = 101 <=> x= 101 - 4 = 97
:)
\(\left(2+x\right)+\left(4+x\right)+\left(6+x\right)+...+\left(52+x\right)=780\)
\(\Leftrightarrow26x+\left(2+4+6+...+52\right)=780\)
\(\Leftrightarrow26x+\left[\left(52+2\right).26:2\right]=780\)
\(\Leftrightarrow26x+702=780\)
\(\Leftrightarrow26x=78\)
\(\Leftrightarrow x=\frac{78}{26}=3\)
a,[(4x+28):3+55]:5=35
(4x+28):3+55=175
(4x+28):3=120
4x+28=360
4x=332
x=83
a) [( 4x + 28 ) : 3 + 55] : 5 = 35
( 4x + 28 ) : 3 + 55 = 175
( 4x + 28 ) : 3 = 120
4x + 28 = 360
4x = 332
x = 83
b) \(\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\right)\cdot x=\frac{9}{7}\)
\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\cdot x=\frac{9}{7}\)
\(\left(\frac{1}{3}-\frac{1}{21}\right)\cdot x=\frac{9}{7}\)
\(\frac{2}{7}\cdot x=\frac{9}{7}\)
\(x=\frac{9}{2}\)