tìm X biết : X +3/4xX=6/-11+-5/2
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Theo đầu bài ta có:
\(A=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{1}{6}\)
\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{6}\)
\(x+3=1:\frac{1}{30}\)
\(x=30-3\)
\(x=27\)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{1}{6}\)
\(\Rightarrow3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=\frac{3}{6}\)
\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\Rightarrow-3.\left(x+3\right)=10\)
=>-3x-9=10
=>-3x=19
=>x=-19/3
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
2.(x-3)+5=11
2.(x-3)=11-5
2.(x-3)=6
x-3=6:2
x-3=3
x=3+3
x=6
vậy :x=6
toán này lớp 6 mà phải lớp 7 đâu còn nếu lớp 7 thì em mới học lớp 6 mà vẫn làm đc
a) \(\frac{X}{2}-\frac{11}{5}=\frac{7}{8}\cdot\frac{64}{49}\)
\(\Rightarrow\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)
\(\Rightarrow\frac{x}{2}=\frac{8}{7}+\frac{11}{5}\)
\(\Rightarrow\frac{x}{2}=\frac{117}{35}\)\(\Rightarrow x=\frac{117}{35}\cdot2=\frac{234}{35}\)
b)\(\frac{x}{5}+\frac{9}{2}=\frac{6}{7}\cdot\frac{36}{49}\)
\(\Rightarrow\frac{x}{5}+\frac{9}{2}=\frac{216}{343}\)
\(\Rightarrow\frac{x}{5}=\frac{216}{343}-\frac{9}{2}=\frac{-2655}{686}\)
\(\Rightarrow x=\frac{-2655}{686}\cdot5=\frac{-13275}{686}\)
x/2 - 11/5 = 7/8 . 64/49
x/2 - 11/5 = 8/7
x/2 =8/7 + 11/5
x/2 = 117/35
x = 2.117/35 ( tính chất phân số bằng nhau)
x = 234/35
x/5 + 9/2 = 6/7 . 36/49
x/5 + 9/2 = 216/343
x/5 = 216/343 - 9/2
x/5 = -2655/686
x = 5.(-2655)/686
x = -13275/686
a, 11/12 - ( 2/5 + x ) = 2/3
<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)
b, 2x . ( x - 1/7 ) = 0
<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
vậy x={\(0;\frac{1}{7}\)}
c, 3/4 + 1/4 : x = 2/5
<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
vậy x=-5/7
a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)
\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)
\(\Leftrightarrow x=-\frac{3}{20}\)
b) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow4x=\frac{-20}{7}\)
\(\Leftrightarrow x=-\frac{5}{7}\)
\(x\) + \(\dfrac{3}{4}\)\(\times\) \(x\) = \(\dfrac{6}{-11}\) + \(\dfrac{-5}{2}\)
\(x\times\) (1 + \(\dfrac{3}{4}\)) = \(\dfrac{-67}{22}\)
\(x\) \(\times\) \(\dfrac{7}{4}\) = \(\dfrac{-67}{22}\)
\(x\) = \(\dfrac{-67}{22}\)
Vậy \(x=\) \(\dfrac{-67}{22}\)