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\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+3\right)}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\)
\(=>x=98\)
\(D=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(D=1-\frac{1}{x+3}=\frac{100}{101}\)
\(D=\frac{1}{x+3}=1-\frac{100}{101}\)
\(D=\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\Rightarrow x=98\)
Ủng hộ mk nha ^_^
đặt VT là A ta đc:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}\)
\(A=\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc:\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
3. ( 1/1.4 +1/4.7 +1/7.10 +...+ 1/x.(x+3)
3/1.4 +1/4.7+1/7.10 + ...+ 3/ x . (x+3)
1/1 - 1/4 + 1/4 - 1/6 + 1/7 - 1/10 + ...+ 1/x-1/x+3
1/1 - 1/x+3
x+3/x+3 - 1/x+3
x+2/x+3
\(2x-\left(21.3.105.61\right)=\left(-11\right).26\)
\(2x-403515=-286\)
\(2x=\left(-286\right)+403515\)
\(2x=403229\)
\(x=403229:2\)
\(x=\frac{403229}{2}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}\)
\(=\frac{1}{1}+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{91}-\frac{1}{91}\right)-\frac{1}{94}\)
\(=\frac{1}{1}-\frac{1}{94}\)
\(=\frac{93}{94}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{18}{19}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{18}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
...............
đặt VT là A ta có:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{6}{19}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}\)
\(\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19
1/3.(1-1/x+3)=6/19
1-1/x+3=6/19:1/3
1-1/x+3=18/19
1/x+3=1-18/19
1/x+3=1/19
=> x+3=19
=>x=19-3
x=16
Đặt biểu thức là A, ta có:
3A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{x\left(x+3\right)}\)
3A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
3A=1-\(\frac{1}{x+3}\)
A=\(\frac{1}{3}-\frac{3}{x+3}\)
=>\(\frac{1}{3}-\frac{3}{x+3}\) =\(\frac{6}{19}\) =>x=168
a) \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)
Vậy....
b) \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\) vô lí do \(\left|a\right|\ge0\)
Vậy pt vô nghiệm
c) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)
Vậy..
d) \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do \(\left|a\right|\ge0\)với mọi a
Vậy pt vô nghiệm
e) \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
A=1-1/4+1/4-1/7+1/7-1/10+....+1/n-1/(n+3)
A=1-1/(n+3)
vì 1/(n+3)lớn hơn 0 nên 1-1/(n+3)<1
=>A<1
(x-1+3)/9=1/y+2
(x+2)/9=1/(y+2)
tích chéo:x.y+2x+2y=5
phân phối ra rồi tìm ước của 5 sau đó lập bảng là ra
\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(=1-\frac{1}{x+3}\)
\(=\frac{x+2}{x+3}=\frac{100}{101}\)
\(\Rightarrow x=98\)
\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(A=1-\frac{1}{x+3}=\frac{100}{101}\)
\(\frac{1}{x+3}=1-\frac{100}{101}=\frac{1}{101}\)
=> x + 3 = 101
=> x = 101 - 3
=> x = 98
Vậy x = 98
Ủng hộ mk nha ^_-