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Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
Bài 1:
1: 7/20-|x+2/5|=10/21
=>|x+2/5|=-53/420(vô lý)
2: \(\left|\dfrac{3}{7}-x\right|-\left(-\dfrac{2}{3}\right)=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-\dfrac{3}{7}\right|=\dfrac{3}{2}-\dfrac{2}{3}=\dfrac{5}{6}\)
=>x-3/7=5/6 hoặc x-3/7=-5/6
=>x=53/42 hoặc x=-17/42
X+1/3 = 3/ 4
X = 3/4 -1/3
X = 5/12
X - 2/5 = 5/7
X = 5/7 -2/5
X = 9/35
-X - 2/3 = 6/7
-X = 6/7 - 2/3
-X = 4/21
4/7 - X = 1/3
X = 4/7 - 1/3
X = 5/21
k mk nha bn!!!! thank bn nhìu nha
x + 1/3 = 3/4
x = 3/4 - 1/3
x = 5/12
x-2/4=5/7
x=5/7+2/5=
x=39/35
tườn tự nhé
giúp tớ nhé
tớ bị trừ 590 điểm
cảm ơn trước
1) \(\left(x-5\right)\left(x+7\right)-7x\left(x+3\right)\)
\(=x^2+7x-5x-35-7x^2-21x\)
\(=-6x^2-19x-35\)
2) \(\left(x+5\right)\left(x+7\right)-\left(x-4\right)\left(x+3\right)\)
\(=x^2+5x+7x+35-\left(x^2+3x-4x-12\right)\)
\(=x^2+12x+35-x^2+x+12\)
\(=13x+47\)
3) \(\left(2x-3\right)\left(x+4\right)+\left(-x+1\right)\left(x-2\right)\)
\(=2x^2+8x-3x-12-x^2+2x+x-2\)
\(=x^2+8x-14\)
b) \(|x-7|=\frac{1}{4}+|\frac{-5}{3}+\frac{1}{3}|\)
\(\Leftrightarrow|x-7|=\frac{1}{4}+\frac{4}{3}\)
\(\Leftrightarrow|x-7|=\frac{19}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=\frac{19}{12}\\x-7=\frac{-19}{12}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{103}{12}\\x=\frac{65}{12}\end{cases}}\)
Phần a bạn nhầm lẫn rồi nhé có 2 dấu bằng
\(c,|x+5|=\frac{1}{7}-|\frac{4}{3}-\frac{1}{6}|\)
\(\Rightarrow|x+5|=\frac{1}{7}-\frac{7}{6}\)
\(\Rightarrow|x+5|=-\frac{43}{42}\)
mà \(|x+5|\ge0\)
\(\Rightarrow\text{không tìm thấy giá trị của x}\)