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Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
Ta có " (x - 5)7 = (x - 5)4
=> (x - 5)7 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)3 - 1] = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^3-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
a) Ta có: \(\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)
TH1: \(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
TH2: \(2x+\frac{3}{5}=\frac{-3}{5}\Rightarrow2x=\frac{-6}{5}\Rightarrow x=\frac{-3}{5}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
Mà \(\frac{-1}{27}=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\Leftrightarrow3x=\frac{1}{6}\Rightarrow x=\frac{1}{18}\)
c) \(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{2}{3}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{2}{3}x+\frac{5}{6}=0\)
\(\Rightarrow-\frac{37}{6}x=\frac{-1}{6}\Rightarrow x=\frac{1}{37}\)
d) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3-x-\frac{1}{5}=0\)
\(\Rightarrow-3x=\frac{47}{10}\Rightarrow x=\frac{-47}{30}\)
a. 1440 : [ 120 - (3x + 9 ) ] = 120
120 - (3x + 9) = 1440 : 120 = 12
3x + 9 = 120 - 12 = 108
3x = 108 - 9 = 99
x = 99 : 3 = 33
b. 120 + [ ( 999 - 9x ) : 60 ] . 24 = 480
[ ( 999 - 9x ) : 60 ] . 24 = 480 - 120 = 360
( 999 - 9x ) : 60 = 360 : 24 = 15
( 999 - 9x ) = 15 . 60 = 900
9x = 999 - 900 = 99
x = 99 : 9 = 11
Bài 1:
a) A=1+22+24+.................+2100
2A=(1+22+24+.................+2100)
2A=2+23+...+2101
2A-A=(2+23+...+2101)-(1+22+24+.................+2100)
A=2101-1
b)bạn tự làm
c) C=-1/90-1/72-1/50-1/42-1/30-1/20-1/12-1/6-1/2
\(=-\left(\frac{1}{90}+\frac{1}{72}+...+\frac{1}{2}\right)\)
\(=-\left(\frac{1}{10.9}+\frac{1}{9.8}+...+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{10}-1\right)\)
\(=-\left(-\frac{9}{10}\right)=\frac{9}{10}\)
Bài 2:
cứ tính lần lượt là ra