tìm x
72x -3- 2. 72= 7 2 . 5
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72x-2 +42 = 18- 12020
<=> 72x-2 + 16 = 18 -1
<=> 72x-2 = 1
<=> 2x-2 = 0
<=> x = 1
\(\dfrac{5}{7}=\dfrac{70}{98};\dfrac{2}{5}=\dfrac{14}{35};\dfrac{1}{8}=\dfrac{9}{72};\dfrac{120}{80}=\dfrac{3}{2}\)
\(\dfrac{5}{7}=\dfrac{70}{98}\)
\(\dfrac{2}{5}=\dfrac{14}{35}\)
\(\dfrac{1}{8}=\dfrac{9}{72}\)
\(\dfrac{120}{80}=\dfrac{3}{2}\)
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
b)
\(3\left(2x^2-7\right)=33\)
\(\Leftrightarrow2x^2-7=11\)
\(\Leftrightarrow2x^2=18\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
a) -2(2x - 8) + 3(4 - 2x) = -72 - 5(3x - 7)
=> -4x + 16 + 12 - 6x = -72 - 15x + 35
=> -10x + 28 = -37 - 15x
=> -10x + 15x = -37 - 28
=> 5x = -65
=> x = -65 : 5
=> x = -13
b) 3(2x2 - 7) = 33
=> 2x2 - 7 = 33 : 3
=> 2x2 - 7 = 11
=> 2x2 = 11 + 7
=> 2x2 = 18
=> x2 = 18 : 2
=> x2 = 9
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy ...
c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
1: 243^5=(3^5)^5=3^25
3*27^8=3*3^24=3^25=243^5
3: 3^300=27^100
2^200=4^100
mà 27>4
nên 3^300>2^200
4: 15^2=3^2*5^2
81^3*125^3=3^12*5^9
=>15^2<81^3*125^3
6: 125^5=5^15
25^7=5^14
mà 15>14
nên 125^5>25^7
\(7^{2x-3}=7^2\cdot5+7^2\cdot2\)
\(7^{2x-3}=7^2\left(5+2\right)\)
\(7^{2x-3}=7^2\cdot7\)
\(7^{2x-3}=7^3\)
\(2x-3=3\)
\(2x=6\)
\(x=3\)
\(7^{2x-3}-2.49=49.5\)
\(7^{2x-3}-2.49=245\)
\(7^{2x-3}-2=245:49\)
\(7^{2x-3}-2=5\)
\(7^{2x-3}=5+2\)
\(7^{2x-3}=7\)
2x-3=1
2x=1+3
2x=4
x=4:2
x=2
vậy x=2