4.12-3.(x-3)=33
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Lời giải:
\(\left\{x^2-[8^2-(5^2-8.3)^3-7.9]^3-4.12\right\}^3=1\\ \Rightarrow x^2-[8^2-(5^2-8.3)^3-7.9]^3-4.12=1\\ \Rightarrow x^2-(64-1-63)^3-48=1\\ \Rightarrow x^2-48=1\\ \Rightarrow x^2=49=7^2=(-7)^2\\ \Rightarrow x=\pm 7\)
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\(\left\{x^2-\left[8^2-\left(5^2-8.3\right)^3-7.9\right]^3-4.12\right\}^3=8000.\)
\(\Rightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-7.9\right]^3-4.12\right\}^3=8000\)
\(\Rightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-63\right]^3-48\right\}^3=8000\)
\(\Rightarrow\left\{x^2-\left[64-1-63\right]-48\right\}^3=8000\)
\(\Rightarrow\left(x^2-48\right)^3=8000\)\(\Rightarrow x^2-48=20\)
\(\Rightarrow x^2=68\)\(\Rightarrow x=2\sqrt{17}\)
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1/4.12/13+1/4.1/13-3/25
1/4.(12/13+1/13)-3/25
1/4.1-3/25
1/4-3/25
1/8
\(\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-12\%=\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-\frac{3}{25}=\frac{1}{4}\cdot\left(\frac{12}{13}+\frac{1}{13}\right)-\frac{3}{25}\)
\(=\frac{1}{4}\cdot1-\frac{3}{25}=\frac{1}{4}-\frac{3}{25}=\frac{13}{100}\)
Nhớ bài đây sửa đi sửa lại cũng vì do cái số " % " :(((
a) \(\left|\frac{2}{5}:x\right|=\frac{1}{4}\)
Trường hợp 1 : \(\frac{2}{5}\) : x = \(\frac{1}{4}\)
=> x = \(\frac{2}{5}:\frac{1}{4}=\frac{2}{5}\cdot4=\frac{8}{5}\)
Trường hợp 2 : \(\frac{2}{5}:x=-\frac{1}{4}\)
=> \(x=\frac{2}{5}:\left(-\frac{1}{4}\right)=\frac{2}{5}\cdot\left(-4\right)=-\frac{8}{5}\)
Vậy \(x=\pm\frac{8}{5}\)
b) \(\frac{x}{24}=-\frac{1}{3}-\frac{1}{8}=-\frac{11}{24}\)
=> x = -11
c) \(\frac{3}{x+3}=\frac{-7}{21}\)
=> \(\frac{3}{x+3}=\frac{-1}{3}\)
=> -1(x + 3) = 9
=> -x - 3 = 9
=> -x = 12
=> x = -12
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\(30-\left[4\left(x-2\right)+15\right]=3\) \(\left(8x-120:4\right).3^3=3^6\)
\(4\left(x-2\right)+15=27\) \(\left(8x-120:4\right)=27\)
\(4\left(x-2\right)=12\) \(8x-30=27\)
\(x-2=3\) \(8x=57\)
\(x=5\) \(x=\frac{57}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chia 2 vế cho \(27^x\) ta được:
\(3\left(\dfrac{8}{27}\right)^x+4\left(\dfrac{12}{27}\right)^x-\left(\dfrac{18}{27}\right)^x-2=0\)
\(\Leftrightarrow3\left(\dfrac{2}{3}\right)^{3x}+4\left(\dfrac{2}{3}\right)^{2x}-\left(\dfrac{2}{3}\right)^x-2=0\)
Đặt \(\left(\dfrac{2}{3}\right)^x=t>0\)
\(\Rightarrow3t^3+4t^2-t-2=0\)
\(\Leftrightarrow\left(t+1\right)^2\left(3t-2\right)=0\)
\(\Rightarrow t=\dfrac{2}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=\dfrac{2}{3}\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
15: \(\Leftrightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-63\right]^3-48\right\}=1\)
\(\Leftrightarrow x^2-48=1\)
=>x=7 hoặc x=-7
4 . 12 - 3 . ( x - 3 ) = 33
48 - 3 . ( x -3 ) = 33
3 . ( x -3 ) = 48 - 33
3 . ( x - 3 ) = 15
x -3 = 15 : 3
x - 3 = 5
x = 5 + 3
x = 8
vậy x = 8
4.12 - 3(x - 3) = 33
=> 48 - 3x + 9 = 33
=> 57 - 3x = 33
=> 3x = 24
=> x = 8