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\(\left(3.x^2-2^4\right)2^3=2^8\\ \left(3.x^2-2^4\right)=2^8:2^3\\ \left(3.x^2-2^4\right)=2^5\\ 3.x^2-2^4=32\\ 3.x^2=32+2^4\\ 3.x^2=48\\ x^2=48:3\\ x^2=16\\ x=+-4\)
a,\(\frac{1}{3}x=-2-\frac{2}{3}=\frac{-8}{3}\)
\(x=\frac{-8}{3}:\frac{1}{3}=\frac{-8}{3}.\frac{3}{1}=-8\)
\(b,\left[x+\frac{1}{1}\right]^2+\frac{5}{6}=\frac{7}{8}\)
\(\Rightarrow\left[x+1\right]^2=\frac{7}{8}-\frac{5}{6}\)
\(\Rightarrow\left[x+1\right]^2=\frac{7\cdot3}{24}-\frac{5\cdot4}{24}\)
\(\Rightarrow\left[x+1\right]^2=\frac{21}{24}-\frac{20}{24}\)
\(\Rightarrow\left[x+1\right]^2=\frac{1}{24}\)
\(\Rightarrow x\in\left\{\varnothing\right\}\)
a: \(\Leftrightarrow\left[\left(3x+14\right):4-3\right]:2=1\)
=>(3x+14):4-3=2
=>(3x+14):4=5
=>3x+14=20
=>3x=6
hay x=2
b: \(\Leftrightarrow\left[\left(x:4+17\right):10+3\cdot16\right]:10=5\)
\(\Leftrightarrow\left(x:4+17\right):10=50-48=2\)
=>x:4+17=20
=>x:4=3
hay x=12
c: \(\Leftrightarrow2\cdot15^2+\left[2\cdot125-\left(2x+4\right)\cdot5\right]:19=453\)
\(\Leftrightarrow250-\left(2x+4\right)\cdot5=\left(453-450\right)\cdot19=57\)
=>5(2x+4)=197
=>2x+4=197/5
=>2x=177/5
hay x=177/10
d: \(\Leftrightarrow\left(19x+50\right):14=5^2-4^2=9\)
=>19x+50=126
=>19x=76
hay x=4
e: \(\Leftrightarrow2\cdot3^x=10\cdot3^{12}+8\cdot3^{12}=18\cdot3^{12}\)
\(\Leftrightarrow3^x=3^2\cdot3^{12}=3^{14}\)
hay x=14
f: \(\Leftrightarrow3\left(x+2\right):7=30\)
=>3(x+2)=210
=>x+2=70
hay x=68
g: \(2480-1570+200-x+5=1010\)
=>1115-x=1010
hay x=105
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
a)
<=> 3x - 3 + x - 2 = 2x - 2 - x + 1
<=> 3x + x - 2x + x = -2 + 1 + 3 + 2
<=> 3x = 4
<=> x = 4/3
Các câu sau làm tương tự
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
a) \(\left(x^2+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\varnothing\\x=\pm2\end{cases}}}\)
Vậy x=\(\pm2\)
b) \(\left(x^3-27\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^3-27=0\\x^3+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^3=27\\x^3=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
Vậy x=3; x=-2
d) \(|3x+8|-|x-4|=0\)
\(\Leftrightarrow|3x+8|=|x-4|\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=x-4\\-3x-8=x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-12\\-4x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\x=-1\end{cases}}}\)
Vậy x=-6; x=-1
a) (3x2-16).8 = 256
3x2-16 = 32
3x2 = 48
x2 = 16 =>x= 4 hoặc (-4)
|x-5| - (-6) = 16
|x-5|+6=16
|x-5|=16-6=10
=> x-5 ∈ {10; -10}
=> x = 15 hoặc (-5)