a)3.2x -3=45
b)4x3 +12=120
c)1500:[(36x+40):x]=30
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a: Ta có: \(\left[\left(6x-39\right):3\right]\cdot28=5628\)
\(\Leftrightarrow\left(6x-39\right):3=201\)
\(\Leftrightarrow6x-39=603\)
\(\Leftrightarrow6x=642\)
hay x=107
b: Ta có: \(4x^3+12=120\)
\(\Leftrightarrow4x^3=108\)
\(\Leftrightarrow x^3=27\)
hay x=3
c: Ta có: \(1500:\left[\left(30x+40\right):x\right]=30\)
\(\Leftrightarrow\left(30x+40\right):x=50\)
\(\Leftrightarrow30x+40=50x\)
hay x=-2
a, \(4x^3+12=120\)
\(\Leftrightarrow4x^3=120-12\)
\(\Leftrightarrow4x^3=108\)
\(\Leftrightarrow x^3=108:4\)
\(\Leftrightarrow x^3=27\)
\(\Leftrightarrow x=3\)
b, \(3.2^x-3=45\)
\(\Leftrightarrow3.2^x=45+3\)
\(\Leftrightarrow3.2^x=48\)
\(\Leftrightarrow2^x=48:3\)
\(\Leftrightarrow2^x=16\)
\(\Leftrightarrow x=4\)
Câu c, bạn làm tương tự nha !
a, 4x^3 + 12 = 120 b,3.2^x-3=45 c, 1500:[(30x +40):x]=30
4x^3=120-12 3.2^x =45+3 (30x +40):x =1500:30
4x^3=108 3.2^x =48 (30x +40):x = 50
x^3=108:4 2^x =48:3 30x:x+40:x = 50
x^3=27 2^x =16 30+40:x=50
=> x^3= 3^3 2^x =2^4 40:x=50-30
=> x = 3 => x =4 40:x= 20
=>x=40:20
=>x=20 ( câu c tui có làm sai thì đừng trách nhé )
a) \(\Leftrightarrow x^2-x-x^2+2x=5\)
\(\Leftrightarrow x=5\)
b) \(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0
\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy x = 0 , x = 3 hoặc x = -3
\(a,\Leftrightarrow x^2-x-x^2+2x=5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow4x\left(x^2-9\right)=0\\ \Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-3^2-72=0\\ \Leftrightarrow\left(x^2-9x+17\right)^2-81=0\\ \Leftrightarrow\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)=0\\ \Leftrightarrow\left(x-8\right)\left(x-1\right)\left(x^2-9x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\\left(x-\dfrac{9}{2}\right)^2+\dfrac{23}{4}=0\left(vô.n_0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a) \(\Rightarrow4x\left(x^2-9\right)=0\)
\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)
\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
ko chia đc bạn nha
nếu chia thì x tương đương -1,33422281521
\(1500:\left(30x+40\right):2=30\)
⇔ \(750:\left(30x+40\right)=30\)
⇒ \(30x+40=750:30=25\)
⇔ \(30x=-15\)
⇔ \(x=-\dfrac{1}{2}\)
\(4x^3+12=120\)
\(\Rightarrow4x^3=108\)
\(\Rightarrow x^3=108:4=27=3^3\)
\(\Rightarrow x=3\)
\(3.2^x-3=45\)
\(\Rightarrow3.2^x=48\)
\(\Rightarrow2^x=16=2^4\Rightarrow x=4\)
\(1500:\left[\left(30x+40\right):x\right]=30\)
\(\Rightarrow\left(30x+40\right):x=50\)
\(\Rightarrow30x:x+40:x=50\)
\(\Rightarrow30+40:x=50\)
\(\Rightarrow40:x=20\Rightarrow x=2\)
3.2x—3=45
3.2x=45+3
3.2x=48
2x=48:3
2x=16
x=16:2
x=8
4x .3+12=120
4x.3=120–12
4x.3=108
4x=108:3
4x=36
x=36:4
x=9
\(3.2x-3=45\)
\(3.2x=45+3\)
\(3.2x=48\)
\(2x=\frac{48}{3}\)
\(2x=16\)
\(x=8\)
Vậy \(x=8\)
\(1500:\left[\left(36x+40\right):x\right]=30\)
\(\left(36x+40\right):x=1500:30\)
\(36x+40=50x\)
\(50x-36x=40\)
\(14x=40\)
\(x=\frac{40}{14}\)
\(x=\frac{20}{7}\)
Vậy \(x=\frac{20}{7}\)