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II3x-3I+2x+1I=3x+2021^0
II3x-3I+2x+1I=3x+1
\(\)ĐK:3x+1\(\ge\)0
3x\(\ge\)-1
x\(\ge\frac{-1}{3}\)
\(\Rightarrow\)I3x-3I+2x+1=3x+1
I3x-3I=x
\(\Rightarrow\)3x-3=\(\pm\)x
TH1:3x-3=x TH2:3x-3=-x
2x=3 4x=3
x=\(\frac{3}{2}\) x=\(\frac{3}{4}\)
Vậy x=\(\frac{3}{2}\); x=\(\frac{3}{4}\)
|5x-3| - x = 7
<=> |5x-3| = 7-x
<=>\(\left[\begin{array}{nghiempt}5x-3=7-x\\5x-3=-7+x\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}6x=10\\4x=-4\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-1\end{array}\right.\)
|5x-3|-x=7
=>5x-3=7+x
=>\(\left[\begin{array}{nghiempt}5x-3=\left(-7\right)+x\\5x-3=7+x\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}5x+x=\left(-7\right)+3\\5x-x=7+3\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}5x-x=\left(-4\right)\\5x-x=10\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}6x=\left(-4\right)\\4x=10\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}x=\left(-4\right):6\\x=10:4\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}x=\frac{-2}{3}\\x=\frac{5}{2}\end{array}\right.\)
Bỏ mấy dấu thẳng phía sau nhé
\(\frac{2x-3}{2}=\frac{32}{2x-3}\)
\(\Rightarrow\left(2x-3\right).\left(2x-3\right)=2.32\)
\(\left(2x-3\right)^2=64=8^2=\left(-8\right)^2\)
=> 2x-3 = 8 => 2x = 11 => x = 11/2
2x -3 = -8 => 2x = -5 => x = -5/2
KL:...
a,\(2x^2-8x=0\)
\(2x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)
\(=2x^2-8x-3x-2x^2\)
=\(-11x\)
c,\(-11x=0\)
\(\Rightarrow x=0\)
\(A\left(x\right)=2x^2-8x\)
\(\Rightarrow2x^2-8x=0\)
\(\Rightarrow x\left(2x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)
\(B\left(x\right)=-3x+2x^2\)
\(B\left(x\right)=2x^2-3x\)
\(2x^2-3x=0\)
\(\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
1)
\(2^{x-1}=16\\ 2^{x-1}=2^4\\ \Rightarrow x-1=4\\ x=4+1\\ x=5\)
5)
\(\left(x-1\right)^2=25\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6)
\(\left|2x-1\right|=5\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
5) (x-1)2 = 25
(x-1)2 = 52
x-1 = 5
x = 5+1
x = 6
6) \(\left|2x-1\right|=5 \)
\(TH1:\) \(2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=6:2\)
\(\Leftrightarrow x=3\)
\(TH2:2x-1=-5\)
\(\Leftrightarrow2x=-5+1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-4:2\)
\(\Leftrightarrow x=-2\)
Vậy x = 3 hoặc x = -2.
Tick nha!
=>|5x-3|=2x+14
Vì 5x-3>=0 với mọi x
=>2x+14>=0
=>2x>=-14
=>x>=-7
Th1:5x-3=2x+14
=>5x-2x=3+14
=>3x=17
=>x=17/3 (thỏa mãn điều kiện x>=-7)
Th2:3-5x=2x+14
=>3-14=2x+5x
=>-11=7x
=>x=-11/7 (thỏa mãn điều kiện x>=-7)