3x+1+/x-3/=22
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b: \(=\left(x^2+3x+1-3x+1\right)^2=\left(x^2+2\right)^2\)
\(a,2\left(x-1\right)+3=x+2\)
\(\Leftrightarrow2x-2+3=x+2\)
\(\Leftrightarrow2x-x=2+2-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
\(b,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)
\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(5+x\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-5;2\right\}\)
+) \(2x\left(x-4\right)-x\left(2x+3\right)+22=0\)
\(\Leftrightarrow2x^2-8x-2x^2-3x+22=0\)
\(\Leftrightarrow-11x+22=0\)
\(\Leftrightarrow-11\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
+) \(\left(2x+3\right)\left(3x+2\right)+2\left(1-3x\right)\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow6x^2+4x+9x+6+\left(2-6x\right)\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow6x^2+13x+6+2x+1-6x^2-3x=1\)
\(\Leftrightarrow12x+7=1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
2x( x - 4 ) - x( 2x + 3 ) + 22 = 0
<=> 2x2 - 8x - 2x2 - 3x + 22 = 0
<=> -11x + 22 = 0
<=> -11x = -22
<=> x = 2
( 2x + 3 )( 3x + 2 ) + 2( 1 - 3x )( x + 1/2 ) = 1
<=> 6x2 + 13x + 6 + 2( -3x2 - 1/2x + 1/2 ) = 1
<=> 6x2 + 13x + 6 - 6x2 - x + 1 = 1
<=> 12x + 7 = 1
<=> 12x = -6
<=> x = -6/12 = -1/2
1:
a: \(35\cdot16+35\cdot28-44\cdot15\)
\(=35\left(16+28\right)-44\cdot15\)
\(=44\left(35-15\right)\)
\(=44\cdot20=880\)
b: \(240-2\left(3\cdot5^2-20:2^2\right)\)
\(=240-2\left(3\cdot25-20:4\right)\)
\(=240-150+10=10+90=100\)
2:
b: \(\left(8-3x\right)^4-1=15\)
=>\(\left(3x-8\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-8=2\\3x-8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=6\end{matrix}\right.\)
=>x=10/3 hoặc x=2
c: \(218-5\left(x-8\right)=2^5:2^2\)
=>\(218-5\left(x-8\right)=2^3=8\)
=>5(x-8)=210
=>x-8=42
=>x=50
d: \(\left(5-3x\right)^4-1=15\)
=>\(\left(3x-5\right)^4-1=15\)
=>\(\left(3x-5\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-5=-4\\3x-5=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)
Gọi a,b,c,... cho dễ nhé!
a,\(7+2x=22-3x\)
\(\Leftrightarrow2x+3x=22-7\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
Vậy...
b,\(x-12+4x=25+2x-1\)
\(\Leftrightarrow x+4x-2x=25-1+12\)
\(\Leftrightarrow3x=36\)
\(\Leftrightarrow x=12\)
Vậy...
c,\(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+x=-4+4-7\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy...
d,\(8x-3=5x+12\)
\(\Leftrightarrow8x-5x=12+3\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy...
e,\(x+2x+3x-19=3x+5\)
\(\Leftrightarrow x+2x+3x-3x=5+19\)
\(\Leftrightarrow3x=24\)
\(\Leftrightarrow x=8\)
Vậy...
f,\(\left(x-1\right)-\left(2x-1\right)=9-x\)
\(\Leftrightarrow x-1-2x+1=9-x\)
\(\Leftrightarrow x-2x+x=9-1+1\)
\(\Leftrightarrow0x=9\) (Vô lý)
Vậy...
a, \(7+2x=22-3x\)
\(\Rightarrow7+2x-22+3x=0\)
\(\Rightarrow5x-15=0\)
\(\Rightarrow5x=15\Rightarrow x=3\)
b, \(x-12+4x=25+2x-1\)
\(\Rightarrow3x-12-24-2x=0\)
\(\Rightarrow x-36=0\Rightarrow x=36\)
c, \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Rightarrow7-2x-4=-x-4\)
\(\Rightarrow3-2x+x+4=0\)
\(\Rightarrow-x=-7\Rightarrow x=7\)
d, \(8x-3=5x+12\)
\(\Rightarrow8x-3-5x-12=0\)
\(\Rightarrow3x-15=0\)
\(\Rightarrow3x=15\Rightarrow x=5\)
e, \(x+2x+3x-19=3x+5\)
\(\Rightarrow6x-19-3x-5=0\)
\(\Rightarrow3x-24=0\)
\(\Rightarrow3x=24\Rightarrow x=8\)
f, \(\left(x-1\right)-\left(2x-1\right)=9-x\)
\(\Rightarrow x-1-2x+1-9+x=0\)
(hình như câu này bị sai đề rồi, bạn xem lại đề nhé)
Chúc bạn học tốt!
\(3x+1+\left|x-3\right|=22\Leftrightarrow\left|x-3\right|=21-3x\)
ĐK : \(21-3x\ge0\Leftrightarrow-3x\ge-21\Leftrightarrow x\le7\)
TH1 : \(x-3=21-3x\Leftrightarrow-4x=-24\Leftrightarrow x=6\)( tm )
TH2 : \(x-3=3x-21\Leftrightarrow2x=18\Leftrightarrow x=9\)( ktm )