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a)\(2x\left(x+1\right)-3-2x=5\)
\(\Leftrightarrow2x^2+2x-3-2x=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)
\(\Rightarrow x=2;-2\)
b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)
\(\Leftrightarrow6x^2+2x+4-2x=7\)
\(\Leftrightarrow6x^2+4=7\)
\(\Leftrightarrow6x^2=3\)
\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow3x^2+15x=0\)
\(\Leftrightarrow3x\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
b) ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0
<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0
<=> ( 2x - 3 )( 1 + x - 1 ) = 0
<=> x( 2x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Vậy .....
a, 25x^2 - 1 - (5x -1)(x+2)=0
=> (5x)^2 - 1 + (5x-1)(x+2) = 0
=> (5x-1)(5x+1) + (5x-1)(x+2) = 0
=> (5x-1)(5x+1+x+2) = 0
=> (5x-1)(6x+3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
Bài 1: Tìm x: (2x-6)^3 + (5-x)^3 + (1-x)^3 = 0
Bài 2: Tìm GTNN :
A= x^2 -2x -4
B= x^2 -x +5
C= 4x^2 +2x -9
D= 2x^2 -4x +7
Giúp tớ với, tớ đang cần gấp
a, \(5\left(2x+1\right)-2x-1=16\)
\(\Leftrightarrow10x+5-2x-1-16=0\Leftrightarrow8x-12=0\Leftrightarrow x=\frac{3}{2}\)
b, \(4x\left(x+5\right)=3\left(x+5\right)\Leftrightarrow4x\left(x+5\right)-3\left(x+5\right)=0\)
\(\Leftrightarrow\left(4x-3\right)\left(x+5\right)=0\Leftrightarrow x=\frac{3}{4};-5\)
c, \(x\left(x-2\right)=3-6\Leftrightarrow x^2-2x+3=0\)
vô nghiệm
a) Theo bài ra ta có bảng xét dấu:
-Xét x\(\le\)2
=>-(x-2)+(-x-5)=2x+5
=>-x+2-x+5=2x+5
=>-2x-3=2x+5
=>-5-3=2x+2x
=>-8=4x
=>x=-2 <t/m>
-Xét 2\(\le\)x\(\le\)5
=>(x-2)+(-x-5)=2x+5
=>x-2-x+4=2x+5
=>2=2x+5
=>2-5=2x
=>.-3=2x
=>x=\(\frac{-3}{2}\)
-Xét x\(\ge\)5
=>(x-2)+(x-5)=2x-5
=>x-2+x-5=2x-5
=>2x-7=2x-5
=>5-7=2x-2x
=>-2=0 <vô lý>
b) Theo bài ra ta có bảng xét dấu:
-Xét x\(\le\)-9
=>(6-x)-(-x+9)=2x+3
=>6-x+x-9=2x+3
=>-3=2x+3
=>-3-3=2x
=>-6=2x
=>x=-3 <ko t/m>
-Xét -9 \(\le\)x\(\le\)6
=>(6-x)-(x+9)=2x+3
=>6-x-x+9=2x+3
=>-2x-3=2x+3
=>-3-3=2x+2x
=>-6=4x
=>x=-6:4=\(\frac{-6}{4}\)=\(\frac{-3}{2}\)<t/m>
-Xét x\(\ge\)6
=>-(6-x)-(x+9)=2x+3
=>-6+x-x+9=2x+3
=>3=2x+3
=>3-3=2x
=>0=2x
=>x=0 <ko t/m>
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