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\(a,=4x\left(x+2\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=x^2\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2+1\right)\)
\(a,=12-3x+4x-x^2+x^2-2x=12-x\\ b,=x^2-2x+1-x^2+4=-2x+5\)
`|5x| = - 3x + 2`
Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :
`5x =-3x+2`
`<=> 5x +3x=2`
`<=> 8x=2`
`<=> x= 2/8=1/4` ( thỏa mãn )
Nếu `5x<0<=>x<0` thì phương trình trên trở thành :
`-5x = -3x+2`
`<=>-5x+3x=2`
`<=> 2x=2`
`<=>x=1` ( không thỏa mãn )
Vậy pt đã cho có nghiệm `x=1/4`
__
`6x-2<5x+3`
`<=> 6x-5x<3+2`
`<=>x<5`
Vậy bpt đã cho có tập nghiệm `x<5`
1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)
2 ) \(2x\left(b-a\right)+a-b=2x\left(b-a\right)-\left(b-a\right)=\left(2x-1\right)\left(b-a\right)\)
3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)
4 ) \(x^2-xy-2x+2y=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
5 ) \(5x^2y+5xy^2+a^2x+a^2y\)
\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)
\(=\left(5xy+a^2\right)\left(x+y\right)\)
6 ) \(2x^2-6xy+5x-15y\)
\(=2x\left(x-3y\right)+5\left(x-3y\right)\)
\(=\left(2x+5\right)\left(x-3y\right)\)
7 ) \(ax^2-3axy+bx-3by\)
\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)
\(=x\left(ax+b\right)-3y\left(ax+b\right)\)
\(=\left(x-3y\right)\left(ax+b\right)\)
8 ) \(x^2+4x-5x-20=0\)
\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
9 ) \(x^2+10x-2x-20=0\)
\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
10 ) \(x^2-6x-4x+24=0\)
\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
:D
c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2=0\)
hay x=1
b) 4x(2-x)+(2x+1)^2=2
8x-4x^2+4x^2+4x+1-2=0
(8x+4x)+(-4x^2+4x^2)+(1-2)=0
12x + 0 -1 =0
12x=1
x=1/12
Vậy x= 1/2
c) (x-3)^3-x^2(x-9)=0
x^3-9x^2+27x-x^3+9x^2=0
(x^3-x^3)+(-9x^2+9x^2)+27x=0
0 + 0 + 27x=0
x= 0
Vậy x=0
\(x\left(1+5x\right)\)
\(\left(x-3\right)\left(x+3\right)\)
\(\left(x^2+1\right)\left(2x+1\right)\)
a x + 5x^2
=x(1+ 5x)
b x^2 – 9
=x^2 – 3^2
=(x-3)(x+3)
c 2x^3 + x^2 + 2x + 1
=(2x^3 + x^2) + (2x + 1)
=x^2(2x + 1)+(2x + 1)
=(2x + 1)(x^2+1)