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\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
a, \(x\left(x+1\right)-x\left(x-5\right)=6\Leftrightarrow x^2+x-x^2+5x=6\)
\(\Leftrightarrow x=1\)
b, \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
c, \(x^2-\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)
d, \(5x^2=20x\Leftrightarrow5x^2-20x=0\Leftrightarrow5x\left(x-4\right)=0\Leftrightarrow x=0;4\)
e, \(4x^2-9-x\left(2x-3\right)=0\Leftrightarrow4x^2-9-2x^2=3x\Leftrightarrow2x^2-9-3x=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{2};3\)
f, \(4x^2-25=\left(2x-5\right)\left(2x+7\right)\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\Leftrightarrow x=-\frac{5}{2}\)
a) x( x + 1 ) - x( x - 5 ) = 6
⇔ x2 + x - x2 + 5x = 6
⇔ 6x = 6
⇔ x = 1
b) 4x2 - 4x + 1 = 0
⇔ ( 2x - 1 )2 = 0
⇔ 2x - 1 = 0
⇔ x = 1/2
c) x2 - 1/4 = 0
⇔ ( x - 1/2 )( x + 1/2 ) = 0
⇔ \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow x=\pm\frac{1}{2}\)
d) 5x2 = 20x
⇔ 5x2 - 20x = 0
⇔ 5x( x - 4 ) = 0
⇔ \(\orbr{\begin{cases}5x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) 4x2 - 9 - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 ) - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 - x ) = 0
⇔ ( 2x - 3 )( x + 3 ) = 0
⇔ \(\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
f) 4x2 - 25 = ( 2x - 5 )( 2x + 7 )
⇔ ( 2x - 5 )( 2x + 5 ) - ( 2x - 5 )( 2x + 7 ) = 0
⇔ ( 2x - 5 )( 2x + 5 - 2x - 7 ) = 0
⇔ ( 2x - 5 )(-2) = 0
⇔ 2x - 5 = 0
⇔ x = 5/2
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
bài 1: <=> 3x2+3x-2x2-2x+x+1=0 <=> x2+2x+1=0 <=>(x+1)2=0<=>x=-1
bài 2: =(x-3)2+1
vì (x-3)2>=0 với mọi x nên (x-3)2+1>=1 => GTNN của x2-6x+10 là 1 khi x=3
a) \(f\left(x\right)+g\left(x\right)+h\left(x\right)\)
\(=6x^7-5x^3+1-3+2x-4x^7-2x^7+2x+7x^2\)
\(=-5x^3+7x^2+4x-2\)
b) \(f\left(x\right)+g\left(x\right)-h\left(x\right)\)
\(=6x^7-5x^3+1-3+2x-4x^7-\left(-2x^7+2x+7x^2\right)\)
\(=2x^7-5x^3+2x-2+2x^7-2x-7x^2\)
\(=4x^7-5x^3-7x^2-2\)