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a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
a, \(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=20x^2-20x+5+4x^2+12x-4x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
b, \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=81x^2-18x+1+1-10x+25x^2+18x-90x^2-2+10x\)
\(=16x^2\)
c;d;e;f tự làm, đầu I giữ lấy còn trường tồn:)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+8x-12-50+60x-18x^2\)
\(=\left(20x^2+4x^2-18x^2\right)+\left(60x+8x-20x\right)+\left(5-12-50\right)\)
\(=6x^2+48x-57\)
Bài làm
a) 4x2 - 6x
= 2x( 2x - 3 )
b) 9x4y3 + 3x2y4
= 3x2y3( 3x2 + y )
c) x3 - 2x2 + 5x
= x( x2 - 2x + 5 )
d) 3x( x - 1 ) + 5( x - 1 )
= ( x - 1 )( 3x + 5 )
e) 2x2( x + 1 ) + 4( x + 1 )
= ( x + 1 )( 2x2 + 4 )
= ( x + 1 )2( x2 + 2 )
= 2( x + 1 )( x2 + 2 )
f) -3x - 6xy + 9xz
= -( 3x + 6xy - 9xz )
= -3x( 1 + 2y - 3z )
# Học tốt #
3)
\(A=\dfrac{5}{x^2-2x+5}\)
ta có x2-2x+5
=x2-2x+1+4
=(x2-2x+1)+4
=(x-1)2+4
=> A=\(\dfrac{5}{\left(x-1\right)^2+4}\)
do \(\left(x-1\right)^2\ge0\forall x\)
=> \(\left(x-1\right)^2+4\ge4\)
=> \(\dfrac{5}{\left(x-1\right)^2+4}\le\dfrac{5}{4}\)
=> A\(\le\dfrac{5}{4}\)
GTLN của A =\(\dfrac{5}{4}\)
khi x-1=0
=> x=1
vậy GTLN của A=\(\dfrac{5}{4}\) khi x=1
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
a) \(x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
b) \(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
c) \(2x^3-x^2-8x+4\)
\(=x^2\left(2x-1\right)-4\left(2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)
d) \(x\left(x-y\right)^2+y\left(x-y\right)^2-xy+x^2\)
\(=\left(x+y\right)\left(x-y\right)^2+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2+x\right)\)
e) \(2x^2-5x+2\)
\(=\left(2x^2-x\right)-\left(4x-2\right)\)
\(=x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
a) (3x-1)(9x2+3x+1)=27x3-1
27x3-1=27x3-1
27x3-1-(27x3-1)=0
27x3-1-27x3+1=0
⇒x=0
b)(x2-5x+25)(x+5)=x3+125
(x+5)(x2-x.5+52)=x3+125
x3+125-(x3+125)=0
x3+125-x3-125=0
⇒x=0
c)(x-3)(x2-6x+9)=(x-3)3
x3-33-(x-3)3=0
x3-27-x3+27=0
⇒x=0
d) Đề phải là thế này chứ \(\left(x-y+4\right).\left(x-y-4\right)\)
\(=\left(x-y\right)^2-4^2\)
\(=\left(x-y\right)^2-16\)
\(=x^2-2.x.y+y^2-16\)
\(=x^2-2xy+y^2-16.\)
Chúc bạn học tốt!
Bài này đề bài phải là khai triển biểu thức, chứ không phải là tính em nhé.
Lời giải:
Ta áp dụng hằng đẳng thức đáng nhớ thôi.
a. $(3+2x)^3=3^3+3.3^2.2x+3.3.(2x)^2+(2x)^3$
$=8x^3+36x^2+54x+27$
b.
$(\frac{1}{2}-y)^3=(\frac{1}{2})^3-3.(\frac{1}{2})^2.y+3.\frac{1}{2}y^2-y^3$
$=-y^3+\frac{3}{2}y^2-\frac{3}{4}y+\frac{1}{8}$
c.
$(x-5)(x^2+5x+25)=(x-5)^2(x^2+5x+5^2)$
$=x^3-5^3=x^3-125$
d.
$(3x+\frac{1}{2})(9x^2-\frac{3}{2}x+\frac{1}{4})$
$=(3x+\frac{1}{2})[(3x)^2-3x.\frac{1}{2}+(\frac{1}{2})^2]$
$=(3x)^3+(\frac{1}{2})^3=27x^3+\frac{1}{8}$