a,x²-8x+16=(x-4)²

b,(x-5y)(x+5y)=x²-25y²

c,4x⁴-16=4(x²-2)(x²+2)

d,x²+4xy+4y²=(x+2y)²

Đề là gì vậy bạn?

Phân tích đa thức thành phân tử ạ

b) \(x^2-8x-4y^2+16=\left(x^2-8x+16\right)-4y^2=\left(x-4\right)^2-4y^2=\left(x-4-2y\right)\left(x-4+2y\right)\)

1: =(16x^2-8x+1)-y^2

=(4x-1)^2-y^2

=(4x-1-y)(4x-1+y)

2: =(x^2-2xy+y^2)-z^2

=(x-y)^2-z^2

=(x-y-z)(x-y+z)

3: =(x^2+4xy+4y^2)-16

=(x+2y)^2-4^2

=(x+2y-4)(x+2y+4)

4: =(x^2-4xy+4y^2)-16

=(x-2y)^2-4^2

=(x-2y-4)(x-2y+4)

a: ĐKXĐ: x>=1/2

\(PT\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

=>\(4\sqrt{2x-1}=12\)

=>\(\sqrt{2x-1}=3\)

=>2x-1=9

=>2x=10

=>x=5(nhận)

b: Sửa đề: \(\sqrt{9x^2-6x+1}=4\)

=>|3x-1|=4

=>3x-1=4 hoặc 3x-1=-4

=>3x=5 hoặc 3x=-3

=>x=-1 hoặc x=5/3

`A(x) =2x-1`

`2x-1=0`

`=> 2x=0+1`

`=>2x=1`

`=>x=1/2`

__

`B(x)  =3 - 6/5x`

`3-6/5x=0`

`=> 6/5x=3-0`

`=> 6/5x=3`

`=> x= 3 : 6/5`

`=> x= 3 xx 5/6`

`=> x=15/6`

__

`C(x) = 4x^2 - 25`

`4x^2 - 25=0`

`=> 4x^2 = 0+25`

`=> 4x^2 =25`

`=> 4x^2 = (+-5)^2`

`=> x= 5/4` hoặc `x=-5/4`

__

`D(x) = ( x + 1/4 )^2 - 16/9`

` ( x + 1/4 )^2 - 16/9=0`

`=> ( x + 1/4 )^2 = 16/9`

`=>( x + 1/4 )^2 =(+-4/3)^2`

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{4}{3}\\x+\dfrac{1}{4}=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

__

`E(x) = 8x^2 + 27`

`8x^2 +27=0`

`=>8x^2=0-27`

`=> 8x^2 =-27`

`->` đề hơi sai;-;.

__

`F(x) = x^2 + 3x`

`x^2 +3x=0`

`=>x(x+3)=0`

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

`@ yl`

1) \(x^2-2x+5+y^2-4y=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)

Để PT bằng 0 thì:

\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)

\(\Rightarrow x=1\)và \(y=2\)

2) \(y^2+2y+5-12x+9x^2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)

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..............<Giải thích như câu đầu>......................

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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)

\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)

3) \(x^2+20+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

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...............<Giải thích như câu đầu>..............

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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)

\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)

1) \(x^2-2x+5+y^2-4y=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)

Để PT bằng 0 thì:

\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)

\(\Rightarrow x=1\)và \(y=2\)

2) \(y^2+2y+5-12x+9x^2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)

..............................................................................

..............<Giải thích như câu đầu>......................

.............................................................................

\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)

\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)

3) \(x^2+20+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

......................................................................

...............<Giải thích như câu đầu>..............

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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)

\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)

\(1,x^2-2x+5+y^2-4y=0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

\(2,y^2+2y+5-12x+9x^2=0\)

\(\Rightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

\(\Rightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(y+1\right)^2=0\\\left(3x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}y=-1\\x=\frac{2}{3}\end{cases}}}\)

\(3,x^2+20+9y^2+8x-12y=0\)

\(\Rightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x+4\right)^2=0\\\left(3y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=\frac{2}{3}\end{cases}}}\)

k cho mik nhá

\(a,x^2-x+1\)

\(x^2-x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(< =>MIN=\frac{3}{4}\)dấu"=" xảy ra khi \(x=\frac{1}{2}\)

\(b,x^2+y^2-4\left(x+y\right)+16\)

\(x^2+y^2-4x-4y+16\)

\(\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+8\)

\(\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)

\(MIN=8\)dấu "=" xảy ra  khi \(x=y=2\)

\(2x^2+8x+9\)

\(\left(x^2+8x+16\right)+x^2-7\)

\(\left(x+4\right)^2+x^2-7\ge-7\)

\(< =>MIN=-7\)dấu "=" xảy ra khi \(x=-4\)