a )\(A=\frac{x^2+4x+4}{x^2-4}=\frac{\left(x+2\right)^2}{x^2-2^2}=\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\frac{x+2}{x-2}=\frac{5}{3}\)

<=> (x + 2).3 = (x - 2).5

<=> 3x + 6 = 5x - 10

<=> 3x - 5x = - 10 - 6

<=> - 2x = - 16

=> x = 8

b ) \(\frac{x+2}{x-2}=\frac{\left(x-2\right)+4}{x-2}=1+\frac{4}{x-2}\)

đến đây tự tìm đc 

Bài 2 lớp 8 ko làm đc thì đi chết đi

Bài 1: 

a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)

\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)

\(\Rightarrow16x-5=x-2\)

\(\Rightarrow16x-x=5-2\)

\(\Rightarrow15x=3\)

\(\Rightarrow x=\dfrac{15}{3}=5\)

b) \(12x^2-4x\left(3x+5\right)=10x-17\)

\(\Rightarrow12x^2-12x^2-20x=10x-17\)

\(\Rightarrow-20x=10x-17\)

\(\Rightarrow-20x-10x=-17\)

\(\Rightarrow-30x=-17\)

\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)

c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)

\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)

\(\Rightarrow-8x=12\)

\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)

Bài 2: 

a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)

\(=x^2-7x+5x-35-7x^2+21x\)

\(=-6x^2+19x-35\)

b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)

\(=x^3-x^2-2x-x^2+x-5x-5\)

\(=x^3-2x^2-6x-5\)

c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)

\(=x^2-7x-5x+35-x^2-3x+4x-12\)

\(=11x+23\)

d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)

\(=x^2-2x-x+2-x^2+2x+5x+10\)

\(=4x+12\)

bài 1

tìm gtng và gtln

d=-4x^2 -4x +3

c= 9x^2 +6x +2

e=25x^2 +16x +4

bài 2 cho đa thức x^4 - x^3 +6x^2 -x +a chia cho x^2 -x +5 tìm a để số dư bằng 0

botay.com.vn

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )²

= 6x² - 4x - ( x² + 4x + 4 )

= 6x² - 4x - x² - 4x - 4

= 5x² - 8x - 4

b) ( x + 2 )( x² - 2x + 4 ) - 2( x + 1 )( 1 - x )

= x³ + 8 + 2( x + 1 )( x - 1 )

= x³ + 8 + 2( x² - 1 )

= x³ + 8 + 2x² - 2

= x³ + 2x² + 6

c) ( 2x - 1 )² - 2( 4x² - 1 ) + ( 2x + 1 )²

= 4x² - 4x + 1 - 8x² + 2 + 4x² + 4x + 1

= 4

d) x² - 3x + xy - 3y

= x( x - 3 ) + y( x - 3 )

= ( x - 3 )( x + y )

Bài 2.

a) 4x² - 4xy + y² = ( 2x - y )²

b) 9x³ - 9x²y - 4x + 4y

= 9x²( x - y ) - 4( x - y )

= ( x - y )( 9x² - 4 )

= ( x - y )( 3x - 2 )( 3x + 2 )

c) x³ + 2 + 3( x³ - 2 )

= x³ + 2 + 3x³ - 6

= 4x³ - 4

= 4( x³ - 1 )

= 4( x - 1 )( x² + x + 1 )

Bài 3.

2( x - 2 ) = x² - 4x + 4

⇔ ( x - 2 )² - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4

\(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-8x+16=4\)

\(\Leftrightarrow\left(x-2\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

Vậy...

\(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow x^2-13x+22=0\)

\(\Leftrightarrow\left(x+\frac{13}{2}\right)^2=\frac{81}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{-21}{2}\end{cases}}\)

Vậy...

b/ \(x^2-4=8\left(x-2\right)\)

<=> \(x^2-4=8x-16\)

<=> \(x^2-4-8x+16=0\)

<=> \(x^2-8x+12=0\)

<=> \(x^2-8x+16-4=0\)

<=> \(\left(x-4\right)^2-4=0\)

<=> \(\left(x-4\right)^2=4\)

<=> \(\orbr{\begin{cases}x-4=2\\x-4=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=2\end{cases}}\)

c/ \(x^2-4x+4=9\left(x-2\right)\)

<=> \(\left(x-2\right)^2-9\left(x-2\right)=0\)

<=> \(\left(x-2\right)\left(x-2-9\right)=0\)

<=> \(\left(x-2\right)\left(x-11\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\x=11\end{cases}}\)

d/ \(4x^2-12x+9=\left(5-x\right)^2\)

<=> \(\left(2x-3\right)^2-\left(5-x\right)^2=0\)

<=> \(\left(2x-3-5+x\right)\left(2x-3+5-x\right)=0\)

<=> \(\left(3x-8\right)\left(x+2\right)=0\)

<=> \(\orbr{\begin{cases}3x-8=0\\x+2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=8\\x=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1

=>-16x-34=x-1

=>-17x=33

=>x=-33/17

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6

=>4x^2+16x-20-4x^2-10x+4=0

=>6x=16

=>x=8/3

a: A=x+3+|x-3|

=x+3+3-x(x<=3)

=6

b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)

\(=\left|x+2\right|-\left|x\right|\)

=x+2-x=2

c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)

\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

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

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a) \(x^2+2x^2+x=x\left(x+2x+1\right)=x\left(x+1\right)^2\)

b) \(xy+y^2-x-y=\left(xy-x\right)+y^2-y=x\left(y-1\right)+y\left(y-1\right)=\left(y-1\right)\left(x+y\right)\)mấy câu sau bạn làm tương tự nhé, đặt biến x với x và y với y là được. có gì ib face cho mình

có gì sai xót mong m.n bỏ qua và nhắc nhở ạ