Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

A) \(\left(x-3\right)^2-\left(x+2\right)^2\)

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

\(=-5.\left(2x-1\right)\)

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

\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

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

\(=\left(x+3-1\right)\left(x+3+1\right)\)

\(=\left(x+2\right)\left(x+4\right)\)

bài 3 A) \(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

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

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

x=0

x=2

Bài 1 :

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

\(A=x^2-6x+9-x^2+25\)

\(A=34-6x\)

b) \(B=2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

Dễ thấy đây là HĐT thứ 1

\(B=\left(x+y+x-y\right)^2\)

\(B=\left(2x\right)^2\)

\(B=4x^2\)

Bài 2 :

a) \(2x\left(x+5\right)-x^2-5x=0\)

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

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

\(2\left(x+5\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)

b) \(4x\left(x-2013\right)-x+2013=0\)

\(4x\left(x-2013\right)-\left(x-2013\right)=0\)

\(\left(x-2013\right)\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2013=0\\4x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2013\\x=\frac{1}{4}\end{cases}}}\)

Bài 6:

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)

\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)

\(\Leftrightarrow-14x+2=30\)

\(\Leftrightarrow-14x=28\)

\(\Leftrightarrow x=-2\)

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

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

\(\Leftrightarrow2x+16=0\)

\(\Leftrightarrow2x=-16\)

\(\Leftrightarrow x=-8\)

Em cần gấp bây h ạ :<

Bài 1:

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

b) \(x^4+x^3+2x^2+x+1\)

\(=x^4+x^2+x^3+x+x^2+1\)

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

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

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p

Ôi trời sao lắm thế ít thôi bạn nên tách ra mà bạn cần gấp lắm à

đúng rồi pn. giúp mik đc bài nào cũng đc

Bài 1 :

a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)

\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)

\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)

\(\)\(=2y^2-10xy\)

Câu b tương tự

Bài 2 :

a ) \(x^2-9+\left(x-3\right)^2\)

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

\(=\left(x-3\right)\left(x+3+x-3\right)\)

\(=2x\left(x-3\right)\)

b ) \(x^3-4x^2+4x-xy^2\)

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

\(=x\left[\left(x-2\right)^2-y^2\right]\)

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

c ) \(x^3-4x^2+12x-27\)

\(=x^3-9x^2+5x^2+27x-15x-3^3\)

\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)

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

\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)

\(=\left(x-3\right)\left(x^2-6x+14\right)\)

d ) \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(3x\left(x+1\right)-10x\left(x+1\right)\)

\(=-7x\left(x+1\right)\)

Bài 1:

a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)

\(\Rightarrow4\left(x-2\right)-3x+4=0\)

\(\Rightarrow4x-8-3x+4=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)

\(\Rightarrow10x+35-15x-6=25\)

\(\Rightarrow-5x+29=25\)

\(\Rightarrow-5x=25-29\)

\(\Rightarrow-5x=-4\)

\(\Rightarrow x=\dfrac{4}{5}\)

c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Rightarrow x+4=0\)

\(\Rightarrow x=-4\)

d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Rightarrow-x-21=0\)

\(\Rightarrow x=-21\)

Bài 2:

a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(P=8x^2y-6y^2-9x^2y+12y^2\)

\(P=-x^2y+6y^2\)

Thay x = -1 ; y = 2 vào P ta được

\(P=-\left(-1\right)^2.2+6.2^2\)

\(P=-2+24=22\)

b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(Q=20x^3-12x^2y-4x^3-x^2y\)

\(Q=16x^3-13x^2y\)

Thay x = -1 ; y = 2 vào Q ta được

\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)

\(Q=-16-26\)

\(Q=-42\)

c) \(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)

\(H=2xy\)

Thay x = 1/4 ; y = 2012 vào H ta được

\(H=2.\dfrac{1}{4}.2012\)

\(H=1006\)

1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Leftrightarrow8x-16-6x+8=2\)

\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)

b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Leftrightarrow30x-20-15x-6+55-20x=25\)

\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)

\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)

2.

a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)

\(\Leftrightarrow x^2y-18y^2\)

tại x=-1 , y=2

ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)

vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2

b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)

\(\Leftrightarrow17x^3-13x^2y\)

tại x=-1,y=2

ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)

vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)

c)\(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)

\(\Leftrightarrow x^4+2xy-x^3\)

tại x=1/4 và y=2012

ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)

a. 8(x-2)-2(3x-4)=2

8x-16-6x-8 =2

(8x-6x)-(16-8)=2

2x-2 =2

2x =2+2

2x =4

x =\(\dfrac{4}{2}\)=2