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) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)

\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)

b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)

\(\Leftrightarrow x=\frac{-3}{2}\)

c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)

\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)

d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)

a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

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

\(6x-9+4-2x=-3\)

\(4x=-2\)

\(x=-\frac{1}{2}\)

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

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

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)

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

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

\(=\left(x-3\right)\left(x-4\right)\)

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

\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)

\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

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

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

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

\(=\left(x-5\right)\left(4x+1\right)\)

Còn lại bạn tự làm nhá

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a)

A= ( 5- x ). (x + 5) - 2.( x - 1). ( x - 3) - 3.( x - 2)2

= 25 - x² - 2(x²- 3x - x + 3) - 3(x² - 4x + 4)

= 25 - x² - 2x² + 6x + 2x - 6 - 3x² + 12x - 12

= 7 - 6x² + 20x

câu b kết quả là:

-14x - 27 +2x²

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

\(=\left(5-x\right)\left(5+x\right)-\left(2x-2\right)\left(x-3\right)-3\left(x^2-2.2x+2^2\right)\)

\(=\left(5^2-x^2\right)-\left[2x\left(x-3\right)-2\left(x-3\right)\right]-3\left(x^2-4x+4\right)\)

\(=25-x^2-\left[\left(2x^2-6x\right)-\left(2x-6\right)\right]-3x^2+12x-12\)

\(=25-x^2-\left(2x^2-6x-2x+6\right)-3x^2+12x-12\)

\(=25-x^2-2x^2+6x+2x-6-3x^2+12x-12\)

\(=7+20x-6x^2\)

b/ \(B=\left(3-2x\right)\left(x-2\right)+\left(2x-5\right)^2-\left(x-4\right)\)

\(=3\left(x-2\right)-2x\left(x-2\right)+\left[\left(2x\right)^2-2.2x.5+5^2\right]-x+4\)

\(=3x-6-2x^2+4x+4x^2-20x+25-x+4\)

\(=23-14x+2x^2\)

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

\(=x\left(x-2\right)-4\left(x-2\right)-\left(3x-6\right)\left(3-2x\right)-\left[\left(2x\right)^2+2.2x.1+1^2\right]\)

\(=x^2-2x-4x+8-\left[3x\left(3-2x\right)-6\left(3-2x\right)\right]-4x^2-4x-1\)

\(=x^2-2x-4x+8-\left(9x-6x^2-18+12x\right)-4x^2-4x-1\)

\(=x^2-2x-4x+8-9x+6x^2+18-12x-4x^2-4x-1\)

\(=25-31x+3x^2\)

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

\(=2.\left(x^2-2x+1\right)-\left(3x-3\right)\left(x+2\right)-\left[\left(2x\right)^2+2.2x+1\right]\)

\(=2x^2-4x+2-\left[3x\left(x+2\right)-3\left(x+2\right)\right]-\left(4x^2+4x+1\right)\)

\(=2x^2-4x+2-\left(3x^2+6x-3x-6\right)-\left(4x^2+4x+1\right)\)

\(=2x^2-4x+2-3x^2-6x+3x+6-4x^2-4x-1\)

\(=7-11x-5x^2\)

P/s: Ko chắc ạ!

chúc bạn học giỏi

a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)

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

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

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

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

hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)

b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)

\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)

=>6=0(vô lý)

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

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

=>x=-2 hoặc x=2

đ: \(\Rightarrow2x^2-2x-5x+5=0\)

=>(x-1)(2x-5)=0

=>x=1 hoặc x=5/2