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

^^

Mình làm từ ý b nhá : 

b) Ta có : |x| = 2x - 1 

\(\Leftrightarrow\orbr{\begin{cases}x=2x-1\\x=1-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2x=-1\\x+2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-1\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}\)

c) Ta có : |2x - 1| = x + 4 

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+4\\2x-1=-x-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x=4+1\\2x+x=-4+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\3x=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

đăng ít 1 thôi bn, nhiều nhất 3 câu 1 lần thôi đăng lắm ngại làm

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

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

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

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

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

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

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

f) |-2x| = 3x+4

⇒\(\left[{}\begin{matrix}-2x=3x+4\\-2x=-3x-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-4}{5}\\x=-4\end{matrix}\right.\)

g) |2x-1| = 6-x

⇒\(\left[{}\begin{matrix}2x-1=6-x\\2x-1=x-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-5\end{matrix}\right.\)

h) |-1+5x| = 8-x

⇒\(\left[{}\begin{matrix}-1+5x=8-x\\-1+5x=x-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-7}{4}\end{matrix}\right.\)

i) |-2x+1| = x+3

⇒\(\left[{}\begin{matrix}-2x+1=x+3\\-2x+1=-x-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=4\end{matrix}\right.\)

k) |-2-5x| = -4x+7

⇒\(\left[{}\begin{matrix}-2-5x=-4x+7\\-2-5x=4x-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-9\\x=\frac{5}{9}\end{matrix}\right.\)

a) |x-2| = 3

⇒\(\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

b) |x+1| = |2x+3|

⇒\(\left[{}\begin{matrix}x+1=2x+3\\x+1=-2x-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-4}{3}\end{matrix}\right.\)

c) |3x| = x+6

⇒\(\left[{}\begin{matrix}3x=x+6\\3x=-x-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3}{2}\end{matrix}\right.\)

d) |x-5| = 13-2x

⇒\(\left[{}\begin{matrix}x-5=13-2x\\x-5=2x-13\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=8\end{matrix}\right.\)

e) |5x-1| = x-12

⇒\(\left[{}\begin{matrix}5x-1=x-12\\5x-1=12-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-11}{4}\\x=\frac{13}{6}\end{matrix}\right.\)