\(2\left(2x-5\right)^2-\left(2x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=70\) 70

\(2\left[\left(2x\right)^2-2.2x.5+5^2\right]-\left[\left(2x\right)^2-2.2x.3+3^2\right]-\left[\left(2x^2\right)-1^2\right]=70\)

\(8x^2-40x+50-4x^2+12x-9-4x^2+1=70\)

\(-28x+42=70\)

\(-28x=70-42\)

\(-28x=28\)

=> x = -1 

a: \(575-\left(2x+70\right)=445\)

=>\(2x+70=575-445=130\)

=>\(2x=130-70=60\)

=>x=60/2=30

b: \(575-2\left(x+70\right)=445\)

=>\(2\left(x+70\right)=575-445=130\)

=>x+70=130/2=65

=>x=65-70=-5

c: \(x^5=32\)

=>\(x^5=2^5\)

=>x=2

d: \(\left(3x-1\right)^3=8\)

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

=>3x-1=2

=>3x=3

=>\(x=\dfrac{3}{3}=1\)

e: \(\left(x-2\right)^3=27\)

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

=>x-2=3

=>x=5

f: \(\left(2x-3\right)^2=9\)

=>\(\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)

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

g: \(2x+5=3^4:3^2\)

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

=>2x+5=9

=>2x=9-5=4

=>x=4/2=2

h: \(\left(4x-5^2\right)\cdot7^3=7^4\)

=>\(4x-25=\dfrac{7^4}{7^3}=7\)

=>4x=25+7=32

=>\(x=\dfrac{32}{4}=8\)

a, 36:(x–5) =  2 2

(x–5) = 9

x = 14

b, [3.(70–x)+5]:2 = 46

[3.(70–x)+5] = 92

70–x = 29

x = 41

c, 450:[41–(2x–5)] =  3 2 .5

41–(2x–5) = 10

2x–5 = 31

2x = 36

x = 18

d, 230+[ 2 4 +(x–5)] = 315. 2018 0

16+(x–5) = 315–230

x–5 = 85–16

x = 69+5

x = 74

e,  2 x + 2 x + 1  = 48

2 x .(2+1) = 48

2 x = 16 =  2 4

x = 4

f,  3 x + 2 + 3 x  = 2430

3 x . 3 2 + 1 = 2430

3 x = 2430:10 = 243 =  3 5

x = 5

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

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

\(\Leftrightarrow-5\left(4x+3\right)=0\)

\(\Leftrightarrow4x+3=0\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=\frac{-3}{4}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)

h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)

\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)

\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)

\(\Leftrightarrow-9x+2x-3-10x=30\)

\(\Leftrightarrow-17x-3=30\)

\(\Leftrightarrow-17x=33\)

\(\Leftrightarrow x=\frac{-33}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)

k) \(\left(2x-1\right)\left(8x+5\right)-\left(4x+3\right)^2=20\)

\(\Leftrightarrow8x\left(2x-1\right)+5\left(2x-1\right)-\left(16x^2+24x+9\right)=20\)

\(\Leftrightarrow16x^2-8x+10x-5-16x^2-24x-9=20\)

\(\Leftrightarrow-8x+10x-5-24x-9=20\)

\(\Leftrightarrow-22x-14=20\)

\(\Leftrightarrow-11x-7=10\)

\(\Leftrightarrow x=\frac{-11}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-11}{17}\right\}\)

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

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

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

a) 6x(3x +5)-2x(9x-2)=17

6x3x+6x5-2x9x-2x(-2)=17

\(18x^2\)+30x-\(18x^2\)+4x=17

\(18x^2-18x^2\)+ 34x=17

0 +34x=17

x=17:34

x=0.5

b)2x(3x-1)-3x(2x+11)-70=0

2x3x-2x1-3x2x+3x11-70=0

\(6x^2-2x-6x^2+33x-70=0\)

-2x+33x-70=0

31x-70=0

31x=0+70

31x=70

x=\(\frac{70}{31}\)

(trong câu c dấu . của mình là nhân nha)

c)5x(2x-3)-4(8-3x)=2(3+5x)

5x2x-5x3-4.8+4.3x=2.3+2.5x

\(10x^2-15x-32+12x=6+10x\)

\(10x^2-15x+12x-10x=6+32\)

\(10x^2-13x=38\)

tạm thời mình bí chổ này thông cảm nha bạn

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) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)