g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

\(A,5x-25=0\)

\(\Leftrightarrow5x-5^2=0\)

\(\Leftrightarrow5\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\)

Chúc bạn học tốt !

* 4x - 1 = 3x - 2

⇔ 4x - 3x = -2 + 1

⇔ x = -1

Vậy tập nghiệm của pt là S = {-1}

* \(\frac{3}{4}-3x=0\)

⇔ \(\frac{3}{4}-\frac{3x.4}{4}=0\)

⇒ 3 - 12x = 0

⇔ 12x = 3

⇔ x = \(\frac{3}{12}=\frac{1}{4}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{1}{4}\right\}\)

* 3x - 2 = 2x + 3

⇔ 3x - 2x = 3 + 2

⇔ x = 5

Vậy tập nghiệm của pt là S = {5}

* 2(x - 3) = 5(x + 4)

⇔ 2x - 6 = 5x + 20

⇔ 2x - 5x = 20 + 6

⇔ -3x = 26

⇔ x = \(\frac{-26}{3}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{-26}{3}\right\}\)

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

bạn ơi !!!

đăng từng câu thôi thế này nhìn loạn cả mắt luôn á

rối mắt quá

a, 2/3+1/3:x=-7

1/3:x=-7-2/3

1/3:x=-23/3

x=1/3:-23/3

x=-1/23

Vậy x=-1/23

c, (2x-3)(6-2x)=0

*TH1: 2x-3=0

2x=3

x=3/2

*TH2: 6-2x=0

2x=6

x=6/2

x=3

Vậy x=3/2 hoặc x=3

d,x:3/4+1/4=-2/3

x:3/4=-2/3-1/4

x:3/4=-11/12

x=-11/12*3/4

x=-11/16

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) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)

b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)

c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)

d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\\ \dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}\\ \dfrac{1}{2}x=2\\ x=2:\dfrac{1}{2}\\ x=4\)

a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)

b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)

c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)

d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)

e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)

\(\Leftrightarrow-8x+20-2=18x-45\)

\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)

f. \(\left(3x-1\right)\left(\dfrac{-1}{2x+5}\right)=0\)

\(\left[{}\begin{matrix}3x-1=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\\\dfrac{-1}{2x+5}=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=\dfrac{-5}{2}\end{matrix}\right.\)

g. \(\dfrac{6x}{12x}-\dfrac{8}{12x}=\dfrac{7x}{12x}\Leftrightarrow-x=8\Leftrightarrow x=-8\)

h. \(\dfrac{45}{60x}+\dfrac{12x}{6x}=\dfrac{10x}{60x}\Leftrightarrow2x=-45\Leftrightarrow x=\dfrac{-45}{2}\)

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)

=>7/6x=2/3

hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)

hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)