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

\(x-\frac{3}{4}=\frac{1}{2}\)

\(x=\frac{1}{2}+\frac{3}{4}\)

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

\(x=\frac{5}{4}\)

\(x-\frac{3}{4}=\frac{1}{2}\)

\(x=\frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}\)

Vậy \(x=\frac{5}{4}\)

\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\)\(:16\frac{2}{3}=0\)

\(\Rightarrow\)\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)=0\)

\(\Rightarrow\)\(5\frac{3}{8}+x-7\frac{5}{24}=2\)

\(\Rightarrow5\frac{3}{8}+x=9\frac{5}{24}\)

\(\Rightarrow x=3\frac{5}{6}\)

( Cách làm thì đúng rồi nhưng đáp án tớ ko chắc chắn )

\(\frac{7-x}{2}-\frac{2x-3}{4}-\frac{x+2}{8}=\frac{-1}{2}\\ \frac{4\left(7-x\right)}{8}-\frac{2\left(2x-3\right)}{8}-\frac{x+2}{8}=\frac{-1}{2}\\ \frac{28-4x}{8}-\frac{4x-6}{8}-\frac{x+2}{8}=\frac{-1}{2}\\ \frac{28-4x-\left(4x-6\right)-\left(x+2\right)}{8}=\frac{-1}{2}\\ \frac{28-4x-4x+6-x-2}{8}=\frac{-1}{2}\\ \frac{34-x}{8}=\frac{-1}{2}\\ \Rightarrow2\left(34-x\right)=8\cdot\left(-1\right)\\ 68-2x=-8\\ \Rightarrow2x=76\\ \Rightarrow x=38\)

Vậy x = 38

\(\frac{1}{4}:x=-\frac{7}{20}\)

\(x=\frac{1}{4}:-\frac{7}{20}\)

\(x=-\frac{5}{7}\)

\(\frac{3}{4}+\frac{1}{4}\div x=\frac{2}{5}\)

\(\Rightarrow\frac{1}{4}\div x=-\frac{7}{20}\)

\(\Rightarrow x=\frac{1}{4}\div\left(-\frac{7}{20}\right)\)

\(\Rightarrow x=-\frac{5}{7}\)

1/4 : x=2/5 - 3/4

1/4 : x= - 7/20

x=1/4 : - 7/20

x=-5/7

k ủng hộ mk nha mọi người

k mk , mk k lại

Do pt có 2 nghiệm \(x_1,x_2\) nên ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{2}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)

Ta có :

\(P=x_1\left(3+x_2\right)+x_2\left(3+x_1\right)+3x^2_1+3x^2_2-10\)

\(=3x_1+x_1x_2+3x_2+x_1x_2+3\left(x_1^2+x_2^2\right)-10\)

\(=3\left(x_1+x_2\right)+2x_1x_2+3\left(x^2_1+x^2_2\right)-10\)

\(=3S+2P+3\left(S^2-2P\right)-10\)

\(=3.\left(-\dfrac{5}{2}\right)+2.\left(-\dfrac{1}{2}\right)+3\left(\left(-\dfrac{5}{2}\right)^2-2\left(-\dfrac{1}{2}\right)\right)-10\)

\(=\dfrac{13}{4}\)

Vậy \(P=\dfrac{13}{4}\)

Đề bài yêu cầu gì?

a, \(3\left(2x-1\right)-3x\left(-x+2\right)=5x-\left(1-3x\right)\cdot x\\ 6x-3+3x^2-6x=5x-x+3x^2\\ 3x^2-3=4x+3x^2\\ 3x^2-3x^2=4x+3\\ 4x+3=0\\ 4x=-3\\ x=\frac{-3}{4}\)

Vậy \(x=\frac{-3}{4}\)

b, \(x-\frac{x-3}{4}=3-\frac{x-3}{12}\\ \frac{4x-x-3}{4}=\frac{36-x-3}{12}\\ \frac{3x-3}{4}=\frac{33-x}{12}\\ \Rightarrow12\left(3x-3\right)=4\left(33-x\right)\\ 36x-36=132-4x\\ 36x+4x=132+36\\ 40x=168\\ x=\frac{168}{40}=\frac{21}{5}\)

Vậy \(x=\frac{21}{5}\)

\(\dfrac{x}{15}\)+\(\dfrac{x}{12}\)=4/1+1/2=9/2

=>x(\(\dfrac{1}{15}\)+\(\dfrac{1}{12}\))=9/2

=>x\(\cdot\)\(\dfrac{3}{20}\)=9/2

=>x=9/2:3/20=30

Vậy x=30

\(\dfrac{x}{15}+\dfrac{x}{12}=\dfrac{9}{2}\Rightarrow\left(\dfrac{1}{15}+\dfrac{1}{12}\right)x=\dfrac{9}{2}\)

\(\Rightarrow\left(\dfrac{12+18}{180}\right)x=\dfrac{9}{2}\Rightarrow\dfrac{30}{180}x=\dfrac{9}{2}\Rightarrow\dfrac{1}{6}x=\dfrac{9}{2}\Rightarrow x=\dfrac{9}{2}.6=27\)