a) \(\left(2x+3\right).\left(\frac{1}{2}.x-\frac{3}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\\frac{1}{2}.x-\frac{3}{2}=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=-3\\\frac{1}{2}.x=\frac{3}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\frac{3}{2}:\frac{1}{2}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=3\end{cases}}\)

Vậy x = \(-\frac{3}{2}\) hoặc x = 3

b)\(\left(\frac{1}{2}-x\right)^2=\frac{64}{49}\)

\(\Rightarrow\left(\frac{1}{2}-x\right)^2=\left(\frac{8}{7}\right)^2\) hoặc \(\left(\frac{1}{2}-x\right)^2=\left(-\frac{8}{7}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{8}{7}\\\frac{1}{2}-x=-\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{8}{7}\\x=\frac{1}{2}+\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{9}{14}\\x=\frac{23}{14}\end{cases}}\)

Vậy x = \(-\frac{9}{14}\) hoặc x = \(\frac{23}{14}\)

c) \(\frac{1}{2}.\left(x-4,5\right)=\frac{3}{4}.x=\frac{5}{12}\) ( câu này mik ko hiểu cho lắm)

k mik nha mn!

doi mik sua

la cong 3/4

a)\(\frac{x-1}{x+2\times0}=\frac{x-1}{x}\)

b)\(\frac{x+1}{x-3}=\frac{x-3+4}{x-3}=1+\frac{4}{x-3}>1\)

cau nay kho that ai giai duoc minh h

a, 2n-1/n+1  tim n          b, x-1/x+2   >0

nhanh len minh can gap lam ai dung va nhanh minh h va ket bn nhe

Xx4+(1/2+1/4+1/8+1/16)=23/16

Xx4=23/16-(1/2+1/4+1/8+1/16)

Xx4=1/2

X=1/2:4

X=1/8

1/8 nha bn

ako lop may mik lop 4 neu ban lop 4 thi ket ban nha

ta co 3 truong hop 

truong hop 1

x-1=0

x=0+1

x=1

truong hop 2

x+2=0

x=0-2

x=-2

truong hop 3

x-3=0

x=0-3

x=-3

mik chac chan dung nha

TA CÓ 3 TRƯỜNG HỢP XẢY RA 

TH1:X-1=0

       X  =0+1

       X  =1

TH2:X+2=0

      X    =0-2

      X    =-2

TH3 :X-3=0

        X   =0+3

        X   =3

VẬY X {1;-2;3}

CHẮC CHẮN 100% K NHA

RONANLDO LÀM SAI TH3 X = 0+3 CHỨ KHÔNG PHẢI X= 0 -3 NHA

\(x^2-2x+3=t\left(t\ge0\right)\)

\(pt\Leftrightarrow\frac{1}{t-1}+\frac{1}{t}=\frac{9}{2\left(t+1\right)}\)

\(\Leftrightarrow\frac{2t\left(t+1\right)}{2t\left(t^2-1\right)}+\frac{2\left(t^2-1\right)}{2t\left(t^2-1\right)}-\frac{9t\left(t-1\right)}{2t\left(t^2-1\right)}=0\)

\(\Leftrightarrow-5t^2+11t-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)