= 3/4 + 1/3 = 13/12

5/4 x X = 5/8

X = 5/8 : 5/4

X = 1/2

vậy X = ...

\(\dfrac{3}{5}:\dfrac{4}{5}+\dfrac{1}{2}\times\dfrac{2}{3}=\dfrac{3}{4}+\dfrac{1}{3}+\dfrac{9}{12}+\dfrac{4}{12}=\dfrac{11}{12}\)

\(\dfrac{5}{4}\times x=\dfrac{5}{8}\)

\(x=\dfrac{5}{8}:\dfrac{5}{4}\)

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

2,3 x X - 1,4 : ( 1 và 1/2 - 3/4 ) = 0,25 x 2/3

        2,3 x X - 1,4 : ( 3/2 - 3/4 ) = 1/4 x 2/3

                     2,3 x X - 1,4 : 3/4 = 1/6

                   2,3 x X - 1,4 : 0,75 = 0,17

                           2,3 x X - 1,87 = 0,17

                                 2,3  x  X   = 0,17 + 1,78

                                 2,3  x  X   = 1,95  

                                            X   = 1,95 : 2,3

                                            X   = 0,85

( cực chi tiết rồi nha bạn )

2/3 . x - 1/4 :(3/2 - 3/4)=1/6

2/3 . x -1/4 : 3/4=1/6

2/3 .x - 1/3=1/6

2/3 .x=1/6 - 1/3

2/3.x= -1/6

x= -1/6 : 2/3

x= -1/4

Vậy x= -1/4

Tìm x biết:

a,x-1/2 =2/15

x=2/15 + 1/2

x =19/30

Vậy x =19/30

b,5 và 3/4 +x =15 - 2 và 1/2

23/4 + x  =27/2

x =27/2 - 23/4

x=31/4

Vậy x =31/4

c,x : 14,1 =0,25

x =0,25.14,1

x=3,525

Vậy x = 3.525

d,x + 1/2.1/3 =3/4

x + 1/6 =3/4

x =3/4 - 1/6

x =7/12

Vậy x = 7/12

a)\(x=\frac{2}{15}+\frac{1}{2}=\frac{4+1}{15x2}=\frac{5}{30}=\frac{1}{6}\)

b)anh chưa hiểu lắm hỗn số à em

c)\(x=\frac{1}{4}x\frac{141}{10}=\frac{141}{40}=3,525\)

d)\(x+\frac{1}{6}=\frac{3}{4}\)

suy ra: \(x=\frac{3}{4}-\frac{1}{6}=\frac{18-4}{24}=\frac{14}{24}=\frac{7}{12}\)

a: \(0.4\cdot\sqrt{0.25-\sqrt{\dfrac{1}{4}}}=0.4\cdot\sqrt{0.25-0.5}\)(đề này sai rồi bạn)

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

\(\Leftrightarrow2\left(x-1\right)=\dfrac{-11}{2}-\dfrac{3}{2}=-7\)

\(\Leftrightarrow x-1=\dfrac{-7}{2}\)

hay \(x=-\dfrac{5}{2}\)

a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow-2x+3+x+4=0\)

\(\Leftrightarrow-x+7=0\)

\(\Leftrightarrow-x=-7\)

hay x=7

Vậy: S={7}

b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)

\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)

\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)

\(\Leftrightarrow8+4x-10x=5-10x+5\)

\(\Leftrightarrow-6x+8=-10x+10\)

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

\(\Leftrightarrow4x-2=0\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)

\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)

nên x-60=0

hay x=60

Vậy: S={60}

a)

\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);                                                   

b)

\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);

c)

\(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) \\= 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 \\= 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);                                                               

d)

\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

2/3xX-1/4:(11/2-3/4)=1/6

2/3xX:19/4=1/6

2/3xX=1/6x19/4

2/3xX=19/24

       X=19/24:2/3

       X=19/16

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