a, \(\dfrac{x}{3}=\dfrac{8}{12};12:3=4\)\(;8:4=2\)

⇒\(x=\dfrac{2}{3}\)

a: \(\dfrac{x}{3}=\dfrac{8}{12}\)

nên x/3=2/3

hay x=2

b: \(x+\dfrac{1}{15}=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)

c: \(x\cdot\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{5}{7}+\dfrac{2}{7}=1\)

hay x=7/4

d: \(\Leftrightarrow x\cdot\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{1}{2}\)

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

hay x=1

( 1/2 + 1/3 ) x 2/5

c1 = 5/6 x 2/5 = 1/3

c2 = 1/2 x 2/5 + 1/3 x 2/5

= 1/5 + 2/15 

= 1/3

3/5 x 17/21 x 2/5

c1 := 17/35 x 2/5 = 34/175

c2 : = (3/5 x 2/5) x 17/21 

= 6/25 x 17/21 

= 34/175?

( 1/3 - 1/5 ) x 1/2

c1 : = 2/15 x 1/2 

= 1/15

c2 : = 1/3 x 1/2 - 1/5 x 1/2

= 1/6 - 1/10

= 1/15

a: \(M=\left(\dfrac{-3}{7}x^3y\right)\cdot\dfrac{7xy^3}{12}-x^2y^2\cdot\left(-\dfrac{3}{4}x^2y^2\right)\)

\(=\dfrac{-1}{4}x^4y^4+\dfrac{3}{4}x^4y^4\)

\(=\dfrac{1}{2}x^4y^4\)

b: Hệ số là 1/2

Biến là \(x^4;y^4\)

bậc là 4+4=8

c: Thay x=-1 và y=-2 vào M, ta được:

\(M=\dfrac{1}{2}\left(-1\right)^4\cdot\left(-2\right)^4=\dfrac{1}{2}\cdot16=8\)

1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)

2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)

\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)

\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)

3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)

\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)

A. (Theo mình là -4xy thì mới rút gọn được)

B = (6x + y)^2

C = (2x)^2 = 4x^2

2/3 + 7/1 = 2/3 + 7 = 2/3 + 21/3 = 23/3

25/48 + 11/24 = 25/48 + 22/48 = 47/48

5/7 + 3/8 = 40/56 + 21/56 = 61/56

15/24 + 12/6 = 5/8 + 2 = 5/8 + 16/8 = 21/8

5/6 + 4/3 = 5/6 + 8/6 = 13/6

3/8 + 7/12 = 9/24 + 14/24 = 23/24

= 23/3

= 47/48

=61/56

=21/8

=13/6

= 23/24

Cảm ơn mn nhìu nhaaaaaa

Ta có: \(P=\frac{4^{12}}{3\cdot2^{24}-16\cdot8^7}\)

\(=\frac{\left(2^2\right)^{12}}{3\cdot2^{24}-2^4\cdot\left(2^3\right)^7}\)

\(=\frac{2^{24}}{3\cdot2^{24}-2^4\cdot2^{21}}=\frac{2^{24}}{2^{24}\left(3-2\right)}=\frac{2^{24}}{2^{24}}\)

=1

\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{1}{2}\)
\(\dfrac{1}{3}:x=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{6}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{1}{6}\right)\)
\(x=-2\)
Vậy ...
#AvoidMe

`2/3 +1/3 : x=1/2`

`=> 1/3 : x= 1/2 -2/3`

`=> 1/3 : x= 3/6 -4/6`

`=> 1/3 : x=-1/6`

`=> x=1/3 :(-1/6)`

`=> x=1/3 xx (-6)`

`=> x= -2`

Vậy `x=-2`

(x+2).5=210

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

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

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

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

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

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

Mà: \(\left|x+1\right|\ge0\forall x\Rightarrow\left|x+1\right|+2\ge2>0\)

\(\Leftrightarrow\left|x+1\right|+2=0\) (vô lí)

Vậy phương trình vô nghiệm:

\(x\in\varnothing\)

=>|x+1|-2|x+1|=2

=>-|x+1|=2

=>|x+1|=-2(vô lý)

Vậy: \(x\in\varnothing\)

\(TH_1:\left\{{}\begin{matrix}x+1\ge0\\-2x-2\ge0\end{matrix}\right.\) 

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

\(\Leftrightarrow x+1+2x+2=2\)

\(\Leftrightarrow3x+3=2\)  

\(\Leftrightarrow x=-\dfrac{1}{3}\left(l\right)\)

\(TH_2:\left\{{}\begin{matrix}x+1< 0\\-2x-2< 0\end{matrix}\right.\) 

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

\(\Leftrightarrow-x-1+2x+2=2\) 

\(\Leftrightarrow x+1=2\)

\(\Leftrightarrow x=1\left(l\right)\) 

Vậy tập nghiệm rỗng.