ok nhưng nhớ tích nhé

\(a,8:\left(\dfrac{1}{4}x\right)=2:0,02\)

\(8:\left(\dfrac{1}{4}x\right)=100\)

\(\dfrac{1}{4}x=100.8\)

\(\dfrac{1}{4}x=800\)

\(\Rightarrow x=...\)

X x 3 - (2 + X)=8

3X - 2 - X=8

2X =8+2

X=10:2

=>X=5

kb vs mik nếu thấy đúng

X x 3 - (2 + X)=8

3X - 2 - X=8

2X =8+2

X=10:2

=>X=5

X x 3 - (2 + x) = 8

X x 3 - X = 8 -2

X x3 - X  = 6

X = 3 x (6-1) : 3

X = 5

k mik nha

a)

`9/2-3`

`=9/2-6/2`

`=3/2`

b)

`8/5xx25/12`

`=10/3`

c)

`4:3/7`

`=4xx7/3`

`=28/3`

thank you bạn

(x+2)^3-(x-2)^3=12x(x-1)-8

<=>x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8

<=>12x^2+16=12x^2-12x-8

<=>12x+24=0

<=>x=-24/12=-2

Vậy S={-2}

tick nha các bạn

(x+2)^3-(x-2)^3=12x(x-1)-8

<=>x³+6x²+12x+8-x³+6x²-12x+8=12x²-12x-8

<=>12x²+16=12x²-12x-8

<=>12x+24=0

<=>x=-24/12=-2

Vậy S={-2}

(x+2)^3-(x-2)^3=12x(x-1)-8

<=>x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8

<=>12x^2+16=12x^2-12x-8

<=>12x+24=0

<=>x=-24/12=-2

Vậy S={-2}

`7/8 - 3/5 =-x`

`=> x=-7/8 + 3/5`

`=> x= -35/40 + 23/40`

`=> x= -12/40`

`=> x= -3/10`

Vậy `x=-3/10`

\(\sqrt[3]{x^3+8}=x+2\\ \Rightarrow x^3+8=\left(x+2\right)^3\\ \Rightarrow x^3+8=x^3+6x^2+12x+8\\ \Rightarrow6x^2+12x=0\\ \Rightarrow6x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

\(PT\Leftrightarrow\sqrt[3]{\left(x+2\right)\left(x^2-2x+4\right)}-\sqrt[3]{\left(x+2\right)^3}=0\\ \Leftrightarrow\sqrt[3]{x+2}\sqrt[3]{x^2-2x+4-x^2-4x-4}=0\\ \Leftrightarrow\sqrt[3]{x+2}\sqrt[3]{-6x}=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\-6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow3^x=5.3^{12}+4.\left(3^3\right)^4=5.3^{12}+4.3^{12}.\)

\(\Leftrightarrow3^x=9.3^{12}=3^2.3^{12}=3^{14}\Leftrightarrow x=14\)

bang 14

tính như bth thôi ráng lên bạn nhá

99/32

\(y x 4/9 - 5/8 = 1/2 : 2/3\)

\(y x 4/9 - 5/8 = 3/4\)

\(y x 4/9 = 3/4 + 5/8\)

\(y x 4/9 = 11/8\)

\(y = 11/8 : 4/9\)

\(y = 99/32\)