đề kiểu gì vậy bn

(2,x+10) á

xàm lone :))

đề là (2x+10)chia hết cho (x+3) à bn

a, \(\frac{6^2+3.6^2+3^2}{-13}=\frac{6^2\left(3+1\right)+3^2}{-13}=\frac{144+9}{-13}=\frac{-153}{13}\)
b, \(\frac{81}{\left(-3\right)}=\frac{-243}{x}\)
< Chắc ý bạn là như vậy nhỉ >
=> x = [ ( -243 ) . ( -3 ) ] : 81 
=> x = 9 

\(a,\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\frac{2^3.3^3+3^3.2^2+3^3}{-13}\)

\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=-3^3=-27\)

câu b tam sàm ba láp

\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)

<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)

<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)

<=> \(\left(x-2\frac{1}{5}\right)\) =    \(\frac{5}{6}\) : \(\frac{5}{12}\)

,<=> \(\left(x-2\frac{1}{5}\right)\)=   2 

<=. x = 2 + \(\frac{11}{5}\)

<=> x = \(\frac{21}{5}\)

a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)

\(\Leftrightarrow x+3=7x+35\)

\(\Leftrightarrow-6x=32\)

\(\Leftrightarrow x=-\frac{16}{3}\)

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

\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)

\(\Leftrightarrow6x-3=-6x-10\)

\(\Leftrightarrow12x=-7\)

\(\Leftrightarrow x=-\frac{7}{12}\)

c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)

\(\Leftrightarrow\left(x+1\right)^2=6^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)

d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)

\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)

\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)

\(\Leftrightarrow2x=2\Leftrightarrow x=1\)

e)\(\frac{x+1}{x-3}=\frac{x+5}{x-2}\Leftrightarrow\left(x+1\right)\left(x-2\right)=\left(x-3\right)\left(x+5\right)\)

\(\Leftrightarrow x^2-2x+x-2=x^2+5x-3x-15\)

\(\Leftrightarrow-3x=-13\)

\(\Leftrightarrow x=\frac{13}{3}\)

f)\(\frac{x+4}{x+5}=\frac{x+6}{x+7}\Leftrightarrow\left(x+4\right)\left(x+7\right)=\left(x+5\right)\left(x+6\right)\)

\(\Leftrightarrow x^2+11x+28=x^2+11x+30\)

\(\Leftrightarrow0x=2\)

\(\Leftrightarrow x\in\varnothing\)

X-1/4.x=6/11.(6/12+9/12+-4/12)

X-1/4.x=6/11.11/12

X-1/4.x=1/2

X.(1-1/4)=1/2

X.(4/4+-1/4)=1/2

X.3/4=1/2

X=1/2:3/4

X=1/2.4/3

X=2/3

Giai thich trong mot bai toan neu co hai dau x ta se ko tinh duoc trong bai toan nay x-25℅.x=6/11.(1/2+3/4+1/3) co hai dau x nen ta phai dua x ra ngoai lam h chung x dung mot minh se la x.1 khi ta bo x ra ngoai thi ben trong con 1-25℅ roi ta tinh ra sau do lam binh thuong

a) (x + 1/2)^2 = 1/16

=> (x + 1/2)^2 = (1/4)^2 hoặc (x + 1/2)^2 = (-1/4)^2

=> x + 1/2 = 1/4 hoặc x + 1/2 = -1/4

* x + 1/2 = 1/4

   x           = 1/4 - 1/2

   x           = -1/4

* x + 1/2 = -1/4

   x           = -1/4 - 1/2

   x           = -3/4

Vậy x = -1/4 hoặc x = -3/4

b) 2^x+2 - 2^x      = 9^6

=> 2^x . 2^2 - 2^x = 9^6

=> 2^x . (2^2 - 1) = 9^6

=> 2^x . (4 - 1)     = 9^6

=> 2^x . 3             = (3^2)^6

=> 2^x . 3             = 3^12

=> 2^x                  = 3^12 : 3

=> 2^x                  = 3^11

Vì 3^11 không chia hết cho 2

=> Không có giá trị nào của x thõa mãn đề bài

c)  (3^x)^2 : 3^3  = 1/243

=>   3^2x = 1/243 . 3^3

=>   3^2x = 1/243 . 27

=>   3^2x = 1/9

=>   3^2x . 9 = 1

=>  3^2x . 3^2 = 1

=>  3^2x+2 = 1

=>  3^2x+2 = 3^0

=> 2x + 2 = 0

=> 2x   = 0 - 2

=> 2x   = -2

=> x = -2 : 2

=> x = -1

a\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2 \)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{4}-\frac{1}{2}\)
\(\Rightarrow x=\frac{-1}{4}\)
 

12x+3.2³=2³.x-4.3²

12x+3.8=8.x-4.9

12x+24=8x-36

12x-8x=36-24

4x=12

x=12:4=3

      \(12x+3\cdot2^3=2^3x-4\cdot3^2\)

\(\Rightarrow12x+24=8x-36\)

\(\Rightarrow12x-8x=-36-24\)

\(\Rightarrow4x=-60\)

\(\Rightarrow x=-15\)

Vay x=-15

k mk nhA