A

lộn

A

b và d nhé

đúng thì nhớ và kb với mik nhé

Đáp án đúng là:

(600 + 30) + 5 = 600 + (30 + 5)

(600 x30) x 5 = 600 x ( 30 x 5)

(600 x 30) x 5 = 600 x (30 x 5)

(600 + 30) + 5 = 600 +(30 + 5)

30x2+30x5+30x3

30x(5+3+2)

30x 10

300

   30 x 2 + 30 x 5 + 30 x 3 

= 30 x ( 2 + 5 + 3 )

= 30 x 10

= 300

30 x 2+ 30 x 5 +30 x3

30x[2+5+3]

30x 10

300

30 + 30 * 20 + 30 * 10 

= 30 * 1 + 30 * 20 + 30 * 10

= 30 * ( 1 + 20 + 10 )

= 30 * 31 

= 930

30 + 30 x20 + 30 x 10 

= 30 x 1 + 30 x 20 + 30 x 10

= 30 x ( 1 + 20 + 10 )

= 30 x 31 

= 930

****nha

30 + 30 * 20 + 30 * 10 

= 30 * 1 + 30 * 20 + 30 * 10

= 30 * ( 1 + 20 + 10 )

= 30 * 31 

= 930

\(\dfrac{30-x}{30}=\dfrac{8}{15}\)

\(\Rightarrow\left(30-x\right)15=30\times8\)

\(\Rightarrow\left(30-x\right)15=240\)

\(\Rightarrow30-x=240:15\)

\(\Rightarrow30-x=16\)

\(\Rightarrow x=30-16\)

\(\Rightarrow x=14\)

__

\(\dfrac{x+30}{72}=\dfrac{5}{8}\)

\(\Rightarrow\left(x+30\right)8=5\times72\)

\(\Rightarrow\left(x+30\right)8=360\)

\(\Rightarrow x+30=360:8\)

\(\Rightarrow x+30=45\)

\(\Rightarrow x=45-30\)

\(\Rightarrow x=15\)

= 5 + 30 - 30 + 900 

= 5 + 900

= 905

5+5*6-30+30*30

=5+30-30+900

=905

5 + 5 x 6 - 30 + 30 x 30 = 5 + 30 - 30 + 900 = 905

30+30-30+30 x 30=930

tick nha

30+30-30+30 x 30=930

a.

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(a=x^2+x-1\) , ta có pt:

\(\left(a+1\right)\left(a-1\right)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)

*Với a = 5 ta được:

\(x^2+x-1=5\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

*Với a = -5 ta được:

\(x^2+x-1=-5\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)

Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)

c)(ĐKXĐ: x khác 30;29)

\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)

\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)

\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)

\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)

\(\Leftrightarrow1741-59x=0\)

\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)

Vậy S={0;\(\dfrac{1741}{59}\);59}

b)(ĐKXĐ:x khác 2;3;4;5;6)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-6\right)\left(x-2\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=32\)

\(\Leftrightarrow x^2-8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow x=-2\) or x=10(đều thỏa)

Vậy ...

\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)

Chọn A

mik nghĩ A