n = 2

n.(n + 1) = 6

n² + n = 6

n² + n - 6 = 0

n² + 3n - 2n - 6 = 0

(n² + 3n) - (2n + 6) = 0

n . (n + 3) - 2 . (n + 3) = 0

(n + 3) . (n - 2) = 0

\(\Rightarrow\) n + 3 = 0 hoặc n - 2 = 0

Ta có:

\(\left\{{}\begin{matrix}n+3=0\\n-2=0\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}n=-3\\n=2\end{matrix}\right.\)

Mà n \(\in\) N

\(\Rightarrow\) n = 2

Bạn học số âm r thì sẽ có cả x = -3

1+1

Là sao vậy ạ ?

1*100000

ok ,tk câu ok đi rùi trả lời cho

a Vậy ( 2.x-15) phải bằng 1 hoặc 0 thì 1^5=1^2 hoặc 0^5=0^2

Trường hợp 1:

2.x-15=0

2.x=0+15

2.x=15

x=15:2

Mà x thuộc N nên không hợp lí.

Trường Hợp 2

2.x-15=1

2.x=1+15

2.x=16

x=16:2

x=2

2 thuộc N

<=> x=2

Ta có:\(\left(2x-15\right)^5=\left(2x-15\right)^2\)

         \(\Leftrightarrow\left(2x-15\right)^2\left(2x-15\right)^3=\left(2x-15\right)^2\)

         \(\Leftrightarrow\left(2x-15\right)^2\left(2x-15\right)^3-\left(2x-15\right)^2=0\)

        \(\Leftrightarrow\left(2x-15\right)^2\left[\left(2x-15\right)^3-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^2=0\\\left(2x-15\right)^3-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^3=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{15}{2}\left(koTM\right)\\x=8\left(TM\right)\end{cases}}\)

           Vậy số tự nhiên cần tìm là 8

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

Câu 11:

=>4,6x=6,21

=>x=1,35

12: \(A=-\left(1.4-x\right)^2-1.4< =-1.4\)

=>x=-1,4

Câu 9:

\(\Leftrightarrow\dfrac{10a+b}{100c+90+d}=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{92}-\dfrac{1}{97}=\dfrac{1}{2}-\dfrac{1}{97}=\dfrac{95}{194}\)

=>a=9; b=5; c=1; d=4

=>a+b+c+d=9+5+1+4=19

\(a,\) \(N=\dfrac{2.L}{3,4}=3000\cdot\left(nu\right)\)

\(b,\) Ta có : \(\%A=\%T=20\%\) \(\Rightarrow\%G=\%X=30\%\)

\(c,\) \(\left\{{}\begin{matrix}A=T=20\%N=600\left(nu\right)\\G=X=30\%N=900\left(nu\right)\end{matrix}\right.\)

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)