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

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

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

a) 15 x 21 - 16 : 4 + 7 

=    315    -      4    + 7

=           311   +     7

=             318

b) 123 - 11 x 6 + 150 : 30

=  123 -    66    +      5

=      57      +     5

=        62

Mình chỉ làm đến đó thôi chứ mấy câu còn lại khó hiểu quá!Ở lớp 5 mình chưa học bao giờ.

a)

=315-4+7=318

b)

=123-66+5=62

c)

=2017+5x[300-700:7]

=2017+5x[300-100]

=2017+5x200

=2017+1000

=3017

d)

=1560:[5x79-370+5x21]

=1560:[395-370+105]

=1560:130

=12

e)

=50-[12:2+34]

=50-40

=10

g)

=10-[16x5+88]:28

=10-168:28

=10-6

=4

a) 15x21 -16:4 +7

= 315-4 +7

=311+7

=318

b) 123 -11x6+150:30

=123-66+5

=57+5

=62

c)2017+5x[300-(715-15):7

=2017+5x[300-(700:7)

=2017+5x[300-100]

=2017+5x200

=2017+1000

=3017

bon gà

giúp em với ạ , em chuẩn bị phải nộp bài r ạ

hai đoạn liên tiếp nhau 

với x < 2 

pt có dạng \(\Leftrightarrow5-x+2-x=5x\Leftrightarrow7x=7\Leftrightarrow x=1\left(tm\right)\)

với 2 =< x < 5 

pt có dạng \(\Leftrightarrow x-2+5-x=5x\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{3}{5}\)(ktmđk) 

với x >= 5 

pt có dạng \(x-5+x-2=5x\Leftrightarrow3x=-7\Leftrightarrow x=-\dfrac{7}{3}\left(ktmđk\right)\)

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

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

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

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

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

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

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

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

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

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

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

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

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

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

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

a)      \(2\left(x+5\right)-x^2-5x=0\)

  \(\Leftrightarrow2x+10-x^2-5x=0\)

 \(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

 \(\Leftrightarrow x^2-2x+5x-10=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

b) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)

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

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

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

c)\(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

d) \(x^3+x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

e)\(x^2-2x-3=0\)

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

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

Cảm ơn bạn nha