ủa đây là dạng toán lớp 6 mà

Bài 2:

c: \(\left(\dfrac{3}{2}+x\right):1\dfrac{2}{5}=\dfrac{1}{2}\cdot\dfrac{3}{5}+0,2\)

=>\(\left(x+\dfrac{3}{2}\right):\dfrac{7}{5}=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{5}{10}=\dfrac{1}{2}\)

=>\(x+\dfrac{3}{2}=\dfrac{1}{2}\cdot\dfrac{7}{5}=\dfrac{7}{10}\)

=>\(x=\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{7}{10}-\dfrac{15}{10}=-\dfrac{8}{10}=-\dfrac{4}{5}\)

f: \(-\dfrac{7}{5}-\left(\dfrac{2}{3}+x\right)=\dfrac{3}{10}\)

=>\(-\dfrac{7}{5}-\dfrac{2}{3}-x=\dfrac{3}{10}\)

=>\(x=-\dfrac{7}{5}-\dfrac{2}{3}-\dfrac{3}{10}=\dfrac{-42-20-9}{30}=\dfrac{-71}{30}\)

g: \(\left|x-\dfrac{1}{6}\right|-\dfrac{7}{12}=\dfrac{1}{4}\)

=>\(\left|x-\dfrac{1}{6}\right|=\dfrac{1}{4}+\dfrac{7}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{5}{6}\\x-\dfrac{1}{6}=-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{6}=1\\x=-\dfrac{5}{6}+1=-\dfrac{4}{6}=-\dfrac{2}{3}\end{matrix}\right.\)

l: \(x:\left(\dfrac{1}{5}-\dfrac{7}{10}\right)=-2+\left(-1\dfrac{2}{3}\right)\)

=>\(x:\left(\dfrac{2}{10}-\dfrac{7}{10}\right)=-2+\dfrac{-5}{3}=\dfrac{-11}{3}\)

=>\(x:\dfrac{-1}{2}=-\dfrac{11}{3}\)

=>\(x=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)

k: \(x:\left(\dfrac{1}{7}-\dfrac{3}{14}\right)=-3+\left(-1\dfrac{2}{3}\right)\)

=>\(x:\left(\dfrac{2}{14}-\dfrac{3}{14}\right)=-3-\dfrac{5}{3}=\dfrac{-14}{3}\)

=>\(x:\dfrac{-1}{14}=\dfrac{-14}{3}\)

=>\(x=\dfrac{14}{3}\cdot\dfrac{1}{14}=\dfrac{1}{3}\)

Bài 3:

a: \(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

b: \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{5-2}{20}=\dfrac{3}{20}\)

4Al+3O2-to>2Al2O3

0,1----0,075---0,05 mol

n Al=0,1 mol

=>m Al2O3=0,05.102=5,1g

`=>` Ta có:

`4Al + 3O2 -to> 2Al_2O3

`0,1----0,075---0,05` `mol`

`n` `Al` `= 0,1` mol`

 => m Al_2O3 = 0,05 . 102 =5,1g`

    Tóm tắt

`m_[Al] = 2,7 g`

`M_[Al] = 27 g // mol , M_[O_2] = 16 g // mol`

_____________________________

`m_[Al_2 O_3] = ? g`

                     Giải

`PTHH: 4Al + 3O_2` $\xrightarrow[]{t^o}$ `2Al_2 O_3`

`n_[Al] = [ 2,7 ] / 27 = 0,1 (mol)`

Theo `PTHH` có: `n_[Al_2 O_3] = 1 / 2 n_[Al] = 1 / 2 . 0,1 = 0,05 (mol)`

        `-> m_[Al_2 O_3] = 0,05 . 102 = 5,1 (g)`

a. \(\sqrt{12^2}\)

= 12

b. \(\sqrt{\left(-7\right)^2}\)

= 7

c. \(\sqrt{\left(2-\sqrt{5}\right)^2}\)

= 2 - \(\sqrt{5}\)

ở đây không phân biệt giỏi hay dốt cả bn nha

g: Ta có: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=2+\sqrt{3}+2-\sqrt{3}\)

=4

j: Ta có: \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3}-1+\sqrt{3}+1\)

\(=2\sqrt{3}\)

1, đổi 100 g = 0,1 kg 

P = m . g = 1 N 

điềm 1 

2, điền 

\(a=\dfrac{v-v_0}{t-t_0}=\dfrac{5}{2}\)

3, đổi 500g = 0,5 kg 

điền F = ma = 1 N 

Bài \(2\)

\(a,\) Theo bài ta có : \(1800=G\left(2^2-1\right)\rightarrow G=X=600\left(nu\right)\)

\(\rightarrow A=T=\dfrac{2}{3}.600=400\left(nu\right)\)

\(\rightarrow N=2A+2G=2000\left(nu\right)\)

\(\rightarrow L=3,4.\dfrac{2000}{2}=3400\overset{o}{A}=0,34\left(micromet\right)\)

\(b,\) \(H=2A+3G=\)\(2600(lk)\)

\(LKHT=2\left(N-1\right)=3998\left(lk\right)\)

\(M=300.N=600000\left(dvC\right)\)

\(C=\dfrac{N}{20}=100\)

Bài \(3\)

\(a,\) Theo bài ta có : \(30\%N=900\rightarrow N=3000\left(nu\right)\)

Gọi số lần \(gen\) tự nhân đôi là \(y\) \(\left(y>0,y\in N\right)\)

Ta có : \(N_{mt}=N\left(2^y-1\right)\)\(\rightarrow y=2\left(tm\right)\)

2.

\(cosx+cos3x=1+\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow2cos2x.cosx=1+cos2x+sin2x\)

\(\Leftrightarrow2cos2x.cosx=2cos^2x+2sinx.cosx\)

\(\Leftrightarrow cosx\left(cos2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cos^2x-sin^2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cosx+sinx\right)\left(cosx-sinx-1\right)=0\)

\(\Leftrightarrow cosx.\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right).\left[\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin\left(x+\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x=k2\pi\end{matrix}\right.\)