\(\dfrac{2}{3x}\)

\(=\dfrac{2x+6}{3x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{3x\left(x+3\right)}=\dfrac{2}{3x}\)

\(\dfrac{2}{3x+9}+\dfrac{2}{x^2+3x}\)

=\(\dfrac{2}{3\left(x+3\right)}+\dfrac{2}{x\left(x+3\right)}\)

=\(\dfrac{4}{3+x}\)

Refer

1 my sister is old. she can driver a car
=> My sister is old enough to drive a car

2 the ladder wasn't verry long .it didn't reach the ceiling
=> The ladder wasn't long enough to reach the ceiling

3 the fire isn't very hot. it won't boil a kettle
=> The fire isn't hot enough to boil a kettle

4 he is strong. he can carry that suicase
=> He is strong enough to carry that suitcase

5 Lan isn't strong . she can't swim across the rive
=> Lan isn't strong enough to swim across the river

6 she is beautiful and intelligent . she can become Miss World
=> She is beautiful and intelligent enough to become Miss World

7 MrRobert isn't rich he can't buy a house
=> Mr Robert isn't rich enough to buy a house

8 the worker in very clever . he can make nice things from wood
=> The worker is clever enough to make nice things from wood

9 our team is very good . we win the foodball match very often 
=> Our team is good enough to win the football match very often

10 the radio isn't small . it can't be put it in your pocket
=> The radio isn't small enough to put in your pocket

=>x+1/2+x+1/6+...+x+1/90=99,9

=>\(\left(x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=99.9\)

=>\(9x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=99.9\)

=>9x+(1-0,1)=99,9

=>9x=99,9+0,1-1=100-1=99

=>x=11

Bài nào ạ. Ảnh bị lỗi.

you đang thi:)?

đang k.tra à:>

vào qanda mà hoỉ 

\(\text{8.C.So le trong}\)

\(\text{9.C.a trùng b}\)

\(\text{10.B.}60^0\)

\(\text{11.C.}150^0\)

\(\text{12.B.A=P}\)

Bài 3: 

Diện tích là:

\(15\cdot6=90\left(m^2\right)\)

Bài 3:

Gọi cd,cr lần lượt là a,b(m;a,b>0)

Áp dụng tc dtsbn:

\(\dfrac{b}{a}=\dfrac{2}{5}\Rightarrow\dfrac{a}{5}=\dfrac{b}{2}=\dfrac{2a+2b}{10+4}=\dfrac{42}{14}=3\\ \Rightarrow\left\{{}\begin{matrix}a=15\\b=6\end{matrix}\right.\\ \Rightarrow S_{hcn}=ab=90\left(m^2\right)\)

Bài 4:

Gọi cd,cr lân lượt là a,b(m;a,b>0)

Đặt \(\dfrac{a}{4}=\dfrac{b}{3}=k\Rightarrow a=4k;b=3k\)

\(ab=300\left(m^2\right)\\ \Rightarrow12k^2=300\\ \Rightarrow k^2=25\Rightarrow k=5\left(k>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a=20\\b=15\end{matrix}\right.\)

Vậy ...

Bài 5:

Gọi số hs 7A,7B,7C,7D ll là a,b,c,d(hs;a,b,c,d∈N*)

Áp dụng tc dtsbn:

\(\dfrac{a}{11}=\dfrac{b}{12}=\dfrac{c}{13}=\dfrac{d}{14}=\dfrac{2b-a}{24-11}=\dfrac{39}{13}=3\\ \Rightarrow\left\{{}\begin{matrix}a=33\\b=36\\c=39\\d=42\end{matrix}\right.\)

Vậy ...

Bài 10:

$-A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$

$=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}$

$=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{10-9}{9.10}$

$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}$

$=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}$

$\Rightarrow A=\frac{-3}{20}$

Bài 11:

$A=\frac{2n}{n+3}=\frac{2(n+3)-6}{n+3}=2-\frac{6}{n+3}$
Để $A$ nguyên thì $\frac{6}{n+3}$ nguyên.

Với $n$ nguyên thì điều trên xảy ra khi $6\vdots n+3$

$\Rightarrow n+3\in\left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$

$\Rightarrow n\in\left\{-4; -2; -1; -5; -6; 0; -9; 3\right\}$

Bài 9:

Gọi d là ƯCLN của $n+1, n+2$

$\Rightarrow n+1\vdots d; n+2\vdots d$

$\Rightarrow (n+2)-(n+1)\vdots d$ hay $1\vdots d$

$\Rightarrow d=1$

Vậy $n+1, n+2$ nguyên tố cùng nhau, suy ra $\frac{n+1}{n+2}$ là phân số tối giản.