gấp lắm mn ơi
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\(=\dfrac{2x+6}{3x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{3x\left(x+3\right)}=\dfrac{2}{3x}\)
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
\(\text{8.C.So le trong}\)
\(\text{9.C.a trùng b}\)
\(\text{10.B.}60^0\)
\(\text{11.C.}150^0\)
\(\text{12.B.A=P}\)
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 $6\vdots n+3$
$\Rightarrow n+3\in\left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$
$\Rightarrow n\in\left\{-4; -2; -1; -5; -6; 0; -9; 3\right\}$
Câu 29: D
Câu 30: A
Câu 31: C