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Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
a) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{1-4}{5}=-\dfrac{3}{5}\)
b) \(\dfrac{3}{5}+\dfrac{4}{-7}=\dfrac{3}{5}-\dfrac{4}{7}=\dfrac{21-20}{35}=\dfrac{1}{35}\)
c, d) giống a, b.
e) \(\dfrac{4}{9}-\dfrac{-5}{6}=\dfrac{4}{9}+\dfrac{5}{6}=\dfrac{8+15}{18}=\dfrac{23}{18}\)
f) \(\dfrac{6}{7}+\dfrac{1}{7}\cdot\dfrac{2}{7}+\dfrac{1}{7}\cdot\dfrac{5}{7}=\dfrac{6}{7}+\dfrac{2}{49}+\dfrac{5}{49}=1\)
g) \(\dfrac{4}{9}\cdot\dfrac{13}{3}-\dfrac{4}{3}\cdot\dfrac{40}{9}=\dfrac{52}{27}-\dfrac{160}{27}=-4\)
h) \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)=\dfrac{16}{7}-\left(\dfrac{4}{3}+\dfrac{8}{7}\right)=\dfrac{16}{7}-\dfrac{52}{21}=-\dfrac{4}{21}\)
i) \(\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}=\left(\dfrac{20}{9}+\dfrac{6}{5}\right)-2\cdot\dfrac{2}{3}=\dfrac{154}{45}-\dfrac{4}{3}=\dfrac{94}{45}\)
k) \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}-\dfrac{26}{19}=\dfrac{56}{209}+\dfrac{21}{209}-\dfrac{26}{19}=-1\)
a)\(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-36}{45}=\dfrac{9}{45}+\dfrac{-36}{45}=\dfrac{-27}{45}\)
b)
\(\dfrac{3}{5}+\dfrac{4}{-7}=\dfrac{3}{5}+\dfrac{-4}{7}=\dfrac{21}{35}+\dfrac{-20}{35}=\dfrac{1}{35}\)
c)\(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-36}{45}=\dfrac{9}{45}+\dfrac{-36}{45}=\dfrac{-27}{45}\)
d)
\(\dfrac{3}{5}+\dfrac{4}{-7}=\dfrac{3}{5}+\dfrac{-4}{7}=\dfrac{21}{35}+\dfrac{-20}{35}=\dfrac{1}{35}\)
(chiều mình làm tiếp!!!)
1.ĐK: n khác 2
Để A nguyên thì \(\dfrac{9}{n-2}\)nguyên <=> 9 chia hết cho n-2 hay n-2 là Ư(9) và n là số tự nhiên
Mà Ư(9)={-9;-3;-1;1;3;9}
Ta có bảng sau:
| n-2 | -9 | -3 | -1 | 1 | 3 | 9 |
| n | -7(L) | -1(L) | 1(TM) | 3(TM) | 5(TM) | 11(TM) |
Vậy n={1;3;5;9} thì A nguyên.
2.Ta xét tích:
(102016+2)(102016-3)
=104032-102016-6
(102016-1)102016
=104032-102016
104032-102016-6<104032-102016
=>(102016+2)(102016-3)<(102016-1)102016
Chia cả 2 vế cho (102016-1)(102016-3)
=>\(\dfrac{10^{2016}+2}{10^{2016}-1}< \dfrac{10^{2016}}{10^{2016}-3}\)
=>A<B
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
\(\Leftrightarrow1-11< =3m< =\left(9-9\right)\cdot A=0\)
=>-10<=3m<=0
hay \(m\in\left\{-3;-2;-1;0\right\}\)
a) \(-1< \dfrac{5x}{13}< 0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5x}{13}>-1\\\dfrac{5x}{13}< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>-\dfrac{13}{5}\\x< 0\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{13}{5},0\right\}\)
b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(\Rightarrow3\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{27}\)
\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)
\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}\)
\(\Leftrightarrow3x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{18}\)
Vậy \(x=\dfrac{1}{18}\)
Xử câu b trc =)
b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}:3=-\dfrac{1}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=\left(-\dfrac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)
\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{6}:3=\dfrac{1}{18}\)
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).