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a) I x-7I + I x-3I =0
có Ix-7I\(\ge\)0\(\forall\)x \(\in\)R
I x -3 I \(\ge0\forall x\in R\)
=> \(\hept{\begin{cases}Ix-7I=0\\Ix-3I=0\end{cases}}\)
=>\(\hept{\begin{cases}x-7=0\\x-3=0\end{cases}}\)
=>\(\hept{\begin{cases}x=7\\x=3\end{cases}}\)
vậy x\(\in\){3,7}
phần B chắc bạn biết lm rùi đúng không
nếu không lm dc ib mik lm tiếp nha
áp dụng công thức lm tiếp như vậy nha
Phạm Minh Hiếu ơi , hình như cậu làm sai rồi đó . Thử lại ko có đúng . Cậu xem lại hộ mình với ...... !!! ^-^
Bài 3:
g: \(\Leftrightarrow3\cdot4^x-4^x\cdot16=-80\)
\(\Leftrightarrow4^x=\dfrac{80}{13}\)
\(\dfrac{-3}{7}\)(\(\dfrac{5}{9}\)+\(\dfrac{4}{9}\)) + 0
=\(\dfrac{-3}{7}\)
*Lần sau bạn viết rõ đề ra, để thế này nhiều ng sẽ không hiểu!
\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(2022\right)^0\)
\(=\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+1\)
\(=\dfrac{-3}{7}.1+1\)
\(=\dfrac{-3}{7}+1\)
\(=\dfrac{-3+7}{7}\)
\(=\dfrac{4}{7}\)
2/
a/
\(\dfrac{4n+2}{n+1}=\dfrac{4n+4-2}{n+1}=\dfrac{4\left(n+1\right)-2}{n+1}=4-\dfrac{2}{n+1}\)
\(\Rightarrow4n+2⋮n+1\) Khi \(n+1=\left\{-2;-1;1;2\right\}\Rightarrow n=\left\{-3;-2;0;1\right\}\)
b/
\(\Rightarrow a+2=\dfrac{8}{b-1}\left(b\ne1\right)\) (1)
a nguyên => a+2 nguyên \(\Rightarrow8⋮\left(b-1\right)\)
\(\Rightarrow b-1=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Rightarrow b=\left\{-7;-3;-1;0;2;3;5;9\right\}\) Thay các giá trị của b vào (1) để tìm a
3/
Gọi số tiền mẹ cho Chi là A \(\Rightarrow100000\le A\le200000\)
Nếu bớt số tiền mẹ cho Chi đi 5000 đồng thì
\(A-5000⋮15000;A-5000⋮8000\)
\(\Rightarrow A-5000=UC\left(8000;15000\right)\) Và \(95000\le A-5000\le195000\)
\(\Rightarrow A-5000=120000\Rightarrow A=125000\)
/ 2x - 1 / + 3 = 16
/ 2x - 1 / = 16 - 3
/ 2x - 1 / = 13
Xét 2x - 1 = 13
2x = 13 + 1
2x = 14
x = 14 : 2
x = 7
Xét 2x - 1 = - 13
2x = - 13 + 1
2x = - 12
x = - 12 : 2
x = - 6
Vậy x = 7
hoặc x = -6
/2x-1/+3=16
/2x-1/=16-3
/2x-1/=13
2x-1=13
2x=13+1
2x=14
x=14:2
x=7
vậy x=7
\(A=1+3+3^2+3^3+...+3^{20}\)
=> \(3A=3+3^2+3^3+3^4+...+3^{21}\)
=> \(3A-A=3^{21}-1\)
=> \(2A=3^{21}-1\)
=> \(A=\frac{3^{21}-1}{2}\)
Bài 3:
a) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}=\dfrac{4}{3}-\dfrac{-4}{3}=\dfrac{8}{3}\)
b) \(2\dfrac{1}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+0,4.5\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+\dfrac{2}{5}.5\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+2\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\dfrac{8}{3}\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\dfrac{7}{6}\)
\(=\dfrac{7}{3}-\dfrac{7}{18}\)
\(=\dfrac{35}{18}\)
c) \(\left(20+9\dfrac{1}{4}\right):2\dfrac{1}{4}=\left(20+\dfrac{37}{4}\right):\dfrac{9}{4}=\dfrac{117}{4}:\dfrac{9}{4}=13\)
d) \(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)
\(=\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\)
\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{32}{5}\)
\(=10-\dfrac{32}{5}\)
\(=\dfrac{18}{5}\)
e) \(\dfrac{32}{15}:\left(-1\dfrac{1}{5}+1\dfrac{1}{3}\right)=\dfrac{32}{15}:\left(\dfrac{-6}{5}+\dfrac{4}{3}\right)=\dfrac{32}{15}:\dfrac{2}{15}=16\)
g) \(0,2.\dfrac{15}{36}-\left(\dfrac{2}{5}+\dfrac{2}{3}\right):1\dfrac{1}{5}\)
\(=\dfrac{1}{5}.\dfrac{5}{12}-\dfrac{16}{15}:\dfrac{6}{5}\)
\(=\dfrac{1}{12}-\dfrac{8}{9}\)
\(=\dfrac{-29}{36}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{8}{15}+0,25\right).\dfrac{24}{27}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{8}{9}\)
\(=\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{8}{9}\)
\(=\dfrac{7}{5}-\dfrac{94}{135}\)
\(=\dfrac{19}{27}\)
g) \(5:\left(4\dfrac{3}{4}-1\dfrac{25}{28}\right)-1\dfrac{3}{8}:\left(\dfrac{3}{8}+\dfrac{9}{20}\right)\)
\(=5:\left(\dfrac{19}{4}-\dfrac{53}{28}\right)-\dfrac{11}{8}:\dfrac{33}{40}\)
\(=5:\dfrac{20}{7}-\dfrac{5}{3}\)
\(=\dfrac{7}{4}-\dfrac{5}{3}\)
\(=\dfrac{1}{12}\)
\(a\)) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}\)
\(=\dfrac{4}{3}-\dfrac{7}{5}.\dfrac{-20}{21}\)
\(=\dfrac{4}{3}-\dfrac{-4}{3}\)
\(=\dfrac{8}{3}\)
\(=2\dfrac{2}{3}\)