Giúp mình bài 2 Từ câu D thôi
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mọi người giúp mình bài này với câu 1 với câu 2 thôi
15 tháng 10 2021
☆┌─┐ ─┐☆ │▒│ -▒- │▒│-▒- │▒ -▒-─┬─ │▒│▒|▒│▒│ ┌┴─┴─┐-┘─┘ │▒┌──┘▒▒▒│ └┐▒▒▒▒▒▒┌┘ └┐▒▒▒▒┌❣
25 tháng 8 2021
Bài 2:
Ta có: \(3n^3+10n^2-5⋮3n+1\)
\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)
hay \(n\in\left\{0;-1;1\right\}\)
5 tháng 9 2021
Bài 3:
a: Ta có: \(P=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x}{x\sqrt{x}-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}+2-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
NB
1
16 tháng 8 2021
bài 7
A=\(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)}+\dfrac{-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+x+1\right)}\)
A=\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
bài 8
P=\(\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)^2}\right].\dfrac{\left(x-1\right)^2}{4x}\)
P=\(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{4x}\)
P=\(\dfrac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(x-1\right)^2}{4x}\)=\(\dfrac{x-1}{2\sqrt{x}\left(\sqrt{x}-1\right)}\)
P=\(\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
bài 9
P=\(\left[\dfrac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\right].\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{4\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{2\sqrt{xy}-x-y}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{-\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{-\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
bài 10
P=\(\left[\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\left(\sqrt{x}+2\right)^2}\right]:\left[\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\)
P=\(\dfrac{\sqrt{x}+2-2}{\left(\sqrt{x}+2\right)^2}:\dfrac{2-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P=\(\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{-\sqrt{x}}\)=\(\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}\)
16 tháng 12 2021
c: \(C=\sqrt{\left(a+5b\right)^2-20ab}=\sqrt{4^2+100}=4\sqrt{29}\)
16 tháng 12 2021
d) D = a4 + 625b4
=> D = a4 + 20a3b + 150a2b2 + 500ab3 625b4 - (20a3b + 150a2b2 + 500ab3)
=> D = (a + 5b)4 - 5ab(4a2 + 40ab + 100b2 - 10ab)
=> D = (a + 5b)4 - 5ab[(2a + 10b)2 - 10ab]
=> D = (a + 5b)4 - 5ab[4(a + 5b)2 - 10ab]
=> D = 44 - 5 . (- 5)[4 . 42 - 10 . (- 5)]
=> D = 256 + 25(64 + 50)
=> D = 256 + 25 . 114
=> D = 256 + 2850 = 3106
Cái này mình dùng HĐT mở rộng bạn cóa thể tham khảo tam giác Pascal
D=\(\frac{-21}{9}-\frac{4}{11}.\frac{-7}{2}+\frac{2}{3}.\left(\frac{-5}{2}\right)\)
=\(\left(-3\right)-\frac{-14}{11}+\frac{-5}{3}=\frac{-99}{33}-\frac{-42}{33}+\frac{-55}{33}=\frac{-112}{33}\)
E=\(\frac{5}{36}:\frac{-1}{6}+\frac{7}{12}.\frac{5}{-2}-\frac{11}{3}.\frac{-1}{4}\)
=\(\frac{-5}{6}+\frac{-35}{24}-\frac{-11}{12}=\frac{-20}{24}+\frac{-35}{24}-\frac{-22}{24}=\frac{-33}{24}=\frac{-11}{8}\)
G=\(\frac{-3}{4}+\frac{-1}{3}.\left(\frac{1}{2}-\frac{5}{18}\right)\)
=\(\frac{-3}{4}+\frac{-1}{3}.\frac{2}{9}=\frac{-3}{4}+\frac{-2}{27}=\frac{-89}{108}\)
I=\(\frac{-19}{9}.\left(2-\frac{4}{-11}-\frac{-2}{3}\right)\)
=\(\frac{-19}{9}.\left(\frac{26}{11}-\frac{-2}{3}\right)=\frac{-19}{9}.\frac{100}{33}=\frac{-1900}{297}\)
J=\(\frac{-5}{6}:\left(\frac{1}{2}.\frac{3}{5}+\frac{-7}{12}-\frac{1}{3}\right)\)
=\(\frac{-5}{6}:\left(\frac{3}{10}+\frac{-11}{12}\right)=\frac{-5}{6}:\frac{-37}{60}=\frac{-5}{6}.\frac{60}{-37}=\frac{50}{37}\)
K=\(\frac{-3}{4}:\left(\frac{11}{3}.\frac{-9}{2}-\frac{5}{18}.\frac{-9}{3}\right)\)
=\(-\frac{3}{4}:\left(-\frac{33}{2}-\frac{-5}{6}\right)=\frac{-3}{4}:\frac{-47}{3}=\frac{-3}{4}.\frac{3}{-47}=\frac{9}{188}\)