Đặt A = 1.3.5 + 3.5.7 + 5.7.9 + ...... + 47.49.51

8A = 1.3.5.8 + 3.5.7.8 + ...... + 47.49.51.8

= 1.3.5(7 + 1) + 3.5.7.(9 - 1) + ...... + 47.49.51(53 - 45)

= 1.3.5.7 + 1.3.5 + 3.5.7.9 - 1.3.5.7 + ......... + 47.49.51.53 - 47.47.49.51

= 1.3.5 + 47.49.51.53

=> A = \(\frac{1.3.5+47.49.51.53}{8}=778128\)

628414

chi tiết là ko biết bởi vì e mới học lớp 5

Các bạn giải nhanh lên nhé, mai mình phải nộp cô rồi

Chọn C

\(0,0\left(3\right)+0,0\left(16\right)\)\(=\frac{1}{30}+\frac{16}{990}\)\(=\)\(\frac{33}{990}+\frac{16}{990}=\frac{49}{990}\)

0,0(3) +0,0(16)=0,0(49)

0,0(3) + 0,0(16) = 1/10 x 1/3 + 1/10 x 16/99

= 1/30 + 16/297

= 777/8910

Lấy (32,43 + 85,94)x 0,01-0,0136

            118,37          x0.01 - 0,0136

                          1.1837 - 0,0136

                                  1,701

=[(-4,9)+4,9]+[5,5+(-5,5)]

=         0     +        0

=                0

(-4,9)+5,5+4,9+(-5,5)

=[(-4,9)+4,9]+[5,5+(-5,5)]

=(4,9-4,9)+(5,5-5,5)

=0+0

=0

b) (-4,9) + 5,5 + 4,9 + (-5,5) = ((-4,9) + 4,9) + ( 5,5 + (-5,5)) = 0 + 0 = 0

Bài 18 

a, Với \(a>0;a\ne1;4\)

\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\left(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

b, Thay a = 9 => căn a = 3 

\(A=\dfrac{3-2}{3.3}=\dfrac{1}{9}\)

c, Ta có : \(A.B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}.\dfrac{3\sqrt{a}}{\sqrt{a}+1}=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}< 0\)

Vì \(\sqrt{a}+1>\sqrt{a}-2\)

\(\left\{{}\begin{matrix}\sqrt{a}+1>0\\\sqrt{a}-2< 0\end{matrix}\right.\Leftrightarrow a< 4\)

Kết hợp với đk vậy \(0< a< 4;a\ne1\)

Bài 18:

1) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

2) Thay a=9 vào B, ta được:

\(B=\dfrac{3\cdot3}{3+1}=\dfrac{9}{4}\)

Bài 19:

1) Ta có: \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

a, \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)ĐK : \(x>0;x\ne1\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b, \(A=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Rightarrow3\sqrt{x}-3=\sqrt{x}\Leftrightarrow2\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\)

c, \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=\dfrac{\sqrt{x}-1-9x}{\sqrt{x}}\)

\(=1-\dfrac{1}{\sqrt{x}}-9\sqrt{x}\)Đặt \(\sqrt{x}=t^2\left(t>0\right)\)

\(1-t-9t^2=-\left(9t^2-t-1\right)=-\left(9t^2-2.3.\dfrac{1}{6}.t+\dfrac{1}{36}-\dfrac{37}{36}\right)\)

\(=-\left(3t-\dfrac{1}{6}\right)+\dfrac{37}{36}\le\dfrac{37}{36}\)

Dấu ''='' xảy ra khi t = 1/18 => t^2 = 1/324 => \(\sqrt{x}=\dfrac{1}{324}\Rightarrow x=\dfrac{1}{104876}\)

Vậy GTLN P là 37/36 khi x = 1/104876

\(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)=\(\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

1) Ta có: \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

2) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-3-\sqrt{x}=0\)

\(\Leftrightarrow2\sqrt{x}=3\)

hay \(x=\dfrac{9}{4}\)(thỏa ĐK)

d. \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)

\(\Rightarrow x^2+5x-14=x^2+3x-4\)

\(\Rightarrow x^2+5x-x^2-3x=-4+14\)

\(\Rightarrow2x=10\) \(\Rightarrow x=\dfrac{10}{3}\) \(\Rightarrow x=5\)

\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

⇔ \(\dfrac{\left(x-2\right)\left(x+7\right)}{\left(x-1\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+7\right)\left(x-1\right)}\)

⇔ (x - 2)(x + 7) = (x + 4)(x - 1)

⇔ x² + 7x - 2x - 14 = x² - x + 4x - 4

⇔ x² - x² + 7x - 2x + x - 4x = 14 - 4

⇔ 2x = 10

⇔ x = 10/2 = 5

d: Ta có: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x=10\)

hay x=5

\(\dfrac{\sqrt{3}-3}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-1\right)}{2}=\dfrac{3-\sqrt{3}-3\sqrt{3}+3}{2}=\dfrac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\)