Đặt biểu thức trên là A.

Ta có: A=1+2+2²+2³+....+2¹⁰

2A=2+2²+2³+....+2¹⁰+2¹¹

2A-A=A=(2+2²+2³+2⁴+....+2¹⁰+2¹¹)-(1+2+2²+2³+.....+2¹⁰)

=2¹¹-1

Độ dài đường chéo thứ hai là:

            4 x 2 : 3/5 = 40/3 (dm)

                                Đáp số : 40/3 dm

15/12

=15/12   *3/3

=5/4

Số các số hạng là:

(2012 - 2) :3 +1=671

Tổng: (2012+2) x 671 : 2=675697

Giá trị của biểu thức là 675697

a) \(C=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

Nếu x\(\ge8\) thì C=\(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

Nếu \(4\le x< 8\) thì \(C=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

b) Ta có \(x=\sqrt{15+\sqrt{6}}\approx4,18\)

\(\Rightarrow4\le x< 8\Rightarrow C=4\)

Vậy khi x=\(\sqrt{15+\sqrt{6}}\) thì C=4

Giả sử \(\dfrac{1}{1+a}+\dfrac{1}{1+b}\ge\dfrac{2}{1+\sqrt{ab}}\Leftrightarrow\dfrac{1}{1+a}-\dfrac{1}{1+\sqrt{ab}}+\dfrac{1}{1+b}-\dfrac{1}{1+\sqrt{ab}}\ge0\Leftrightarrow\dfrac{1+\sqrt{ab}-1-a}{\left(1+a\right)\left(1+\sqrt{ab}\right)}+\dfrac{1+\sqrt{ab}-1-b}{\left(1+b\right)\left(1+\sqrt{ab}\right)}\ge0\Leftrightarrow\dfrac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)}+\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(1+b\right)\left(1+\sqrt{ab}\right)}\ge0\Leftrightarrow\dfrac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(1+b\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)\left(1+b\right)}+\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\left(1+a\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)\left(1+b\right)}\ge0\Leftrightarrow\dfrac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(1+b\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\left(1+a\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)\left(1+b\right)}\ge0\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(-\sqrt{a}-\sqrt{a}.b+\sqrt{b}+a\sqrt{b}\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)\left(1+b\right)}\ge0\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left[\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)\right]}{\left(1+a\right)\left(1+\sqrt{ab}\right)\left(1+b\right)}\ge0\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{ab}-1\right)}{\left(1+a\right)\left(1+\sqrt{ab}\right)\left(1+b\right)}\ge0\)(1)

Vì \(\left(\sqrt{a}-\sqrt{b}\right)\left(1+a\right)\left(1+b\right)\ge0\)(2)

\(\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{ab}-1\right)\ge0\)(vì ab\(\ge1\))(3)

Từ (1),(2),(3)\(\Rightarrow\) điều giả sử đúng

Vậy \(\dfrac{1}{1+a}+\dfrac{1}{1+b}\ge\dfrac{2}{1+\sqrt{ab}}\)

số h/s nữ = 40x2/5=16 h/s

số lớn =  1989x5/9=1105

36/45=4/5 

đặt p/s =A/b=>5a=4b (1)

mà a+7=b-7=>4a+28=4b-28=>4b=4a+56(2)

từ (1) và (2)=>4a+56=5a

=>a=56=>b=70

=>p/s cần tìm là 56/70

Chẳng hiểu mà đây học toán chứ đâu học TV

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

Đặt A=\(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\Leftrightarrow A^2=\sqrt{2}-1+\sqrt{2}+1-2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}-2\sqrt{1}=2\sqrt{2}-2\Leftrightarrow A=\pm\sqrt{2\sqrt{2}-2}\)

Ta có \(\sqrt{\sqrt{2}-1}< \sqrt{\sqrt{2}+1}\Leftrightarrow\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}< 0\Leftrightarrow A< 0\)

Vậy A=\(-\sqrt{2\sqrt{2}-2}\)

(1)\(=\sqrt{2}.\left(-\sqrt{2\sqrt{2}-2}\right)=-\sqrt{4\sqrt{2}-4}\)

b) \(\sqrt{4-2\sqrt{3}}+\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{27}=\sqrt{3-2.\sqrt{3}.1+1}+\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{9.3}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\dfrac{4+2\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}}-3\sqrt{3}=\left|\sqrt{3}-1\right|+\sqrt{4+2\sqrt{3}}-3\sqrt{3}=\sqrt{3}-1-3\sqrt{3}+\sqrt{3+2\sqrt{3}+1}=-2\sqrt{3}-1+\sqrt{\left(\sqrt{3}+1\right)^2}=-2\sqrt{3}-1+\sqrt{3}+1=-\sqrt{3}\)

c) \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-3\sqrt{x}-2\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left[3\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3\sqrt{x}-2\right)}{\sqrt{x}+3}=\dfrac{2-3\sqrt{x}}{\sqrt{x}+3}\)

có số cây được trồng trên đoạn đường là

                    40 : 4 = 10 (cây)

                              đáp số : 10 cây