a) \(A=\frac{1}{a^2+a}+\frac{1}{a^2+3a+2}+\frac{1}{a^2+5a+6}+\frac{1}{a^2+7a+12}+\frac{1}{a^2+9a+20}\)

\(A=\frac{1}{a\left(a+1\right)}+\frac{1}{\left(a+1\right)\left(a+2\right)}+\frac{1}{\left(a+2\right)\left(a+3\right)}+\frac{1}{\left(a+3\right)\left(a+4\right)}+\frac{1}{\left(a+4\right)\left(a+5\right)}\)

\(A=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}+\frac{1}{a+3}-\frac{1}{a+4}+\frac{1}{a+4}-\frac{1}{a+5}\)

\(A=\frac{1}{a}-\frac{1}{a+5}=\frac{a+5-a}{a\left(a+5\right)}=\frac{5}{a^2+5a}\)

b) Điều kiện: \(a\ne0;-1;-2;-3;-4;-5\)

\(A>\frac{5}{6}\) \(\Leftrightarrow\frac{5}{a^2+5a}>\frac{5}{6}\) \(\Leftrightarrow\frac{5}{a^2+5a}-\frac{5}{6}>0\) \(\Leftrightarrow\frac{30-5a^2-25a}{30\left(a^2+5a\right)}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}-6< a< -5\\0< a< 1\end{matrix}\right.\)

Kết luận: ....

ĐKXĐ: ...

a/ \(A=\frac{1}{a\left(a+1\right)}+\frac{1}{\left(a+1\right)\left(a+2\right)}+\frac{1}{\left(a+2\right)\left(a+3\right)}+\frac{1}{\left(a+3\right)\left(a+4\right)}+\frac{1}{\left(a+4\right)\left(a+5\right)}\)

\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+...+\frac{1}{a+4}-\frac{1}{a+5}\)

\(=\frac{1}{a}-\frac{1}{a+5}=\frac{5}{a\left(a+5\right)}\)

\(A>\frac{5}{6}\Rightarrow\frac{5}{a\left(a+5\right)}>\frac{5}{6}\)

\(\Leftrightarrow\frac{1}{a\left(a+5\right)}-\frac{1}{6}>0\Leftrightarrow\frac{6-a^2-5a}{a\left(a+5\right)}>0\)

\(\Leftrightarrow\frac{\left(1-a\right)\left(a+6\right)}{a\left(a+5\right)}>0\Rightarrow\left[{}\begin{matrix}-6< a< -5\\0< a< 1\end{matrix}\right.\)

\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)

\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)

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

Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.

\(1,ĐK:x\ne0;x\ne\pm6\)

\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right].\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}.\frac{1}{12\left(x^2+1\right)}\)

\(=\frac{12\left(x^2+1\right)}{x}.\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

\(2,A=\frac{1}{x}=\frac{1}{\frac{1}{\sqrt{9+4\sqrt{5}}}}=\sqrt{9+4\sqrt{5}}\)

Cho tam giác ABC vuông tại B có góc B₁=B₂  ; Â=60^o, kẻ BH vuông góc với AC (H thuộc AC). Qua B kẻ đường thẳng d song song với AC.

a) Tính góc ABH.

b) Chứng minh đường thẳng d vuông góc với BH.

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

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

\(A=\dfrac{4}{x-1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{2}-\sqrt{x^2-1}\)

\(A=2\left(x+1\right)-\sqrt{\left(x+1\right)\left(x-1\right)}\)

\(A=\sqrt{x+1}\left(2\sqrt{x+1}-\sqrt{x-1}\right)\)

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

a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)

b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)

c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)

\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)

Câu 1 Tính 

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)

Câu 2 Tính 

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\)

\(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

Câu 3 

a) Ta có : M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹ (1)

=> 3M = 3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰  (2)

Lấy (2) trừ (1) theo vế ta có : 

3M - M = (3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰) - ( M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹)

=>  2M = 3¹²⁰ - 1

=>    M = \(\frac{3^{120}-1}{2}\)

b) M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

        = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

        = (1 + 3 + 3²) + 3³(1 + 3 + 3²) + ... + 3¹¹⁷(1 + 3 + 3²)

        = 13 + 3³.13 + ... + 3¹¹⁷.13

        = 13(1 + 3³ + ... + 3¹¹⁷) \(⋮\)13

=> M \(⋮\)13

M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

= (1 + 3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶ + 3⁷) + ... + (3¹¹⁶ + 3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

= (1 + 3 + 3² + 3³) + 3⁴(1 + 3 + 3² + 3³) + ... + 3¹¹⁶(1 + 3 + 3² + 3³)

= 40 + 3⁴.40 + ... + 3¹¹⁶.40

= 40(1 + 3⁴ + ... + 3¹¹⁶) 

= 5.8.(1 + 3⁴ + ... + 3¹¹⁶)  \(⋮\)5

4) Tính 

A = 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1

=> 2A =  2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1

Lấy 2A trừ A theo vế ta có : 

2A - A = (2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1) - (2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1)

=>   A = 2¹⁰¹ - 2¹⁰⁰ - 2¹⁰⁰ + 1

=>   A = 2¹⁰¹ - (2¹⁰⁰ + 2¹⁰⁰) + 1

=>   A  = 2¹⁰¹ - 2¹⁰⁰ . 2 + 1

=>   A = 1

Câu 5 a) C = 1.2 + 2.3 + 3.4 + ... + 99.100

=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

          = 99.100.101 

=> C = 99.100.101 : 3 =  333300

b) Ta có : D = 2² + 4² + 6² + ... + 98²

                    = 2²(1² + 2²  + 3² + ... + 49²

                    =  2² .(1² + 2²  + 3² + ... + 49²)

                    = 2².(1.1 + 2.2 + 3.3 + ... + 49.49)

                    = 2².[1.(2 - 1) + 2..(3 - 1) + 3(4 - 1) + ... + 49(50 - 1)]

                    = 2².[(1.2 + 2.3 + 3.4 + ... + 49.50) - (1 + 2 + 3 + 4 + ... + 49)]

Đặt E = 1.2 + 2.3 + 3.4 + ... + 49.50

=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

          = 49.50.51 

=> E = 49.50.51/3 = 41650

Khi đó D = 2².[41650 - (1 + 2 + 3 + 4 + ... + 49)]

               = 2².[41650 - 49(49 + 1)/2]

               = 2².[41650 - 1225 

               = 2².40425

               = 161700

=> D = 161700

\(\begin{array}{l}A = \left( {7 - \frac{2}{5} + \frac{1}{3}} \right) - \left( {6 - \frac{4}{3} + \frac{6}{5}} \right) - \left( {2 - \frac{8}{5} + \frac{5}{3}} \right)\\A = 7 - \frac{2}{5} + \frac{1}{3} - 6 + \frac{4}{3} - \frac{6}{5} - 2 + \frac{8}{5} - \frac{5}{3}\\A = \left( {7 - 6 - 2} \right) + \left( { - \frac{2}{5} - \frac{6}{5} + \frac{8}{5}} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)

Chú ý:

Trong phép tính chỉ có phép cộng trừ, ta có thể đổi chỗ các số hạng tùy ý kèm theo dấu của chúng.

Lời giải:

a) ĐKXĐ: $x\neq 0; x\neq \pm 2$

\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)

\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)

b)

Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)

\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)

\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)

Lời giải:

a) ĐKXĐ: $x\neq 0; x\neq \pm 2$

\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)

\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)

b)

Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)

\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)

\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)