-20/147

=\(\frac{-4}{15}\)

\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\cdot\left(\frac{-4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right)\cdot\frac{5}{7}}\)

\(=-\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\cdot\frac{4}{15}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right)\cdot\frac{5}{7}}\)

\(=-\frac{\frac{4}{15}\cdot\frac{5-4\sqrt{2}}{70}}{\frac{5}{7}\cdot\frac{5-4\sqrt{2}}{50}}\)

\(=-\frac{4\left(5-4\sqrt{2}\right)}{15\left(5-4\sqrt{2}\right)}\)

\(=-\frac{4}{15}\)

a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)

\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)

\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)

\(=\left(4-8+25\right)\sqrt{x^2+1}\)

\(=21\sqrt{x^2+1}\)

b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)

\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)

\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)

\(B=\sqrt{3}\)

Dạ đậy ạ,mong dc gp

1. A = 75(4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 25

    A = 25 . [3 . (4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 1]

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 3 + 1)

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 4)

    A = 25 . 4 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1) \(⋮\) 100

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

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

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

2.

A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\

A=0

4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)

\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)

Tìm z thì dễ rồi

Bài 1:

\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)

\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)

\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)

\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)

\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)

\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)

\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)

\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)

\(\Rightarrow C=\sqrt{14}\)

\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)

\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)

\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)

\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)

\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)

Bài 2:

a) Bạn xem lại đề.

b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)

c)

\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)

\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)

Bài 3:

a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)

\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)

\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

b)

Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)

c)

Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)

Kết hợp với ĐKXĐ suy ra $x>4$