Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

Sửa đề:

\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+12\sqrt{5}-8}}{\sqrt{5}+\sqrt{9-6\sqrt{5}+5}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\left(3-\sqrt{5}\right)}=\dfrac{1}{3}\)

Thế vô A ta được

\(A=\left(3.\dfrac{1}{3^3}+8.\dfrac{1}{3^2}+2\right)^{2018}=3^{2018}\)

Thay 1=ab+bc+ca vào M

\(\Rightarrow M=a+b-\sqrt[]{\dfrac{\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)}{c^2+ab+bc+ca}}\)

\(\Leftrightarrow M=a+b-\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)}{\left(b+c\right)\left(a+c\right)}}\)

\(\Rightarrow M=a+b-a-b\)

\(\Rightarrow M=0\)

bạn viết công thức đi

viết này ai hiểu

\(=4\sqrt{x}-5\cdot2\sqrt{x}-5\sqrt{x}-3\sqrt{x}-5=4\sqrt{x}-10\sqrt{x}-5\sqrt{x}-3\sqrt{x}-5=-14\sqrt{x}-5\)

a) \(\dfrac{2}{x-3}\sqrt{\dfrac{x^2-6x+9}{4y^4}}=\dfrac{2}{x-3}.\dfrac{3-x}{2y^2}=\dfrac{2.2y^2}{\left(x-3\right)\left(3-x\right)}=-\dfrac{4y^2}{x^2-6x+9}=-\dfrac{2y}{x-3}\)

=\(\dfrac{2}{2x-1}\sqrt{5}x\sqrt[]{\left(1-2x\right)^2}\)

=\(\dfrac{2\sqrt{5}x\left(1-2x\right)}{2x-1}\)

=\(\dfrac{-2\sqrt{5}x\left(2x-1\right)}{2x-1}\)​​

=\(-2\sqrt{5}x\)

a)\(4\sqrt{x}-5\sqrt{4x}-\sqrt{25x}-3\sqrt{x}-5\)

=\(4\sqrt{x}-10\sqrt{x}-5\sqrt{x}-3\sqrt{x}-5\)

=\(-14\sqrt{x}-5\)

b)\(\sqrt{16x}-5\left(\sqrt{x}-2\right)\sqrt{79x}-5\)

=\(4\sqrt{x}-\left(5\sqrt{x}-10\right)\sqrt{79x}-5\)

=\(4\sqrt{x}-\left(5\sqrt{79}x-10\sqrt{79}x\right)-5\)

=\(4\sqrt{x}+5\sqrt{79}x-5\)

a) 4 \(\sqrt{x}\)-5\(\sqrt{4x}\)-\(\sqrt{25x}\)- 3\(\sqrt{x}\)-5

=4\(\sqrt{x}\)- 10\(\sqrt{x}\)-5\(\sqrt{x}\)-3\(\sqrt{x}\)-5

= (4-10-5-3-5)\(\sqrt{x}\)

=-19\(\sqrt{x}\)

=\(4\sqrt{^{ }x}-5\sqrt[]{4x}-\sqrt{25x}-3\sqrt{x}-5\)

=\(\sqrt{x}\left(4-10-5-3\right)-5\)

=\(-14\sqrt{x}-5\)

Đặt \(\sqrt{2x^2+7x+10}=a;\sqrt{2x^2+x+4}=b\left(a,b>0\right)\)

pt <=> a + b = 3(x + 1)

Mà a² - b² = 2x² + 7x + 10 - 2x² - x - 4 = 6x + 6

nên pt <=> a + b = \(\dfrac{a^2-b^2}{2}\)

<=> (a - b)(a + b) = 2(a + b)

Vì a;b > 0 nên a + b khác 0. Chia cả 2 vế của pt cho a + b ta có

pt <=> a - b = 2

<=> \(\sqrt{2x^2+7x+10}-\sqrt{2x^2+x+4}=2\)

<=> \(\sqrt{2x^2+7x+10}=2+\sqrt{2x^2+x+4}\)

Bình phương 2 vế ta có:

pt <=> \(2x^2+7x+10=2x^2+x+8+8\sqrt{2x^2+x+4}\)

<=> \(3x+1=4\sqrt{2x^2+x+4}\)

Bình phương lần nữa rồi làm nốt, làm xong thì thử lại.

3.

\(•x=3+\sqrt{2}\\ x^2=\left(3+\sqrt{2}\right)^2\\ x^2=9+2.3.\sqrt{2}+2\\ x^2=11+6\sqrt{2}\\• y=\sqrt{11+6\sqrt{2}}\\ y^2=\left(\sqrt{11+6\sqrt{2}}\right)^2\\ y^2=11+6\sqrt{2}\)

\(\Rightarrow x^2=y^2=11+6\sqrt{2}\)

1. ta có : \(4\sqrt{7}=\sqrt{112}\)

\(3\sqrt{3}=\sqrt{27}\)

ta thấy : \(\sqrt{112}>\sqrt{27}\) hay \(4\sqrt{7}>3\sqrt{3}\)

2. \(\dfrac{1}{4}\sqrt{82}=\sqrt{\dfrac{41}{8}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

ta thấy :\(\sqrt{\dfrac{41}{8}}< \sqrt{\dfrac{36}{7}}\) hay \(\dfrac{1}{4}\sqrt{82}< 6\sqrt{\dfrac{1}{7}}\)

3. \(x^2=\left(3+\sqrt{2}\right)^2\)

\(y^2=11+6\sqrt{2}\)=\(\left(3+\sqrt{2}\right)^2\)

ta thấy : \(x^2=y^2\Rightarrow x=y\)

Dễ mà----- bạn chỉ càn áp dụng hằng đẳng thức là được

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