5√3 - 4√3 /3 +9/2*2√6/3+6√3

=15√3/3-4√3/3+9√6/3+18√3/3

=17√3/3+9√6/3

\(\left(\frac{2}{\sqrt{3}+1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)

= \(\left[\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{2\left(\sqrt{3}+2\right)}{1}+\frac{15\left(3+\sqrt{3}\right)}{6}\right].\frac{1}{\sqrt{3}+5}\)

= \(\left[\frac{2\left(\sqrt{3}+1\right)-6\left(\sqrt{3}+2\right)+15\left(\sqrt{3}+3\right)}{2}\right].\frac{1}{\sqrt{3}+5}\)

= \(\left[\frac{2\sqrt{3}+2-6\sqrt{3}-12+5\sqrt{3}+15}{2}\right]\).\(\frac{1}{\sqrt{3}+5}\)

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

= \(\frac{1}{2}\)

=( 8 căn 7-5 căn 7+ 6 căn 7-4 căn 7)*căn 7

= 5 căn 7*căn 7

=35

( bấm máy tính là ra mà bạn, hì)

Vũ Minh TuấnLê Thị Thục Hiền@Nk>↑@

\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\) (*) (đk : \(x\ge\frac{5}{2}\))

Đặt \(\sqrt{2x-5}=a\left(a\ge0\right)\)

=> 2x-5=a²

<=> \(x=\frac{a^2+5}{2}\)

Có \(\sqrt{\frac{a^2+5}{2}-2+a}+\sqrt{\frac{a^2+5}{2}+2+3a}=7\sqrt{2}\)

<=> \(\sqrt{\frac{a^2+5-4+2a}{2}}+\sqrt{\frac{a^2+5+4+6a}{2}}=7\sqrt{2}\)

<=>\(\sqrt{\frac{a^2+2a+1}{2}}+\sqrt{\frac{a^2+6a+9}{2}}=7\sqrt{2}\)

<=> \(\frac{\sqrt{\left(a+1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(a+3\right)^2}}{\sqrt{2}}=7\sqrt{2}\)

<=> \(\left|a+1\right|+\left|a+3\right|=7\sqrt{2}.\sqrt{2}\)

<=> \(a+1+a+3=14\)(do a\(\ge\)0)

<=> \(2a=10\) <=> a=5(t/m)

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

<=> \(2x-5=25\) <=> \(x=15\)(tm pt (*))

Vậy pt (*) có tập nghiệm \(S=\left\{15\right\}\)

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

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

\(=\frac{1}{\sqrt{2}}-\frac{2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}+\frac{\left(\sqrt{2}-1\right)\left(-2\sqrt{6}+6\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{1}{\sqrt{2}}\)

\(=\frac{2-\sqrt{6}}{2}-4\sqrt{3}+6\sqrt{2}+2\sqrt{6}-6\)

\(=6\sqrt{2}-4\sqrt{3}+\frac{3\sqrt{6}}{2}-5\)

Kết quả xấu quá, chắc bạn ghi nhầm đề

Đã kiểm tra đáp án bằng casio

So sánh A=(2∛5)³ và B=(1/2)³.(∛311)³

Ta có : A = 40; B = 311/8 < 320/8 = 40.

Suy ra A>B.

Suy ra 2∛5 > (1/2).∛311

\(A=15+12+4\sqrt{45}+12\sqrt{5}=27+24\sqrt{5}\)

\(B=\left(2\sqrt{3}+6\sqrt{3}\right).\frac{\sqrt{3}}{2}-5\sqrt{6}=\frac{8\sqrt{3}.\sqrt{3}}{2}-5\sqrt{6}=12-5\sqrt{6}\)

\(C=4\sqrt{3}+\frac{4}{\sqrt{3}}+10\sqrt{5}-\frac{10}{\sqrt{5}}=\frac{16}{\sqrt{3}}+8\sqrt{5}\)