Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(\left(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}\right)^2=5-3\sqrt{2}+3\sqrt{2}-4+2\sqrt{5-3\sqrt{2}}\sqrt{3\sqrt{2}-4}\)
\(=1+2\sqrt{27\sqrt{2}-38}\)
Áp dụng vào bài toán t được
\(\dfrac{\sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)
\(=\dfrac{\sqrt{\left(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}\right)^2}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)
\(=\dfrac{\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}=1\)
Ta có: \(\frac{-3}{4\sqrt{3}-7}-\frac{3}{4\sqrt{3}+7}+\sqrt{4-2\sqrt{3}}\)
\(=\frac{-3\left(4\sqrt{3}+7\right)-3\left(4\sqrt{3}-7\right)}{\left(4\sqrt{3}\right)^2-7^2}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\frac{-12\sqrt{3}-21-12\sqrt{3}+21}{48-49}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\frac{-24\sqrt{3}}{-1}+\left|\sqrt{3}-1\right|\)
\(=24\sqrt{3}+\sqrt{3}-1\)(vì \(\sqrt{3}>1\))
\(=25\sqrt{3}-1\)
\(=\frac{-3\left(4\sqrt{3}+7\right)-3\left(4\sqrt{3}-7\right)}{\left(4\sqrt{3}\right)^2-49}+\sqrt{4-2\sqrt{3}}\)
Tiếp tục nhé
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
ĐKXĐ: \(x\ge\frac{3}{2}\)
PT (=) \(\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=7\)
(=) \(\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
(=) \(2\sqrt{2x-3}=2\) (=) \(\sqrt{2x-3}=1\)(=) 2x = 4 (=) x = 2 ( Thỏa mãn điều kiện )
Vậy x=2