Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Biến đổi vế trái ta có:
\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\)
\(=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}=\frac{\sqrt{6}}{6}=VP\)
Vậy đẳng thức trên đc chứng minh
b) Biến đổi vế trái ta có:
\(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
\(=\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right)\cdot\frac{1}{\sqrt{6x}}\)
\(=x\sqrt{\frac{6}{x}\cdot\frac{1}{6x}}+\sqrt{\frac{2x}{3}\cdot\frac{1}{6x}}+\sqrt{6x}\cdot\frac{1}{\sqrt{6x}}\)
\(=x\sqrt{\frac{1}{x^2}}+\sqrt{\frac{1}{9}}+1=1+\frac{1}{3}+1=2\frac{1}{3}=VP\)
Vậy đẳng thức trên đc chứng minh
a)\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3}{2}\sqrt{6}+2\frac{\sqrt{6}}{3}-4\frac{\sqrt{6}}{2}\)
\(=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-\frac{4}{2}\right)=\sqrt{6}.\frac{1}{6}\)
b) \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}=\left(x.\frac{\sqrt{6x}}{x}+\frac{\sqrt{6x}}{3}+\sqrt{6x}\right):\sqrt{6x}\)
\(=1+\frac{1}{3}+1=2\frac{1}{3}\)
Bài 1:
a)
\(\frac{\sqrt{2.3}+\sqrt{2.7}}{2\sqrt{3}+2\sqrt{7}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{7})}{2(\sqrt{3}+\sqrt{7})}=\frac{\sqrt{2}}{2}\)
b)
\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{3+2\sqrt{2}}{2-1}=3+2\sqrt{2}\)
Bài 2:
a)
\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}=\sqrt{4}-\sqrt{1}=1\) (đpcm)
b)
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\frac{(\sqrt{3}+1)^2}{2}}+\sqrt{\frac{(\sqrt{3}-1)^2}{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\) (đpcm)
c) Sửa đề:
\(\left(\frac{\sqrt{a}}{\sqrt{a}+2}-\frac{\sqrt{a}}{\sqrt{a}-2}+\frac{4\sqrt{a}-1}{a-4}\right):\frac{1}{a-4}=\left[\frac{a-2\sqrt{a}-(a+2\sqrt{a})}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{a-4}\right].(a-4)\)
\(=\left(\frac{-4\sqrt{a}}{a-4}+\frac{4\sqrt{a}-1}{a-4}\right).(a-4)=-4\sqrt{a}+4\sqrt{a}-1=-1\)
d)
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{(\sqrt{a}+\sqrt{b})^2-(\sqrt{a}-\sqrt{b})^2}{2(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}+\frac{2b}{a-b}=\frac{4\sqrt{ab}}{2(a-b)}+\frac{2b}{a-b}\)
\(=\frac{2\sqrt{ab}+2b}{a-b}=\frac{2\sqrt{b}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)
--> dpcm
b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)
=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(7-5\right)\)
=-1.2
=-2
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
a) Biến đổi vế trái ta có:
\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\)
\(\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}=\frac{\sqrt{6}}{6}=VP\)
Vậy đẳng thức trên được chứng minh
b)Biến đổi vế trái ta được
\(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
\(=\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right)\cdot\sqrt{\frac{1}{6x}}\)
\(=x\sqrt{\frac{6}{x}\cdot\frac{1}{6x}}+\sqrt{\frac{2x}{3}\cdot\frac{1}{6x}}+\sqrt{6x\cdot\frac{1}{6x}}\)
\(=x\sqrt{\frac{1}{x^2}}+\sqrt{\frac{1}{9}}+1=1+\frac{1}{3}+1=2\frac{1}{3}=VP\)
Vậy đẳng thức trên được chứng minh