Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)
\(=6-3b\) (vì b < 2 )
b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\)
\(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{3-2\sqrt{3}+1}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+6\sqrt{2}+2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
c) \(\sqrt{25x^2}-2x=-5x-2x=-7x\)(vì x < 0)
d) \(x-5+\sqrt{25-10x+x^2}=x-5+\sqrt{\left(5-x\right)^2}=x-5+x-5=2x-10\) (vì x > 5)
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(x-1\right)^2}{2}\)
\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}.\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b) Với \(0< x< 1\)\(\Rightarrow0< \sqrt{x}< 1\)
\(\Rightarrow\sqrt{x}-1< 0\)
mà \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-1\right)< 0\)
\(\Rightarrow-\sqrt{x}.\left(\sqrt{x}-1\right)>0\)\(\Rightarrow P>0\)( đpcm )
c) \(P=-x+\sqrt{x}=-x+\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)
\(=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)
\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)( thỏa mãn ĐKXĐ )
Vậy \(maxP=\frac{1}{4}\)\(\Leftrightarrow x=\frac{1}{4}\)
ĐKXĐ \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
a, Ta có \(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\frac{2\sqrt{x}-2x}{\sqrt{2}}\)
\(P=\sqrt{2x}-\sqrt{2}x\)
\(P=\sqrt{2x}\left(1-\sqrt{x}\right)\)
b, Vì \(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow1-\sqrt{x}< 1\)
\(\Rightarrow\sqrt{2x}\left(1-\sqrt{x}\right)>0\)
c, Ta có \(P=-\sqrt{2}\left(x-\sqrt{x}\right)\)
\(P=-\sqrt{2}\left(x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)
\(P=-\sqrt{2x}\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{\sqrt{8}}\le\frac{1}{\sqrt{8}}\)
Dấu = xảy ra \(\Leftrightarrow\)\(\sqrt{x}-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{4}\)
vậy GTLN của P là \(\frac{1}{\sqrt{8}}\)với x=\(\frac{1}{4}\)
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
Mấy bài này dài vật vã ghê =)))))))))))))
1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)
=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b, M = \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))
= \(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)
= \(\sqrt{3}\left(x-1\right)\)
Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:
M = \(\sqrt{3}\left(2+\sqrt{3}-1\right)=\sqrt{3}\left(1+\sqrt{3}\right)=3+\sqrt{3}\)
Vậy với x = \(2+\sqrt{3}\)thì M = \(3+\sqrt{3}\)
2, Mình chỉ giải câu a thôi nhé:
\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)
\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)
\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)
\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)
Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)
\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)
\(\Leftrightarrow2\left(b+c\right)\ge4a\)
\(\Leftrightarrow b+c\ge2a\)
4*. Thật ra cái này mình xài làm trội, làm giảm là được mà
Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)
Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)
\(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)
+ .........................................................
\(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Cộng tất cả vào
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)
\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)
\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)
\(A+1>2\sqrt{n+1}-3+1\)
\(A+1>2\sqrt{n+1}-2\)
\(A+1>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có điều phải chứng minh.
tạm thời chưa nghĩ ra cách dùng \(a^3+b^3\ge a^2b+ab^2=ab\left(a+b\right)\) :'<
Có: \(\sqrt[3]{4\left(a^3+b^3\right)}=\sqrt[3]{2\left(a+b\right)\left(2a^2-2ab+2b^2\right)}\)
\(=\sqrt[3]{2\left(a+b\right)\left[\frac{1}{2}\left(a+b\right)^2+\frac{3}{2}\left(a-b\right)^2\right]}=\sqrt[3]{2\left(a+b\right)\frac{1}{2}\left(a+b\right)^2}=a+b\)
Tương tự cộng lại ta có đpcm
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
ư ư.. ra r :))))))))) cộng thêm Cauchy-Schwarz nữa nhé
Có: \(a^3+b^3\ge a^2b+ab^2\)\(\Leftrightarrow\)\(2\left(a^3+b^3\right)\ge a^3+b^3+a^2b+ab^2=\left(a+b\right)\left(a^2+b^2\right)\)
\(\Rightarrow\)\(\sqrt[3]{4\left(a^3+b^3\right)}\ge\sqrt[3]{2\left(a+b\right)\left(a^2+b^2\right)}\ge\sqrt[3]{2\left(a+b\right).\frac{\left(a+b\right)^2}{2}}=a+b\)
Tương tự cộng lại ra đpcm
đề bài là rút gon biểu thức nhé
\(\sqrt{9\left(b-2\right)^2}=\sqrt{9}.\sqrt{\left(b-2\right)^2}=3.\left|b-2\right|\)
\(\sqrt{a^2\left(a+1\right)^2}=\sqrt{a^2}.\sqrt{\left(a+1\right)^2}=\left|a\right|.\left|a+1\right|\) Nhưng do a > 0
Nên: \(\left|a\right|.\left|a+1\right|=a.\left(a+1\right)=a^2+a\)
\(\sqrt{b^2\left(b-1\right)^2}=\sqrt{b^2}.\sqrt{\left(b-1\right)^2}=\left|b\right|.\left|\left(b-1\right)\right|\)
Em mới lớp 5 thôi sai đừng trách :v
Chúc anh học tốt !!!