Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{9\left(b-2\right)^2}=\sqrt{9}.\sqrt{\left(b-2\right)^2}=3.\left|b-2\right|\)
\(\sqrt{a^2\left(a+1\right)^2}=\sqrt{a^2}.\sqrt{\left(a+1\right)^2}=\left|a\right|.\left|a+1\right|\) 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 !!!
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=\(\frac{1}{4}\)
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
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) \(\sqrt{\frac{3a}{4}}.\sqrt{\frac{4a}{27}}=\frac{\sqrt{3a}}{2}.\frac{\sqrt{4a}}{3\sqrt{3}}=\frac{\sqrt{3}.\sqrt{a}.2.\sqrt{a}}{6\sqrt{3}}=\frac{a.2\sqrt{3}}{6\sqrt{3}}=\frac{a}{3}\)
b) \(\sqrt{15x}.\sqrt{\frac{60}{x}}=\sqrt{15x}.\frac{2\sqrt{15}}{\sqrt{x}}=\frac{30\sqrt{x}}{\sqrt{x}}=30\)
a) \(\sqrt{\frac{3a}{4}}.\sqrt{\frac{4a}{27}}=\sqrt{\frac{3a}{4}.\frac{4a}{27}}=\sqrt{\frac{1}{9}.a^2}=\sqrt{\frac{1}{9}}.\sqrt{a^2}=\frac{1}{3}.a\)( Vì \(a\ge0\)nên \(\sqrt{a^2}=\left|a\right|=a\))
b) \(\sqrt{15x}.\sqrt{\frac{60}{x}}=\sqrt{15x.\frac{60}{x}}=\sqrt{900}=30\)
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)