\(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left[a+2\sqrt{ab}+b\right]^4\)
Áp dụng bđt cô - si, ta có:
\(\left[a+2\sqrt{ab}+b\right]^4\ge\left[2\sqrt{2\left(a+b\right)\sqrt{ab}}\right]^4=2^4.2^2.ab.\left(a+b\right)^2\)
\(=64ab\left(a+b\right)^2\)
Đẳng thức xảy ra khi a = b.
Trl
-Bạn kia làm đúng rồi nhé ~!
Chúc bạn học tốt
#Mưaa
chứng minh với a,b\(\ge0\)
thì: \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\)
CM \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\) Với \(a,b\ge0\)
CM \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\) Với \(a,b\ge0\)
Ta có \(\sqrt{a+b}+\sqrt{a-b}\le2\sqrt{a}\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{a-b}\right)^2\le\left(2\sqrt{a}\right)^2\)\(\Leftrightarrow a+b+a-b+2.\sqrt{\left(a+b\right)\left(a-b\right)}\le4a\)
\(\Leftrightarrow2a+2\sqrt{\left(a+b\right)\left(a-b\right)}\le4a\)
\(\Leftrightarrow-2a+2.\sqrt{\left(a+b\right)\left(a-b\right)}\le0\)\(\Leftrightarrow-\left(2a-2.\sqrt{\left(a+b\right).\left(a-b\right)}\right)\le0\)
\(\Leftrightarrow a+b+a-b-2.\sqrt{\left(a+b\right)\left(a-b\right)}\ge0\)
\(\Leftrightarrow\left(\sqrt{a+b}-\sqrt{a-b}\right)^2\ge0\)( luôn đúng nên suy ra điều phải chứng minh )
Rút gọn biểu thức
a) \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\left(\sqrt{a+\sqrt{b}}\right)^2-4\sqrt{ab}}.\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) \(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)\(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
HELP ME PLSSSSSSSSSS
Chứng minh giúp mình mấy câu bất đẳng thức này nha
a) \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\left(a,b>0\right)\)
b) \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\left(a,b>0\right)\)
c) \(y\left(\frac{1}{x}+\frac{1}{x}\right)+\frac{1}{y}\left(x+z\right)\le\left(\frac{1}{x}+\frac{1}{z}\right)\left(x+z\right)\left(0< x\le y\le z\right)\)
d) \(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a,b,c>0;a+b+c=abc\right)\)
Cho \(a\ge0\), \(b\ge0\). CMR: \(\frac{1}{2}\left(a+b\right)^2+\frac{1}{4}\left(a+b\right)\ge a\sqrt{b}+b\sqrt{a}\)
giải pt sau bằng các định lý : \(f\left(x\right)=g\left(x\right)\Leftrightarrow\left[f\left(x\right)\right]^{2k+1}=\left[g\left(x\right)\right]^{2k+1}\)
\(\sqrt[2k+1]{f\left(x\right)}=g\left(x\right)\Leftrightarrow f\left(x\right)=\left[g\left(x\right)\right]^{2k+1}\)
\(\sqrt[2k+1]{f\left(x\right)}=\sqrt[2k+1]{g\left(x\right)}\Leftrightarrow f\left(x\right)=g\left(x\right)\)
\(\sqrt[2k]{f\left(x\right)}=g\left(x\right)\Leftrightarrow\orbr{\begin{cases}g\left(x\right)>0\\f\left(x\right)=\left[g\left(x\right)\right]^{2k}\end{cases}}\)
\(\sqrt[2k]{f\left(x\right)}=\sqrt[2k]{g\left(x\right)}\Leftrightarrow\hept{\begin{cases}f\left(x\right)\ge0\\g\left(x\right)\ge0\\f\left(x\right)=g\left(x\right)\end{cases}}\)hoặc
a) \(\sqrt{x+1}+\sqrt{4x+13}=\sqrt{3x+12}\)
b)\(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)
c) \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
Cho a, b, c là các số dương thỏa mãn điều kiện a+b+c+\(\sqrt{2abc}=2\)
CMR \(\sqrt{a\left(2-b\right)\left(2-c\right)}+\sqrt{b\left(2-c\right)\left(2-a\right)}+\sqrt{c\left(2-a\right)\left(2-b\right)}=\sqrt{8}+\sqrt{abc}\)
giúp mik vs nhé cảm ơn rất nhìu
