CMR:Với a,b dương thì \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
Cho \(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
CMR:Với mọi a,b>1 thì \(a\sqrt{b-1}+b\sqrt{a-1\le ab}\)
\(D=\sqrt{2+1-2\sqrt{2}}-\sqrt{2+1+2\sqrt{2}}\)
\(D=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(D=\sqrt{2}-1-\left(\sqrt{2}+1\right)\)
\(D=\sqrt{2}-1-\sqrt{2}-1\)
\(D=-2\)
CÂU THỨ 2 NHA !!!!!!
XÉT: \(2VT=2a\sqrt{b-1}+2b\sqrt{a-1}\)
=> \(2VT=a.2.\sqrt{1}.\sqrt{b-1}+b.2.\sqrt{1}.\sqrt{a-1}\)
TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC:
=> \(2VT\le a\left(1+b-1\right)+b\left(1+a-1\right)\)
=> \(2VT\le ab+ab\)
=> \(2VT\le2ab\)
=> \(VT\le ab\)
=> TA CÓ ĐIỀU PHẢI CHỨNG MINH.
\(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{2+2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|-\left|\sqrt{2}+1\right|=\sqrt{2}-1-1-\sqrt{2}=-2\)
CMR:Với mọi a,b>0,a khác b,ta có:
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Bài làm:
Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
=> đpcm
CMR:Với mọi a,b>0,a khác b,ta có:
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Ta có: \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-\left(a-2\sqrt{ab}+b\right)+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4b+4\sqrt{ab}}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(đpcm)
Cho \(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
CMR:Với mọi a,b>1 thì \(a\sqrt{b-1}+b\sqrt{a-1\le ab}\)
\(D=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{2}-1-\sqrt{2}-1=-2\)
___
Ta có: \(\left(\sqrt{a-1}-1\right)^2\ge0\forall a\ge1\)
\(\Leftrightarrow a-2\sqrt{a-1}\ge0\)
\(\Leftrightarrow\frac{\sqrt{a-1}}{a}\le\frac{1}{2}\)
Tương tự: \(\frac{\sqrt{b-1}}{b}\le\frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-1}}{b}\le1\)
\(\Leftrightarrow b\sqrt{a-1}+a\sqrt{b-1}\le ab\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)
CMR nếu a,b dương thì\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
Nhận thấy \(a+b< a+b+2\sqrt{ab}\)<=>\(a+b< \left(\sqrt{a}+\sqrt{b}\right)^2\)
Do a,b đều dương, lấy căn 2 vế ta được:
\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)(đpcm)
Chúc bạn học tốt!
CMR: với a,b dương thì\(\sqrt{a+b<}\sqrt{a}+\sqrt{b}\)
\(0<2\sqrt{ab}\) cộng 2 vế với a+ b
a+b< a+b+ 2.căn(ab)
\(a+b<\left(\sqrt{a}+\sqrt{b}\right)^2\)
lấy căn 2 vế là xong
Cho a,b,c là các số dương. CMR nếu b là trung bình cộng của a và c thì \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{c}+\sqrt{a}}\)
Chứng minh rằng nếu a, b, c là số dương thỏa mãn a+c=2b thì ta luôc có: \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\)
Chứng minh nếu \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\) thì với mọi số nguyên dương lẻ n có \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)