GTNN bt
\(\sqrt{x+8}+\sqrt{x-1}\)
Giải nhanh Nha mk k nha:
Tìm GTNN của bt:
A= \(\sqrt{x+8}+\sqrt{x-1}\)
Tìm GTLN,GTNN của bt sau: \(A=6\sqrt{x-2}+8\sqrt{5-x}\)
+) \(B=6\sqrt{x-2}+6\sqrt{5-x}\Leftrightarrow B^2=\left(6\sqrt{x-2}+6\sqrt{5-x}\right)^2\)
\(=36\left(x-2\right)+36\left(5-x\right)+72\sqrt{\left(x-2\right)\left(5-x\right)}\ge108\Rightarrow B\ge6\sqrt{3}\)
+) \(A=B+2\sqrt{5-x}\ge6\sqrt{3}\)
Vậy \(A_{min}=6\sqrt{3}\)khi x=5
+) Đặt \(a=\sqrt{x-2};b=\sqrt{5-x}\)
+) Ta có: \(a^2+b^2=3\)
+) \(\left(a^2+b^2\right)\left(6^2+8^2\right)\ge\left(6a+8b\right)^2\Leftrightarrow\left(6a+8b\right)^2\le300\Rightarrow6a+8b\le10\sqrt{3}\)
Dấu = xảy ra khi \(\frac{a}{6}=\frac{b}{8}\Leftrightarrow\frac{\sqrt{x-2}}{6}=\frac{\sqrt{5-x}}{8}\Leftrightarrow\frac{x-2}{36}=\frac{5-x}{64}\Leftrightarrow64x-128=180-36x\Leftrightarrow308=100x\)
\(\Leftrightarrow x=3.08\)
Vậy \(A_{max}=10\sqrt{3}\)khi x=3.08
Cho bt: \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
a. Tìm đkxđ, rút gon bt
b. Tìm Gtnn của P
c. Tìm x để bt \(Q=\frac{2\sqrt{x}}{P}\)
\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?
Tìm GTNN của bt: A= \(\sqrt{-x^2+2x+8}-\sqrt{-x^2+x+2}\)
giúp mình vs ^^
Cho bt : \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)
Khi x thỏa mãn dkxd. Hãy tìm GTNN của bt B, với B = A(x - 1)
\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(B=A\left(x-1\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\right)\left(x-1\right)\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2\)
\(=x+\sqrt{x}-2\sqrt{x}+2-2\)
\(=x-\sqrt{x}\)
\(=x-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
\(\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy \(Min_B=-\frac{1}{4}\) khi \(x=\frac{1}{4}\)
Tìm gtnn của bt:
\(A=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}+2019\)
\(DK:x\ge1\)
\(A=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}+2019\)
\(=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|+2019\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|+2019\ge|\sqrt{x-1}+1+1-\sqrt{x-1}|+2019=2021\)
Dau '=' xay ra khi \(\left(\sqrt{x-1}+1\right)\left(1-\sqrt{x-1}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x-1}+1\ge0\\1-\sqrt{x-1}\ge0\end{cases}\Leftrightarrow x=2\left(n\right)}\)
TH2:
\(\hept{\begin{cases}\sqrt{x-1}+1\le0\\1-\sqrt{x-1}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}\le-1\\\sqrt{x-1}\ge1\end{cases}\left(l\right)}}\)
Vay \(A_{min}=2021\)khi \(x=2\)
tìm GTNN; GTLN của bt:
1, A=\(\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
2, B=\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
Tìm đc mỗi GTNN, cách tìm GTLN chưa chắc chắn lắm nên mk ko lm nha :D
1/ \(A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(3-x\right)^2}=\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)
2/ \(B=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\sqrt{\left(1-\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)
Tìm GTNN của bt P=\(\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\left(x>0,x\ne1\right)\)
Cho bt:
\(A=\frac{3\sqrt{x}-2}{\sqrt{x}+1}-\frac{14\sqrt{x}+4}{x+4\sqrt{x+3}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\) với \(x\ge0\)
a. Rút gọn bt A
b. Tìm GTNn của A
đề bài có sai chỗ nào k bn???