Cho x<0<2, tìm GTNN của A= \(\dfrac{9x}{2-x}+\dfrac{2}{x}\)
Cho x<0<2, tìm GTNN của A= \(\dfrac{9x}{2-x}+\dfrac{2}{x}\)
Lời giải:
Ta có:
\(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{-9(2-x)+18}{2-x}+\frac{2}{x}\)
\(=-9+\frac{18}{2-x}+\frac{2}{x}\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{18}{2-x}+\frac{2}{x}\right)(2-x+x)\geq (\sqrt{18}+\sqrt{2})^2\)
\(\Rightarrow \frac{18}{2-x}+\frac{2}{x}\geq\frac{(\sqrt{18}+\sqrt{2})^2}{2}=16\)
Do đó: \(A\geq -9+16=7=A_{\min}\)
Dấu "=" xảy ra khi \(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Leftrightarrow x=\frac{1}{2}\)
RÚT GỌN
\(A=\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2+\sqrt{2}}}\sqrt{2+\sqrt{2+\sqrt{2}}}\)
+) ta có \(A=\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(\dfrac{5+2\sqrt{6}+10-4\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(\dfrac{\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\dfrac{15^2-\left(2\sqrt{6}\right)^2}{5^2-\left(2\sqrt{6}\right)^2}=201\)
+) \(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2+\sqrt{2}}}\sqrt{2+\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}\sqrt{\left(2-\sqrt{2+\sqrt{2}}\right)\left(2+\sqrt{2+\sqrt{2}}\right)}\)
\(=\sqrt{4+\sqrt{8}}\sqrt{2^2-\left(\sqrt{2+\sqrt{2}}\right)^2}=\sqrt{2}\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2}\left(\sqrt{2^2-\left(\sqrt{2}\right)^2}\right)=\sqrt{2}.\sqrt{2}=2\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2+\sqrt{2}}}\sqrt{2+\sqrt{2+\sqrt{2}}}\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{\left(2-\sqrt{2+\sqrt{2}}\right)\left(2+\sqrt{2+\sqrt{2}}\right)}\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{4-2-\sqrt{2}}\)
\(B=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2}}\)
\(B=\sqrt{8-4\sqrt{2}+4\sqrt{2}-4}=\sqrt{4}=2\)
Tìm GTNN của
C=\(\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)
Lời giải:
Ta có:
\(C=\sqrt{(5x-2)^2}+\sqrt{(5x)^2}\)
\(=|5x-2|+|5x|=|2-5x|+|5x|\)
Áp dụng BĐT \(|a|+|b|\geq |a+b|\) ta có:
\(C=|2-5x|+|5x|\geq |2-5x+5x|=2\)
Vậy \(C_{\min}=2\). Dấu "=" xảy ra khi \(5x(2-5x)\geq 0\Leftrightarrow 0\leq x\leq \frac{2}{5}\)
rút gọn biểu thức sau
a. \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-4}\)
b. \(\dfrac{a^2\sqrt{b}-\sqrt{ab^3}}{\sqrt{a^3b^2}-b^2}\)
c. \(\dfrac{a^3-2\sqrt{2}}{a-\sqrt{2}}\)
d. \(18-\sqrt{8}+\dfrac{1}{4}\sqrt{2}\)
\(a.\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-4}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\sqrt{2}-4}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{-2\sqrt{2}\left(\sqrt{2}-1\right)}=-\dfrac{\sqrt{3}}{2}\)
\(b.\dfrac{a^2\sqrt{b}-\sqrt{ab^3}}{\sqrt{a^3b^2}-b^2}=\dfrac{a^2\sqrt{b}-b\sqrt{ab}}{ab\sqrt{a}-b^2}=\dfrac{\sqrt{ab}\left(a\sqrt{a}-b\right)}{b\left(a\sqrt{a}-b\right)}=\sqrt{\dfrac{a}{b}}\left(a;b>0\right)\)
\(c.\dfrac{a^3-2\sqrt{2}}{a-\sqrt{2}}=\dfrac{\left(a-\sqrt{2}\right)\left(a^2+a\sqrt{2}+2\right)}{a-\sqrt{2}}=a^2+a\sqrt{2}+2\left(a\ne\sqrt{2}\right)\)
\(d.\sqrt{18}-\sqrt{8}+\dfrac{1}{4}\sqrt{2}=3\sqrt{2}-2\sqrt{2}+\dfrac{1}{4}\sqrt{2}=\left(\dfrac{1}{4}+1\right)\sqrt{2}=\dfrac{5}{4}\sqrt{2}\)
Bài 1: Tìm các giá trị nguyên của x để các biểu thức sau có giá trị nguyên
a/C=\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\) ; b/D=\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}\)
Bài 2: Chứng minh
a/\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}=\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\) b/\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
bài 2 : chữa đề câu a chút nha
a) ta có : \(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(\sqrt{\left(\dfrac{2}{\sqrt{5}-2}\right)^2}-\sqrt{\left(\dfrac{2}{\sqrt{5}+2}\right)^2}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+4-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\dfrac{8}{5-4}=8\left(đpcm\right)\)
b) ta có : \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\) \(=2\left(9-5\right)=2.4=8\left(đpcm\right)\)
Bài 1 : Mình gợi ý thôi nhé :v
\(C=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}=1+\dfrac{5}{\sqrt{x}-2}\)
\(D=\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{2\left(\sqrt{x}+3\right)-7}{\sqrt{x}+3}=2-\dfrac{7}{\sqrt{x}+3}\)
Thầy đã sửa
bài 1
a/
\(\dfrac{x+2}{x-5}=\dfrac{x-5+7}{x-5}=1+\dfrac{7}{x-5}\)
A là số nguyên khi x-5∈ Ư(7)
⇔x∈{-2;4;6;12}
b/\(\dfrac{3x+1}{2-x}=\dfrac{3x-6+7}{2-x}=\dfrac{-3\left(2-x\right)}{2-x}+\dfrac{7}{2-x}=-3+\dfrac{7}{2-x}\)
B là số nguyên khi 2-xϵ Ư(7)
⇔xϵ {3;1;9;-5}
1. Áp dụng quy tắc khai phương một thương, hãy tính:
a, \(\sqrt{\dfrac{36}{121}}\) b, \(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}\) c, \(\sqrt{0,0169}\)
d,\(\dfrac{\sqrt{15}}{\sqrt{735}}\) e, \(\sqrt{\dfrac{81}{8}:\sqrt{3\dfrac{1}{8}}}\) g, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
2. Tính:
a,\(\sqrt{\dfrac{25}{144}}\) b,\(\sqrt{2\dfrac{7}{81}}\) c,\(\sqrt{\dfrac{2,25}{16}}\) d, \(\sqrt{\dfrac{1,21}{0,49}}\)
3. Áp dụng quy tắc chia hai căn bậc hai, hãy tính:
a, \(\sqrt{18}:\sqrt{2}\) b, \(\sqrt{45}:\sqrt{80}\)
c, (\(\sqrt{20}-\sqrt{45}+\sqrt{5}\) ) : \(\sqrt{5}\) d, \(\dfrac{\sqrt{8^2}}{\sqrt{4^5.2^3}}\)
4. Khẳng định nào sau đây là đúng?
A. \(\sqrt{\dfrac{3}{\left(-5\right)^2}}=-\dfrac{\sqrt{3}}{5}\) B. \(\left(\sqrt{\dfrac{-3}{-5}}\right)^2=\dfrac{3}{5}\)
5. Tính.
a, \(\sqrt{2\dfrac{7}{81}}:\dfrac{\sqrt{6}}{\sqrt{150}}\) b, \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
c, \(\left(\sqrt{\dfrac{1}{5}-\sqrt{\dfrac{9}{5}}+\sqrt{5}}\right):\sqrt{5}\) d, \(\sqrt{\dfrac{2+\sqrt{3}}{\sqrt{2}}}\)
6. So sánh
a, So sánh \(\sqrt{144-49}\) và \(\sqrt{144}-\sqrt{49}\);
b, Chứng minh rằng , với hai số a,b thỏa mãn a> b> 0 thì \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
Bài 2:
a: \(\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)
b: \(\sqrt{2+\dfrac{7}{81}}=\sqrt{\dfrac{169}{81}}=\dfrac{13}{9}\)
c: \(\sqrt{\dfrac{2.25}{16}}=\dfrac{1.5}{4}=\dfrac{3}{8}\)
d: \(\sqrt{\dfrac{1.21}{0.49}}=\sqrt{\dfrac{121}{49}}=\dfrac{11}{7}\)
Bài3:
a: \(=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)
b: \(=\sqrt{\dfrac{45}{80}}=\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}\)
c: \(=\dfrac{2\sqrt{5}-3\sqrt{5}+\sqrt{5}}{\sqrt{5}}=0\)
d: \(=\sqrt{\dfrac{2^6}{2^{10}\cdot2^3}}=\sqrt{\dfrac{1}{2^7}}=\dfrac{1}{8\sqrt{2}}=\dfrac{\sqrt{2}}{16}\)
tính : \(\dfrac{\sqrt{48}+\sqrt{192}-\sqrt{175}}{\sqrt{3}}\)
\(\dfrac{\sqrt{48}+\sqrt{192}-\sqrt{175}}{\sqrt{3}}=\dfrac{\left(\sqrt{48}+\sqrt{192}-\sqrt{175}\right).\sqrt{3}}{\sqrt{3}.\sqrt{3}}=\dfrac{12+24-5\sqrt{21}}{3}=\dfrac{36-5\sqrt{21}}{3}\)
Cách khác :
\(\dfrac{\sqrt{48}+\sqrt{192}-\sqrt{175}}{\sqrt{3}}=\dfrac{4\sqrt{3}+8\sqrt{3}-5\sqrt{7}}{\sqrt{3}}=\dfrac{12\sqrt{3}-5\sqrt{7}}{\sqrt{3}}=12-\dfrac{5\sqrt{7}}{\sqrt{3}}\)
tính : \(\dfrac{\sqrt{2}}{98}\) nhớ làm đầy đủ nha các bạn
\(\dfrac{\sqrt{2}}{98}=\dfrac{\sqrt{2}}{2.49}=\dfrac{\sqrt{2}}{\sqrt{2}.\sqrt{2}.49}=\dfrac{1}{49\sqrt{2}}\)
1/Cho a,b>0.CMR:a+b/2 ≥ √ab
2/So sánh:
√2017+√2019 và 2*√2018
Câu 1:
Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2>=0\)
\(\Leftrightarrow a+b-2\sqrt{ab}>=0\)
=>\(a+b>=2\sqrt{ab}\)
hay \(\dfrac{a+b}{2}>=\sqrt{ab}\)
Cho biểu thức:
\(P=\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\)
a ) Tìm x để P có nghĩa .
b) Rút gọn P
Câu a : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Câu b : \(P=\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\)
\(=x+\sqrt{x}+1+\sqrt{x}\)
\(=x+2\sqrt{x}+1\)
\(=\left(\sqrt{x}+1\right)^2\)
Chúc bạn học tốt !!