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30 tháng 8
Bài 3:
a: \(A=\frac{1}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{2\sqrt{x}}{4-x}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-2-\sqrt{x}-2-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=-\frac{2}{\sqrt{x}-2}\)
b: Thay x=3 vào A, ta được: \(A=-\frac{2}{\sqrt3-2}=\frac{2}{2-\sqrt3}=2\left(2+\sqrt3\right)=4+2\sqrt3\)
Bài 2:
a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)
\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)
b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
Bài 1:
a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)
\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)
30 tháng 8
Bài 2:
a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)
\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)
b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
Bài 1:
a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)
\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)
29 tháng 7 2020
Bài 2:
a, Ta có
\(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}\)
= \(3\left|-2\right|+\left|-5\right|\)
=\(6+5\)
= 11
Vậy \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}=11\)
29 tháng 7 2020
b, Ta có
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}\)
= \(\left|\sqrt{5}+1\right|-\sqrt{5}\)
= \(\sqrt{5}+1-\sqrt{5}=1\)
Vậy \(\sqrt{6+2\sqrt{5}}-\sqrt{5}=1\)
TB
23 tháng 7 2018
Mấy bài này dài vật vã ghê =)))))))))))))
1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)
=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b, M = \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))
= \(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)
= \(\sqrt{3}\left(x-1\right)\)
Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:
M = \(\sqrt{3}\left(2+\sqrt{3}-1\right)=\sqrt{3}\left(1+\sqrt{3}\right)=3+\sqrt{3}\)
Vậy với x = \(2+\sqrt{3}\)thì M = \(3+\sqrt{3}\)
2, Mình chỉ giải câu a thôi nhé:
\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)
\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)
\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)
\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)
Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)
\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)
\(\Leftrightarrow2\left(b+c\right)\ge4a\)
\(\Leftrightarrow b+c\ge2a\)
4*. Thật ra cái này mình xài làm trội, làm giảm là được mà
Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)
Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)
\(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)
+ .........................................................
\(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Cộng tất cả vào
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)
\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)
\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)
\(A+1>2\sqrt{n+1}-3+1\)
\(A+1>2\sqrt{n+1}-2\)
\(A+1>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có điều phải chứng minh.
22 tháng 5 2022
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
Chọn đáp án C.