Biết I = \(\int_2^1\dfrac{dx}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}\)=\(\sqrt{a}-\sqrt{b}-c\) với a,b,c là các số nguyên. Tính P = a+b+c
A. P = 24
B. P = 12
C. P = 18
D. P = 46
Cho \(\int\limits^4_1\sqrt{\dfrac{1}{4x}+\dfrac{\sqrt{x}+e^x}{\sqrt{x}\cdot e^{2x}}}dx=a+e^b-e^c\) với a, b, c là các số nguyên. Tính a + b + c
Note: \(\sqrt{\dfrac{1}{4x}+\dfrac{\sqrt{x}+e^x}{\sqrt{x}.e^{2x}}}=\sqrt{\dfrac{1}{4x}+\dfrac{1}{e^x.\sqrt{x}}+\dfrac{1}{e^{2x}}}=\sqrt{\left(\dfrac{1}{2\sqrt{x}}+\dfrac{1}{e^x}\right)^2}=\dfrac{1}{2\sqrt{x}}+\dfrac{1}{e^x}\)
Vấn đề bây giờ có lẽ đã quá đơn giản
Tìm GTLN của:
a. \(A=x+\sqrt{2-x}\)
b.\(A=x\sqrt{1-x^2}\)
c. \(C=\left|x-y\right|\) với \(x+4y^2=1\)
d. \(D=a^2+b^2+c^2\) với \(-1\le a,b,c\le3,a+b+c=1\)
e. \(E=\left(\sqrt{a}+\sqrt{b}\right)^2\) với \(a>0,b>0,a+b\le1\)
f. \(F=\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}+\sqrt{c}\right)^4+\left(\sqrt{a}+\sqrt{d}\right)^4+\left(\sqrt{b}+\sqrt{c}\right)^4+\left(\sqrt{b}+\sqrt{d}\right)^4+\left(\sqrt{c}+\sqrt{d}\right)^4\)
Với a,b,c,d là các số dương và \(a+b+c+d\le1\)
g. \(G=\dfrac{1}{a^3+b^3+1}+\dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}\) Với a,b,c là các số dương và abc=1
h. \(H=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Với a,b,c là các số dương thỏa mãn \(1\le a\le b\le c\le2\)
i. \(I=x^2\sqrt{9-x^2}\)
1.a,b,c là các số thực dương. CM \(\left(\dfrac{\sqrt{ab}}{\sqrt{a+b}}+\dfrac{\sqrt{bc}}{\sqrt{b+c}}\right)\left(\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}\right)\le2\)
2. x,y là các số nguyên sao cho \(x^2-2xy-y^2\) ;\(xy-2y^2-x\) đều chia hết cho 5Chứng minh \(2x^2+y^2+2x+y\) cũng chia hết cho 5
3. cho \(a_1a_2...a_{50}\) là các số nguyên thoả mãn \(1\le a_1\le a_2...\le a_{50}\le50;a_1+a_2+...+a_{50}=100\) chứng minh rằng từ các số đã cho có thể chọn đc một vài số có tổng là 50
Cho a, b, x là những số dương. Đơn giản các biểu thức sau :
a) \(A=\left[\dfrac{2a+\left(ab\right)^{\dfrac{1}{2}}}{3a}\right]^{-1}\left[\dfrac{a^{\dfrac{3}{2}}-b^{\dfrac{3}{2}}}{a-\left(ab\right)^{\dfrac{1}{2}}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\right]\)
b) \(B=\left(\dfrac{\sqrt{a}-\sqrt{x}}{\sqrt{a+x}}-\dfrac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\right)^{-2}-\left(\dfrac{\sqrt{a}-\sqrt{x}}{\sqrt{a+x}}-\dfrac{\sqrt{a+x}}{\sqrt{a}-\sqrt{x}}\right)^{-2}\)
c) \(C=\sqrt{16^{\dfrac{1}{\log_74}}+81^{\dfrac{1}{\log_69}}+15}\)
d) \(D=49^{1-\log_72}+5^{-\log_54}\)
Bài 1: A= \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
a) RÚt gọn A
b) tính A khi \(a^2\) -3 =0
Bài 2:B= \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a) Rút gọn B
b) C/m rằng: B>0 với mọi x>0 , x khác 1
Bài 3:C = \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{2}{a-1}\right)\)
Rút gọn C
Bài 3:
\(C=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1+2}{a-1}\)
\(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{a-1}{\sqrt{a}+3}\)
\(=\dfrac{\left(a-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
Câu 1: Cho bt: A= \(\left(\dfrac{1}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{1}{\sqrt{1-x^2}}+1\right)\)
a) Tìm x để A có nghĩa
b) Rút gọn
c) Tính A với x =\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
Câu 2: Cho bt B= \(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\)
a) Rút gọn
b) CM B\(\ge\)0
c) So sánh B với \(\sqrt{B}\)
Cho biểu thức
1) A=\(\dfrac{\sqrt{x}+2}{2\sqrt{x}-4}+\dfrac{\sqrt{x}-2}{2\sqrt{x}+4}\)
a. Rút gọn A
b. Tìm x để A=8
2) C= \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
a. Rút gọn C
b.Tìm x để C= -8
3) D=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\dfrac{\sqrt{4x}}{x-4}\)
a.Rút gọn D
b.Tìm x để D>3
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
Bài 1. Tìm x, y, z biết: \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\) (trong đó, a + b + c = 3)
Bài 2.
a) Chứng minh rằng: \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b/ Cho S = \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\). Chứng minh rằng: 18<S<19
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
Bài 1: Rút gọn
a) \(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\) với x>0 x≠4
b)\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}-1}\right).\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
Bài 2: Cho P=\(\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\) với a>0, a≠1, a≠2
a)Rút gọn P
b)Tìm a ∈ Z để P có giá trị nguyên
Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6