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\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2+2+1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)+3}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}\)
\(=1+\dfrac{3}{\sqrt{x}-2}\)
Để A nguyên thì \(\dfrac{3}{\sqrt{x}-2}\) phải nguyên
Do đó 3⋮( \(\sqrt{x}\) -2)
⇒\(\sqrt{x}-2\) ∈ Ư(3)
Mà Ư(-1;1;-3;3)
Nên \(\left[{}\begin{matrix}\sqrt{x}-2=-1\\\sqrt{x}-2=1\\\sqrt{x}-2=-3\\\sqrt{x}-2=3\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=3\\\sqrt{x}=-1\left(v\text{ô}l\text{í}\right)\\\sqrt{x}=5\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=1\left(nh\text{ận}\right)\\x=9\left(nh\text{ận}\right)\\x=25\left(nh\text{ận}\right)\end{matrix}\right.\)
Vậy x=1 hay x=9 hay x=25 thì A nguyên
a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)
\(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)
\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa
a/ Cho x, y ≥ 1. Chứng minh: 1/(1 + x^2) + 1/(1 + y^2) ≥ 2/(1 + xy)
b/ Đề:...Tìm GTLN
Có:
\(\dfrac{1}{4x^2-4x+2}=\dfrac{1}{\left(2x-1\right)^2+1}\le\dfrac{1}{2}\forall x\ge1\)
\(\dfrac{1}{9y^2+6y+2}=\dfrac{1}{\left(3y+1\right)^2+1}\le\dfrac{1}{2}\forall y\ge0\)
\(\Rightarrow A=\dfrac{1}{4x^2-4x+2}+\dfrac{1}{9y^2+6y+2}\le\dfrac{1}{2}+\dfrac{1}{2}=1\)
Vậy MAXA = 1 khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
1: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
2: Để P là số nguyên thì \(2\sqrt{x}+2⋮2\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}=2\)
hay x=1(nhận)
3: \(P-\dfrac{1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}-\dfrac{1}{2}=\dfrac{2\sqrt{x}+2-\sqrt{x}}{2\sqrt{x}}=\dfrac{\sqrt{x}+2}{2\sqrt{x}}>0\)
=>P>1/2
a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+2\sqrt{x}-1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
b: |2x-5|=3
=>2x-5=3 hoặc 2x-5=-3
=>2x=2 hoặc 2x=8
=>x=1(loại) hoặc x=4(nhận)
Khi x=4 thì \(Q=\dfrac{4+5\cdot2}{\left(2+2\right)\left(2-1\right)}=\dfrac{14}{4}=3.5\)
c: Để Q=3 thì \(3x+3\sqrt{x}-6=x+5\sqrt{x}\)
=>\(2x-2\sqrt{x}-6=0\)
hay \(x=\left(\dfrac{1+\sqrt{13}}{2}\right)^2\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
chắc luôn bạn... vì mk làm đi làm lại rồi
\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)
\(\Leftrightarrow3< 1\) ( Vô lý )
\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)
\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)
\(\Leftrightarrow2b-2\sqrt{ab}< 0\)
\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)
Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)
\(\RightarrowĐpcm.\)
\(2a.\) Áp dụng BĐT Cauchy , ta có :
\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)
\(\Leftrightarrow x-4=a^2\)
\(\Leftrightarrow x=a^2+4\left(TM\right)\)
\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)
\(\Leftrightarrow x+4=x^2+4x+4\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
KL....
a)Do A chia hết cho 4 nên\(\dfrac{x^2}{x-1}\) \(\in\) Z
suy ra 1 chia hết cho x-1 suy x\(\in\) \(\left\{0;2\right\}\)
b)Do P thuộc Z nên 3 chia hết cho 2x+1
suy ra x\(\left\{-2;-1;0;1\right\}\)
tớ ko hiểu lắm câu b tsao P thuộc Z thì 3 chia hết cho 2x+1
