Tìm \(x\varepsilon Z\)sao cho \(\sqrt{x^2+x+3}\)có giá trị nguyên
Tìm \(x\varepsilon Z\)sao cho \(\sqrt{x^2+x+3}\) có giá trị nguyên
Để \(\sqrt{x^2+x+3}\) nguyên thì
\(\Rightarrow x^2+x+3=a^2\left(a\in Z\right)\)
\(\Leftrightarrow4x^2+4x+12=4a^2\)
\(\Leftrightarrow4a^2-\left(2x+1\right)^2=11\)
\(\Leftrightarrow\left(2a+2x+1\right)\left(2a-2x-1\right)=11\)
\(\Leftrightarrow\left(2a+2x+1,2a-2x-1\right)=\left(1,11;11,1;-1,-11;-11,-1\right)\)
\(\Leftrightarrow\left(a,x\right)=\left(3,-3;3,2;-3,-3;-3,2\right)\)
Vậy ....
1. Cho A = \(\frac{\sqrt{x}-3}{2}\). Tìm \(x\varepsilon Z\)và \(x< 30\)để A có giá trị nguyên.
2. Cho B = \(\frac{5}{\sqrt{x}-1}\). Tìm \(x\varepsilon Z\)để B có giá trị nguyên.
3. Cho \(x>y>0\). Chứng minh rằng: \(x^3>y^3\)
4. Tìm \(x,y\)biết \(\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}\)và \(x^4y^4=81\)
Bài 1:
a.\(\sqrt{x^2-2x+2}+\sqrt{3x^2-6x+4}=2\)
b. \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=-x^2-2x+5\)
Bài 2: Tìm x \(\varepsilon\)Z sao cho :
\(\sqrt{x^2+x+3}\) có giá trị nguyên
. Dạ mấy ac giúp dùm e bài này nha ^^ E cảm ơn nhiều lắm ạ
a)\(pt\Leftrightarrow\sqrt{x^2-2x+2}+\sqrt{3x^2-6x+4}-2=0\)
\(\Leftrightarrow\sqrt{x^2-2x+2}-1+\sqrt{3x^2-6x+4}-1=0\)
\(\Leftrightarrow\frac{x^2-2x+2-1}{\sqrt{x^2-2x+2}+1}+\frac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}=0\)
\(\Leftrightarrow\frac{x^2-2x+1}{\sqrt{x^2-2x+2}+1}+\frac{3x^2-6x+3}{\sqrt{3x^2-6x+4}+1}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{x^2-2x+2}+1}+\frac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(\frac{1}{\sqrt{x^2-2x+2}+1}+\frac{3}{\sqrt{3x^2-6x+4}+1}\right)=0\)
Dễ thấy: \(\frac{1}{\sqrt{x^2-2x+2}+1}+\frac{3}{\sqrt{3x^2-6x+4}+1}>0\) (loại)
Nên x-1=0 suy ra x=1
b)\(pt\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}+x^2+2x-5=0\)
\(\Leftrightarrow\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+21}-4+x^2+2x+1=0\)
\(\Leftrightarrow\frac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\frac{5x^2+10x+21-16}{\sqrt{5x^2+10x+21}+4}+\left(x+1\right)^2=0\)
\(\Leftrightarrow\frac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\frac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+21}+4}+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(\frac{3}{\sqrt{3x^2+6x+7}+2}+\frac{5}{\sqrt{5x^2+10x+21}+4}+1\right)=0\)
Dễ thấY: \(\frac{3}{\sqrt{3x^2+6x+7}+2}+\frac{5}{\sqrt{5x^2+10x+21}+4}+1>0\) (loại luôn)
Nên x+1=0 suy ra x=-1
BT: Cho:
\(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a) Rút gọn biểu thức A
b) Tìm x \(\varepsilon\)Z để A có giá trị nguyên
c) Tìm x để \(A+\frac{1}{A}+2=0\)
Cho \(B=\dfrac{2}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
Tìm \(x\in Z\) để B có giá trị nguyên
\(B=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
Để B nguyên thì \(\sqrt{x}-3\in\left\{1;-1;5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;8\right\}\)
hay \(x\in\left\{16;4;64\right\}\)
Cho \(A=\frac{x+1}{x-3}\)
1 ) Tìm x\(\varepsilon\)N để A nhỏ nhất . Tính giá trị nhỏ nhất đó
2) Tìm x\(\varepsilon\)Z để A đạt giá trị nguyên
Bài 1:
Cho E = \(\frac{1}{x+\sqrt{x}}\)
Tìm x thuộc Z để E có giá trị nguyên.
Bài 2:
Cho F = \(\frac{3}{x+\sqrt{x}+1}\)
Tìm x thuộc Z để F có giá trị nguyên.
Tìm x thuộc Z để biểu thức có giá trị nguyên: A=\(\dfrac{2\sqrt{x}+3}{3\sqrt{x}-1}\)
Để A nguyên thì \(2\sqrt{x}+3⋮3\sqrt{x}-1\)
\(\Leftrightarrow6\sqrt{x}+9⋮3\sqrt{x}-1\)
\(\Leftrightarrow3\sqrt{x}-1\in\left\{-1;1;11\right\}\)
\(\Leftrightarrow3\sqrt{x}\in\left\{0;12\right\}\)
hay \(x\in\left\{0;16\right\}\)
Cho biểu thức: Q = \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\).
a) Tìm điều kiện xác định và rút gọn Q.
b) Tìm các giá trị của x để Q < -1.
c) Tìm các giá trị của x \(\in\) Z sao cho 2Q \(\in\) Z.
a, đk: \(x\ge0,x\ne9,x\ne4\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2-\sqrt{x}}{-\left(\sqrt{x}-3\right)\left(2-\sqrt{x}\right)}=\dfrac{-1}{\sqrt{x}-3}\)
b,\(Q< -1=>\dfrac{-1}{\sqrt{x}-3}+1< 0< =>\dfrac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(< =>\dfrac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)
\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left[{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\end{matrix}\right.\)\(< =>9< x< 16\)
c, \(=>2Q=\dfrac{-2}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\in Z\)
\(< =>\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)\(=>x\in\left\{16;4\right\}\)(loại 4)
=>x=16
a) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\)
Ta có \(x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>9\\x>2\end{matrix}\right.\) \(\Leftrightarrow x>9\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-3\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(x-4\right)-\left(x-2\sqrt{x}-3\right)-\left(3\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-1}{\left(\sqrt{x}-3\right)}=\dfrac{1}{3-\sqrt{x}}\)
b) \(Q< -1\Leftrightarrow\dfrac{1}{3-\sqrt{x}}< -1\) \(\Leftrightarrow\dfrac{1}{3-\sqrt{x}}+1< 0\) \(\Leftrightarrow\dfrac{4-\sqrt{x}}{3-\sqrt{x}}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-\sqrt{x}>0\\3-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4-\sqrt{x}< 0\\3-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow9< x< 16\)
Vậy để \(Q< -1\) thì \(S=\left\{x/9< x< 16\right\}\)
c) \(2Q\in Z\Leftrightarrow\dfrac{2}{3-\sqrt{x}}\in Z\)
\(\Rightarrow3-\sqrt{x}\inƯ\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}3-\sqrt{x}=2\\3-\sqrt{x}=-2\\3-\sqrt{x}=1\\3-\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=25\\x=4\\x=16\end{matrix}\right.\)
Kết hợp với ĐKXĐ,ta có để \(2Q\in Z\) thì \(x\in\left\{16;25\right\}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-3}{x-5\sqrt{x}+6}\)
\(=\dfrac{x-4-x+2\sqrt{x}+2-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-1}{\sqrt{x}-3}\)
c) Để 2Q là số nguyên thì \(-2⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1\right\}\)