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Vì \(4x^3+14x^2+9x-6\) là số chính phương nên ta có \(4x^3+14x^2+9x-6=k^2\) với \(k\inℕ\)
Ta có \(4x^3+14x^2+9x-6=\left(x+2\right)\left(4x^2+6x-3\right)\)nên ta có \(\left(x+2\right)\left(4x^2+6x-3\right)=k^2\)
Đặt \(\left(x+2;4x^2+6x-3\right)=d\)với \(d\inℕ^∗\)
Ta có \(x+2⋮d\Rightarrow\left(x+2\right)\left(4x-2\right)⋮d\Rightarrow4x^2+6x-4⋮d\)
Ta lại có \(4x^2+6x-3⋮d\Rightarrow\left(4x^2+6x-3\right)-\left(4x^2+6x-4\right)=1⋮d\)
\(\Rightarrow d=1\)(Vì \(d\inℕ^∗\))
Vậy \(\left(x+2;4x^2+6x-3\right)=1\)
mà \(\left(x+2\right)\left(4x^2+6x-3\right)=k^2\)nên ta có:
x + 2 và 4x2 + 6x - 3 là số chính phương\(\Rightarrow\hept{\begin{cases}x+2=a^2\\4x^2+6x-3=b^2\end{cases}}\left(a,b\right)\inℕ^∗\)
Vì x > 0 nên ta có \(4x^2< b^2< 4x^2+12x+9\Leftrightarrow\left(2x\right)^2< b^2< \left(2x+3\right)^2\)
Vì b lẻ nên \(b^2=\left(2x+1\right)^2\Leftrightarrow4x^2+6x-3=4x^2+4x+1\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
Vậy x = 2 thì \(4x^3+14x^2+9x-6\)là số chính phương
\(A=k^4-8k^3+23k^2-26k+10\)
\(=k^2\left(k^2-2k+1\right)-6k\left(k^2-2k+1\right)+10\left(k^2-2k+1\right)\)
\(=\left(k^2-6k+10\right)\left(k-1\right)^2\)
+ TH1 : \(\left(k-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}A=0\\k=1\left(TM\right)\end{matrix}\right.\)
+ TH2 : \(\left(k-1\right)^2\ne0\)
=> A là số cp \(\Leftrightarrow k^2-6k+10\) là số cp
\(\Leftrightarrow k^2-6k+10=n^2\) ( \(n\in N\)* )
\(\Leftrightarrow\left(k-3\right)^2+1=n^2\)
\(\Leftrightarrow\left(n-k+3\right)\left(n+k-3\right)=1\)
Xét các TH rồi tìm đc \(k=3\)
Ta có : 4x3 + 14x2 + 9x - 6 = ( x + 2 ) ( 4x2 + 6x - 3 )
Chứng minh x+2 và 4x2 + 6x - 3 nguyên tố cùng nhau nên để 4x3 + 14x2 + 9x - 6 là số chính phương
thì x + 2 và 4x2 + 6x -3 là số chính phương
đặt x + 2 = a2 ; 4x2 + 6x -3 = b2
\(\Rightarrow x=a^2-2\)
Thay vào ta có : 4 ( a2 - 2 )2 + 6 ( a2 - 2 ) - 3 = b2 hay 4a4 - 10a2 + 1= b2
\(\Rightarrow16a^4-40a^2+4=4b^2\Rightarrow\left(4a^2-2b-5\right)\left(4a^2+2b-5\right)=21\)
Mà 0 < 4a2 - 2b - 5 < 4a2 + 2b - 5
..... tìm được x = 2
a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(A=\frac{x^3-3x^2-2x^2+6x+3x-9+7}{x-3}\)
\(A=\frac{x^2\left(x-3\right)-2x\left(x-3\right)+3\left(x-3\right)+7}{x-3}\)
\(A=\frac{\left(x-3\right)\left(x^2-2x+3\right)+7}{x-3}\)
\(A=x^2-2x+3+\frac{7}{x-3}\)
\(x\in Z,x>0;=>A\in Z<=>\frac{7}{x-3}\in Z\)
x-3 | 1 | -1 | 7 | -7 |
x | 4 | 2 | 10 | -4 |
Vậy \(A\in Z<=>x\in\left\{-4;2;4;10\right\}\)
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
4x3+8x2+6x2+12x-3x-6=0
=> 4x2(x+2)+6x(x+2)-3(x+2)=0
=> (4x2+6x-3)(x+2)=0
=> \(\left[{}\begin{matrix}4x^2+6x-3=0\\x+2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(2x+\dfrac{3}{2}\right)^2=\dfrac{21}{4}\\x=-2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\sqrt{21}-3}{4}\\x=\dfrac{-\sqrt{21}-3}{4}\end{matrix}\right.\\x=-2\end{matrix}\right.\)