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Làm hơi tắt , thông cảm ;))
Từ (1) \(\Rightarrow36=\left(x+y+z\right)^2\Leftrightarrow36=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow36=18+2\left(xy+yz+zx\right)\Leftrightarrow xy+yz+zx=9\)(4)
Từ (3) \(\Rightarrow16=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\Leftrightarrow16=x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=5\Leftrightarrow\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2=25\)
\(\Leftrightarrow xy+yz+zx+2\left(\sqrt{xy^2z}+\sqrt{xyz^2}+\sqrt{x^2yz}\right)=25\)
\(\Leftrightarrow\sqrt{xyz}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=8\Leftrightarrow\sqrt{xyz}=\frac{8}{4}\Leftrightarrow xyz=4\)(5)
Vậy hệ đã cho tương đương với :
\(\hept{\begin{cases}x+y+z=6\left(1\right)\\xy+yz+zx=9\left(4\right)\\xyz=4\left(5\right)\end{cases}}\)
Từ (5) \(\Rightarrow yz=\frac{4}{x}\)(Dễ thấy \(x,y,z>0\))
(4) \(\Leftrightarrow xy+yz+zx+x^2=9+x^2\Leftrightarrow x\left(x+y+z\right)+yz=9+x^2\)
\(\Leftrightarrow x.6+\frac{4}{x}=9+x^2\Leftrightarrow x^3-6x^2+9x-4=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}.}\)
Thế vào ta suy ra hệ có các nghiệm : \(\left(x,y,z\right)=\left(1,1,4\right),\left(1,4,1\right),\left(4,1,1\right).\)
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(x + 1)(x + 2)(x + 3) = x3 - 1
=> x3 + 6x2 + 11x + 6 - x3 + 1 = 0
=> 6x2 + 11x + 7 = 0
Vì 6x2 + 11x + 7 > 0 => vô nghiệm
Vậy \(x\in\phi\)
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1. Tìm x thỏa mản phương trình x nguyên
\(\left|x+1\right|\left(x^2-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left|x+1\right|=0\Rightarrow x=-1\) ( nhận )
Hoặc
\(x^2-5=0\Rightarrow x^2=5\) ( loại )
Hoặc
\(x^2-4=0\Rightarrow x^2=2^2\Rightarrow x=\pm2\)
Vậy: \(x=\left(-2;-1;2\right)\)
Bài 1:
\(\left|x+1\right|\left(x^2-5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+1\right|=0\\x^2-5=0\\x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\sqrt{5}\\x=\pm2\end{matrix}\right.\)
Do \(x\in Z\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Vậy...
Bài 3:
\(x^2-2xy+2y^2=0\)
\(\Rightarrow x^2-2xy+y^2+y^2=0\)
\(\Rightarrow\left(x-y\right)^2+y^2=0\)
Mà \(\left(x-y\right)^2+y^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\y^2=0\end{matrix}\right.\Rightarrow x=y=0\)
Vậy...
Bài 5,6 áp dụng t/c dãy tỉ số bằng nhau là ra
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\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{2\left(x+3\right)}{2x\left(x+3\right)}+\frac{2x}{2x\left(x+3\right)}=\frac{x\left(x+3\right)}{2x\left(x+3\right)}\)
\(\Leftrightarrow2x+6+2x=x^2+3x\)
\(\Leftrightarrow x=3\)
\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{x+x+3}=\frac{1}{2}\)
x+x+3=2
2x=-1
x=-1/2
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Ta có : (x - 1)2 + (x + 3)2 = 2(x - 2)(x + 1) + 38
<=> x2 - 2x + 1 + x2 + 6x + 9 = 2x2 - 2x - 4 + 38
<=> x2 - 2x + 1 + x2 + 6x + 9 - 2x2 + 2x + 4 - 14 = 24
<=> x2 + x2 - 2x2 - 2x + 2x + 6x + 1 + 9 + 4 - 14 = 24
<=> 6x = 24
=> x = 24 : 6
=> x = 4
\(\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(x^2+6x+9\right)=2\left(x^2-x-2\right)+38\)
\(\Leftrightarrow2x^2+4x+10=2x^2-2x+34\)
\(\Leftrightarrow2x^2+4x-2x^2+2x=34-10\)
\(\Leftrightarrow6x=24\)
\(\Rightarrow x=4\)
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\(a.ĐK:x\ne3;1\)
\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)
\(\Leftrightarrow7x-21=7x^2-28x+21\)
\(\Leftrightarrow7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
b.\(ĐK:x\ne2;4\)
\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)
\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)
\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)
=>(x-3)(7x-14)=0
=>x=3(loại) hoặc x=2(nhận)
b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)
\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)
\(\Leftrightarrow2x^2-4x=0\)
=>2x(x-2)=0
=>x=0(nhận) hoặc x=2(loại)
`x+2-2(x+1)=-x`
`x+2-2x-2=-x`
`x-2x+x=2-2`
`0x=0` (LĐ)
Vậy `x in RR`
\(x+2-2\left(x+1\right)=-x\)
\(x+2-2x-2+x=0\)
\(0=0\left(đúng\right)\)
Vậy \(x\in R\)