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16 tháng 10 2019
Tự tìm ĐKXĐ nhé
\(P=\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
c, \(P=\frac{\sqrt{x}+4}{\sqrt{x}+2}=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow1+\frac{2}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1;0\right\}\)
\(\Rightarrow x=\left\{0\right\}\)
Kết hợp với ĐKXĐ =>...
1 tháng 11 2023
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
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
H
1 tháng 11 2023
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
TB
1
CH
Cô Hoàng Huyền
Admin
VIP
6 tháng 8 2018
ĐK: \(x^3+4x^2+5x+6\ge0\)
Ta có: \(x^3+4x^2+5x+6=\left(x+3\right)\left(x^2+x+2\right);x^2+2x+5=\left(x+3\right)+\left(x^2+x+2\right)\)
Đặt \(\hept{\begin{cases}\sqrt{x+3}=u\\\sqrt{x^2+x+2}=v\end{cases}}\)
Vậy nên ta có phương trình: \(\)\(u^2+v^2=\frac{5}{2}uv\)
\(\Leftrightarrow2u^2-5uv+2v^2=0\Leftrightarrow\orbr{\begin{cases}u=2v\\u=\frac{1}{2}v\end{cases}}\)
Với u = 2v ta có: \(\sqrt{x+3}=2\sqrt{x^2+x+2}\Leftrightarrow x+3=4x^2+4x+8\)
\(\Leftrightarrow4x^2+3x+5=0\) (Vô nghiệm)
Với \(u=\frac{1}{2}v\) ta có: \(2\sqrt{x+3}=\sqrt{x^2+x+2}\Leftrightarrow4x+12=x^2+x+2\)
\(\Leftrightarrow x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\left(tmđk\right)\)
Vậy phương trình có nghiệm \(x\in\left\{5;-2\right\}\)
DC
1
G
8 tháng 5 2019
\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)
\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)
\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)
\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)
\("="\Leftrightarrow x=3\)
16 tháng 6 2021
a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)
Vậy ...
8 tháng 11 2023
a:
ĐKXĐ: \(x>=-2\)
\(1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)
=>\(1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)
Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)(ĐK: a>0 và b>0)
Phương trình sẽ trở thành:
1+ab=a+b
=>ab-a-b+1=0
=>a(b-1)-(b-1)=0
=>(b-1)(a-1)=0
=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)
=>\(\left\{{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)
=>\(x\in\varnothing\)
b: \(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2+8x-2\)
=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)
=>\(\sqrt{\left(2x-\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)
=>\(\left|2x-\dfrac{1}{2}\right|=4x^3-x^2+8x-2\)(1)
TH1: x>=1/4
\(\left(1\right)\Leftrightarrow4x^3-x^2+8x-2=2x-\dfrac{1}{2}\)
=>\(4x^3-x^2+6x-\dfrac{3}{2}=0\)
=>\(x^2\left(4x-1\right)+1,5\left(4x-1\right)=0\)
=>\(\left(4x-1\right)\left(x^2+1,5\right)=0\)
=>4x-1=0
=>x=1/4(nhận)
TH2: x<1/4
Phương trình (1) sẽ trở thành:
\(4x^3-x^2+8x-2=-2x+\dfrac{1}{2}\)
=>\(x^2\left(4x-1\right)+2\left(4x-1\right)+0,5\left(4x-1\right)=0\)
=>\(\left(4x-1\right)\cdot\left(x^2+2,5\right)=0\)
=>4x-1=0
=>x=1/4(loại)
14 tháng 10 2021
\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)