\(\sqrt{x-6}=2\)
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NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
1.
ĐKXĐ: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)
AD
15 tháng 8 2023
\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)
\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)
Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)
15 tháng 8 2023
a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)
=>x+6<0
=>x<-6
b: ĐKXĐ: (-2)/(6-x)>=0
=>6-x<0
=>x>6
c: ĐKXĐ: (-x+3)/(-6)>=0
=>-x+3<=0
=>-x<=-3
=>x>=3
d: ĐKXĐ: (7x-1)/-9>=0
=>7x-1<=0
=>x<=1/7
e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0
=>x+2>=0
=>x>=-1
f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0
=>x-2>=0
=>x>=2
N
2 tháng 10 2019
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
N
2 tháng 10 2019
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
PA
19 tháng 7 2017
\(\sqrt{28-6\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=3\sqrt{3}-1\)
\(\sqrt{6-\sqrt{20}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-1\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\left(x+2\right)}}\)
\(=\sqrt{\left(\sqrt{x+2}+\sqrt{x+1}\right)^2}\)
\(=\sqrt{x+2}+\sqrt{x+1}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{\left(x-1\right)\left(x+3\right)}+\left(x+3\right)}\)
\(=\sqrt{\left(\sqrt{x+3}-\sqrt{x-1}\right)^2}\)
\(=\left|\sqrt{x+3}-\sqrt{x-1}\right|\)
\(\sqrt{21-6\sqrt{6}}+\sqrt{21+6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\)
\(=6\sqrt{2}\)
PA
19 tháng 7 2017
\(M=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)\(\left[\dfrac{\left(\sqrt{x}+1\right)-\left(3-\sqrt{x}\right)}{\sqrt{x}+1}\right]\)
\(=\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right]\times\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}\times2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
PB
29 tháng 5 2021
a) a) Biến đổi vế trái thành và làm tiếp.
b) Biến đổi vế trái thành và làm tiếp
18 tháng 1
a: ĐKXĐ: \(x^2-5x-6>=0\)
=>(x-6)(x+1)>=0
=>\(\left[{}\begin{matrix}x>=6\\x< =-1\end{matrix}\right.\)
\(\sqrt{x^2-5x-6}=x-2\)
=>\(\left\{{}\begin{matrix}x-2>=0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=6\\-5x-6=-4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=6\\-x=10\end{matrix}\right.\)
=>\(x\in\varnothing\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-8x+16}=4-x\)
=>\(\sqrt{\left(x-4\right)^2}=4-x\)
=>|x-4|=4-x
=>x-4<=0
=>x<=4
c: ĐKXĐ: \(x^2-2x>=0\)
=>x(x-2)>=0
=>\(\left[{}\begin{matrix}x>=2\\x< =0\end{matrix}\right.\)
\(\sqrt{x^2-2x}=2-x\)
=>\(\left\{{}\begin{matrix}x^2-2x=\left(2-x\right)^2\\x< =2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x=x^2-4x+4\\x< =2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=4\\x< =2\end{matrix}\right.\Leftrightarrow x=2\left(nhận\right)\)
d: ĐKXĐ: x>=-27/2
\(\sqrt{2x+27}-6=x\)
=>\(\sqrt{2x+27}=x+6\)
=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+6\right)^2=2x+27\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-6\\x^2+12x+36-2x-27=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-6\\x^2+10x+9=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+9\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-6\\x\in\left\{-9;-1\right\}\end{matrix}\right.\)
=>x=-1
Kết hợp ĐKXĐ, ta được: x=-1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 1
a.
\(\sqrt{x^2-5x-6}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x=-10\left(ktm\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
b.
\(\sqrt{x^2-8x+16}=4-x\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=4-x\)
\(\Leftrightarrow\left|x-4\right|=-\left(x-4\right)\)
\(\Leftrightarrow x-4\le0\)
\(\Rightarrow x\le4\)
\(\sqrt[]{x-6}=2\left(x\ge6\right)\)
\(\Rightarrow x-6=4\Rightarrow x=10\)
\(\sqrt{x-6}\) =2
Đk \(x\) - 6 ≥ 0 ⇒ \(x\) ≥ 6
\(\sqrt{x-6}=2\)
\(x-6=4\)
\(x=4+6\)
\(x\) = 10