WR
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TN
1
15 tháng 10 2021
\(a,PT\Leftrightarrow\left|3x-1\right|=\left|x-3\right|\Leftrightarrow\left[{}\begin{matrix}3x-1=x-3\\3x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\\ b,PT\Leftrightarrow\left|x-4\right|=4-x\Leftrightarrow\left[{}\begin{matrix}x-4=4-x\left(x\ge4\right)\\x-4=x-4\left(x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
WR
1
11 tháng 8 2016
\(A=\frac{1}{x-1}\sqrt{\frac{3x^2-6x+3}{x}}\)\(=\frac{1}{x-1}\sqrt{\frac{3\left(x^2-2x+1\right)}{x}}=\frac{1}{x-1}\sqrt{\frac{3\left(x-1\right)^2}{x}}=\frac{1}{x-1}\cdot\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x}}=\frac{\sqrt{3}}{\sqrt{x}}\)
4 tháng 7 2021
a)Pt \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\dfrac{1}{3}+\dfrac{1}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Vậy...
b)Đk:\(x\ge3\)
Pt \(\Leftrightarrow\sqrt{x-3}\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
Vậy...
c)Đk:\(x\ge1\)
\(x+\sqrt{x-1}=13\)
\(\Leftrightarrow\sqrt{x-1}=13-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)
Vậy...
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 12 2020
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)
\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)
Pt trở thành:
\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)
\(\Rightarrow\sqrt{1+x}+\sqrt{1-x}=2\)
\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)
\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)
b.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)
Pt trở thành:
\(t=t^2-4-16\Leftrightarrow...\)
30 tháng 8 2021
a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)
\(\Leftrightarrow2x+9=5-4x\)
\(\Leftrightarrow6x=-4\)
hay \(x=-\dfrac{2}{3}\left(nhận\right)\)
b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)
\(\Leftrightarrow2x-1=x-1\)
hay x=0(loại)
c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2+3x=x\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
KK
30 tháng 8 2021
a. \(\sqrt{2x+9}=\sqrt{5-4x}\)
<=> 2x + 9 = 5 - 4x
<=> 2x + 4x = 5 - 9
<=> 6x = -4
<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)
KN
1
7 tháng 5 2020
\(4x^4+4x^3+x^2+3x\ge0\)
\(4x^4+4x^2+1-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)
\(\Leftrightarrow\left(2x^2+1\right)^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)
\(2x^2+1=u;\sqrt{4x^4+4x^3+x^2+3x}=v\left(u>0;v>0\right)\)
\(\hept{\begin{cases}u^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)v\\v^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)u\end{cases}\Rightarrow u^2-v^2=\left(x^2-x+1\right)\left(v-u\right)\Leftrightarrow\orbr{\begin{cases}u=v\\u+v+x^2-x+1=0\end{cases}}}\)
- \(u+v+x^2-x+1=0\Leftrightarrow u+v+\left(x-\frac{1}{2}\right)^2=-\frac{3}{4}\)
- \(u=v\Leftrightarrow4x^4+4x^2+1=4x^4+4x^3+x^2+3x\Leftrightarrow\left(x-1\right)^3=-3x^3\Leftrightarrow x-1=-x\sqrt[3]{3}\Leftrightarrow x=\frac{1}{1+\sqrt[3]{3}}\)Đối chiếu điều kiện ta thu được nghiệm duy nhất \(x=\frac{1}{1+\sqrt[3]{3}}\)
11 tháng 12 2023
a: \(x^2\cdot2\sqrt{3}+x+1=\sqrt{3}\cdot\left(x+1\right)\)
=>\(x^2\cdot2\sqrt{3}+x\left(1-\sqrt{3}\right)+1-\sqrt{3}=0\)
\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}+24=28-10\sqrt{3}=\left(5-\sqrt{3}\right)^2>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\left(5-\sqrt{3}\right)}{2\cdot2\sqrt{3}}=\dfrac{-1+\sqrt{3}-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+5-\sqrt{3}}{2\cdot2\sqrt{3}}=\dfrac{4}{4\sqrt{3}}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
b: \(5x^2-3x+1=2x+31\)
=>\(5x^2-3x+1-2x-31=0\)
=>\(5x^2-5x-30=0\)
=>\(x^2-x-6=0\)
=>(x-3)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c: \(x^2+2\sqrt{2}x+4=3\left(x+\sqrt{2}\right)\)
=>\(x^2+2\sqrt{2}x+4-3x-3\sqrt{2}=0\)
=>\(x^2+x\left(2\sqrt{2}-3\right)+4-3\sqrt{2}=0\)
\(\text{Δ}=\left(2\sqrt{2}-3\right)^2-4\left(4-3\sqrt{2}\right)\)
\(=17-12\sqrt{2}-16+12\sqrt{2}=1\)>0
Do đó, phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(2\sqrt{2}-3\right)-1}{2}=\dfrac{-2\sqrt{2}+3-1}{2}=-\sqrt{2}+1\\x_2=\dfrac{-\left(2\sqrt{2}-3\right)+1}{2}=\dfrac{-2\sqrt{2}+4}{2}=-\sqrt{2}+2\end{matrix}\right.\)
\(A=\frac{1}{x-1}\sqrt{\frac{3x^2-6x+3}{x}}=\frac{1}{x-1}\sqrt{\frac{\left(\sqrt{3}x-\sqrt{3}\right)^2}{x}}\)
\(=\frac{1}{x-1}.\frac{\sqrt{3}x-\sqrt{3}}{\sqrt{x}}=\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\sqrt{\frac{3}{x}}\)