BN
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 11 2019
ĐKXĐ:
\(\Leftrightarrow x\left(\sqrt{1-x}+1\right)=2x\left(\sqrt{1+x}+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{1-x}+1=2\sqrt{1+x}+2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{1-x}=2\sqrt{1+x}+1\)
\(\Leftrightarrow1-x=4x+5+4\sqrt{1+x}\)
\(\Leftrightarrow4\sqrt{1+x}=-5x-4\) (\(x\le-\frac{4}{5}\))
\(\Leftrightarrow16x+16=25x^2+40x+16\)
\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-\frac{24}{25}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 6 2019
Câu 1: ĐKXĐ: ...
\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)
\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)
\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)
\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow...\)
Câu 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)
\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)
\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)
\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)
\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)
28 tháng 6 2019
\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)
\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)
\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)
\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)
\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)
Tìm được mỗi nghiệm thôi à :v
VA
0
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 11 2019
a/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)
\(\Leftrightarrow x-1=8x-3+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
Do \(x\ge1\Rightarrow2-7x< 0\Rightarrow\left\{{}\begin{matrix}VP\ge0\\VT< 0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|=2\)
Mà \(\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=2\)
Dấu "=" xảy ra khi và chỉ khi \(1-\sqrt{x-1}\ge0\Rightarrow x\le2\Rightarrow1\le x\le2\)
Vậy nghiệm của pt là \(1\le x\le2\)
24 tháng 12 2017
C\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}\)
\(=\dfrac{\sqrt{5}-1}{2}+\sqrt{\dfrac{3+\sqrt{5}}{2}}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{\left(3+\sqrt{5}\right).2}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{\left(1+\sqrt{5}\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{1+\sqrt{5}}{2}\)
\(=\dfrac{\sqrt{5}-1+1+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}}{2}\)
\(=\sqrt{5}\)
15 tháng 5 2018
a) \(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)
Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)
\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-1}\right)^2=\left(\sqrt{2x+2}\right)^2\)
\(\Leftrightarrow x+3-2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=2x+2\)
\(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)
\(\Leftrightarrow-2\sqrt{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)
Vậy \(S=\left\{1\right\}\)
x-1=x^2-2x+1 (x>=1)
x^2-3x+2=0
(x-1)(x-2)=0
=> x=1
hoặc x=2
ĐK \(x\supseteq1\)
\(x-1=\left(x-1\right)^2\)