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làm tạm câu này vậy
a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)
\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)
\(\Leftrightarrow\left\{\left(x^2-x+1\right)^2+2x^2\right\}=\left(3x^2\right)^2\)
\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)
\(\Leftrightarrow\left(x^2-x+1\right)=x^2\)
\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)
Vậy...
1. \(x^3-6x^2+10x-4=0\)
<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
<=> \(\left(x-2\right)\left(x^2-4x+2\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)
Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)
=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)
\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)
1) Ta có: \(x^3-6x^2+10x-4=0\)
\(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)
+ \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
+ \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=2\)
\(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,5858;2;3,4142\right\}\)
Điều kiện xác định bạn tự tìm
a) \(\sqrt{x^2-4x+3}=x-2\Leftrightarrow\)\(\left(\sqrt{x^2-4x+3}\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+3=x^2-4x+4\Leftrightarrow0=1\) vô lý
pt vô nghiệm
b) \(\sqrt{x^2-1}-\left(x^2-1\right)=0\Leftrightarrow\sqrt{x^2-1}\left(1-\sqrt{x^2-1}\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-1}=0\\1-\sqrt{x^2-1}=0\end{cases}}\)
<=>\(\orbr{\begin{cases}\\\end{cases}}\begin{matrix}x=\pm1\\x=\pm\sqrt{2}\end{matrix}\)
c)\(\sqrt{x^2-4}-\left(x-2\right)=0\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\left(x-2\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-\sqrt{x-2}\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-\sqrt{x-2}=0\end{cases}}\)
<=>x=2 còn cái kia vô nghiệm
bạn tự trình bày chi tiết nhé
1/ \(\sqrt{x-2}-\sqrt{1-3x}=0\\ đk:\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)
=> pt vô no
2/ \(\sqrt{15-x}+\sqrt{3-x}=6\\ đk\left\{{}\begin{matrix}15-x\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le15\\x\le3\end{matrix}\right.\Leftrightarrow x\le3\)
\(pt\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)
\(\Leftrightarrow2\sqrt{\left(15-x\right)\left(3-x\right)}=2x+36\)
\(\Leftrightarrow4\left(15-x\right)\left(3-x\right)=\left(2x+18\right)^2\left(đk:x\ge-9\right)\)
\(\Leftrightarrow-144x=144\Leftrightarrow x=-1\left(nhan\right)\)
Câu 1: ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ \(\Rightarrow\) pt vô nghiệm
Câu 2:
ĐKXĐ: \(x\le3\)
\(\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)
\(\Leftrightarrow x+9=\sqrt{x^2-18x+45}\) (\(x\ge-9\))
\(\Leftrightarrow x^2+18x+81=x^2-18x+45\)
\(\Leftrightarrow36x=-36\Rightarrow x=-1\)
Câu 3:
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}=2+\sqrt{x+1}\)
\(\Leftrightarrow x-1=4+x+1+4\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{x+1}=-\frac{3}{2}\)
Phương trình vô nghiệm
A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
Giải PT
a) \(3\sqrt{9x}+\sqrt{25x}-\sqrt{4x} = 3\)
\(\Leftrightarrow\) \(3.3\sqrt{x} +5\sqrt{x} - 2\sqrt{x} = 3 \)
\(\Leftrightarrow\) \(9\sqrt{x}+5\sqrt{x}-2\sqrt{x} = 3 \)
\(\Leftrightarrow\) \(12\sqrt{x} = 3\)
\(\Leftrightarrow\) \(\sqrt{x} = 4 \)
\(\Leftrightarrow\) \(\sqrt{x^2} = 4^2\)
\(\Leftrightarrow\) \(x=16\)
b) \(\sqrt{x^2-2x-1} - 3 =0\)
\(\Leftrightarrow\) \(\sqrt{(x-1)^2} -3=0\)
\(\Leftrightarrow\) \(|x-1|=3\)
* \(x-1=3\)
\(\Leftrightarrow\) \(x=4\)
* \(-x-1=3\)
\(\Leftrightarrow\) \(-x=4\)
\(\Leftrightarrow\) \(x=-4\)
c) \(\sqrt{4x^2+4x+1} - x = 3\)
<=> \(\sqrt{(2x+1)^2} = 3+x\)
<=> \(|2x+1|=3+x\)
* \(2x+1=3+x\)
<=> \(2x-x=3-1\)
<=> \(x=2\)
* \(-2x+1=3+x\)
<=> \(-2x-x = 3-1\)
<=> \(-3x=2\)
<=> \(x=\dfrac{-2}{3}\)
d) \(\sqrt{x-1} = x-3\)
<=> \(\sqrt{(x-1)^2} = (x-3)^2\)
<=> \(|x-1| = x^2-2.x.3+3^2\)
<=> \(|x-1| = x-6x+9\)
<=> \(|x-1| = -5x+9\)
* \(x-1= -5x+9\)
<=> \(x+5x = 9+1\)
<=> \(6x=10\)
<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)
* \(-x-1 = -5x+9\)
<=> \(-x+5x = 9+1\)
<=> \(4x = 10\)
<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)
giải pt
ảo ma