giải các phương trình sau:
phương trình chứa ẩn trong dấu giá trị tuyệt đối
1) \(\sqrt{x^2+6x+9}=\left|2x-1\right|\)
2) \(\left|x^2-2x-3\right|=x^2+\left|2x+3\right|\)
3) |2x-5|+\(\left|2x^2-7x+5\right|=0\)
4) |x+3|+|7-x|=10
5) \(x^2-2x+\left|x-1\right|-1=0\)
6)\(4x^2-4x-\left|2x-1\right|-1=0\)
7) \(\left|x^2-4x-5\right|=4x-17\)
8) | x - 1|+|2x+1|=|3x|
phương trình chứa ẩn dưới dấu căn
1) \(\sqrt{2x-3}=x-3\)
2) \(x-\sqrt{2x-5}=4\)
3) \(\sqrt{x^2+x-12}=8-x\)
4) \(\sqrt{x^2+2x+4}=\sqrt{2-x}\)
5) (x-3)\(\sqrt{x^2+4}=x^2-9\)
toán lớp 10 ,giúp mình với mọi người ơi:(((
1.
ĐKXĐ: \(x\in R\)
\(pt\Leftrightarrow x^2+6x+9=4x^2-4x+1\)
\(\Leftrightarrow3x^2-10x-8=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy ...
2.
TH1: \(x\le-\frac{3}{2}\)
\(pt\Leftrightarrow x^2-2x-3=x^2-2x-3\)
\(\Rightarrow\) Phương trình có vô số nghiệm \(x\in(-\infty;-\frac{3}{2}]\)
TH2: \(-\frac{3}{2}< x\le-1\)
\(pt\Leftrightarrow x^2-2x-3=x^2+2x+3\)
\(\Leftrightarrow x=-\frac{3}{2}\left(l\right)\)
TH3: \(-1< x\le3\)
\(pt\Leftrightarrow-x^2+2x+3=x^2+2x+3\)
\(\Leftrightarrow x=0\left(l\right)\)
TH4: \(x>3\)
\(pt\Leftrightarrow x^2-2x-3=x^2+2x+3\)
\(\Leftrightarrow x=-\frac{3}{2}\left(l\right)\)
Vậy \(x\in(-\infty;-\frac{3}{2}]\)
3.
\(pt\Leftrightarrow\left|2x-5\right|+\left|\left(2x-5\right)\left(x-1\right)\right|=0\)
\(\Leftrightarrow\left|2x-5\right|.\left(\left|x-1\right|+1\right)=0\)
\(\Leftrightarrow\left|2x-5\right|=0\left(\text{Vì }\left|x-1\right|+1>0\right)\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy ...
4.
\(\Leftrightarrow\left|x+3\right|+\left|7-x\right|=10\)
TH1: \(x\le-3\)
\(pt\Leftrightarrow-x-3+7-x=10\)
\(\Leftrightarrow x=-3\left(tm\right)\)
TH2: \(-3< x\le7\)
\(pt\Leftrightarrow x+3+7-x=10\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow\) Phương trình có vô số nghiệm \(x\in(-3;7]\)
TH3: \(x>7\)
\(pt\Leftrightarrow x+3-7+x=10\)
\(\Leftrightarrow x=7\left(l\right)\)
Vậy \(x\in\left[-3;7\right]\)
7.
TH1: \(x\le-1\)
\(pt\Leftrightarrow x^2-4x-5=4x-17\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(l\right)\\x=2\left(l\right)\end{matrix}\right.\)
TH2: \(-1< x\le5\)
\(pt\Leftrightarrow-x^2+4x+5=4x-17\)
\(\Leftrightarrow x^2=22\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{22}\left(tm\right)\\x=-\sqrt{22}\left(l\right)\end{matrix}\right.\)
TH3: \(x\ge5\)
\(pt\Leftrightarrow x^2-4x-5=4x-17\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=2\left(l\right)\end{matrix}\right.\Leftrightarrow x=6\)
Vậy phương trình có nghiệm ...
8.
TH1: \(x\le-\frac{1}{2}\)
\(pt\Leftrightarrow-x+1-2x-1=-3x\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow x\in(-\infty;-\frac{1}{2}]\)
TH2: \(-\frac{1}{2}< x\le0\)
\(pt\Leftrightarrow-x+1+2x+1=-3x\)
\(\Leftrightarrow x=-\frac{1}{2}\left(l\right)\)
TH3: \(0< x\le1\)
\(pt\Leftrightarrow-x+1+2x+1=3x\)
\(\Leftrightarrow x=1\left(tm\right)\)
TH4: \(x>1\)
\(pt\Leftrightarrow x-1+2x+1=3x\)
\(\Leftrightarrow0x=0\)
\(\Leftrightarrow x\in\left(1;+\infty\right)\)
Vậy \(x\in S=(-\infty;-\frac{1}{2}]\cup[1;\infty)\)
