\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)

\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)

\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)

\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)

\("="\Leftrightarrow x=3\)

a: ĐKXĐ: x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)

=>\(5-\sqrt{2x-1}=0\)

=>\(\sqrt{2x-1}=5\)

=>2x-1=25

=>2x=26

=>x=13(nhận)

c: \(\sqrt{x^2-10x+25}=2\)

=>\(\sqrt{\left(x-5\right)^2}=2\)

=>\(\left|x-5\right|=2\)

=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

d: \(\sqrt{x^2-14x+49}-5=0\)

=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)

=>\(\sqrt{\left(x-7\right)^2}=5\)

=>|x-7|=5

=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)

Vậy \(S=\left\{9\right\}\)

\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)

Vậy \(S=\left\{13\right\}\)

\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{7;3\right\}\)

\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{12;2\right\}.\)

1, \(\sqrt{4-4x+x^2}=3\)

\(\Leftrightarrow\sqrt{\left(2+x\right)^2}=3\)

\(\Leftrightarrow\left|2+x\right|=3\)

TH1: \(\left|2-x\right|=2-x\) với \(2-x\ge0\Leftrightarrow x\le2\)

Pt trở thành:

\(2-x=3\) (ĐK: \(x\le2\) )

\(\Leftrightarrow x=2-3\)

\(\Leftrightarrow x=-1\left(tm\right)\)

TH2: \(\left|2-x\right|=-\left(2-x\right)\) với \(2-x< 0\Leftrightarrow x>2\)

Pt trở thành:

\(-\left(2-x\right)=3\) (ĐK: \(x>2\))

\(\Leftrightarrow-2+x=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\left(tm\right)\)

Vậy \(S=\left\{-1;5\right\}\)

2, \(\sqrt{x^2-6x+9}=1\)

\(\Leftrightarrow\sqrt{x^2-2\cdot3\cdot x+3^2}=1\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=1\)

\(\Leftrightarrow\left|x-3\right|=1\)

TH1: \(\left|x-3\right|=x-3\) với \(x-3\ge0\Leftrightarrow x\ge3\)

Pt trở thành:

\(x-3=1\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=1+3\)

\(\Leftrightarrow x=4\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x-3< 0\Leftrightarrow x< 3\)

Pt trở thành:

\(-\left(x-3\right)=1\) (ĐK: \(x< 3\))

\(\Leftrightarrow-x+3=1\)

\(\Leftrightarrow-x=1-3\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy \(S=\left\{2;4\right\}\)

1) √(4 - 4x + x²) = 3

⇔ √(2 - x)² = 3

ĐKXĐ: Với mọi x ∈ R

⇔ |2 - x| = 3 (1)

*) |2 - x| = 2 - x ⇔ 2 - x ≥ 0 ⇔ x ≥ 2

(1) ⇔ 2 - x = 3

⇔ x = 2 - 3

⇔ x = -1 (nhận)

*) |2 - x| = x - 2 ⇔ 2 - x < 0 ⇔ x > 2

(1) ⇔ x - 2 = 3

⇔ x = 5 (nhận)

Vậy x = -1; x = 5

Thiếu soát gì mog bạn thông cảm :]

c/ \(C=\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}\)

\(=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}\)

\(=|3-x|+|x+5|\ge|3-x+x+5|=8\)

d/ \(D=\sqrt{x^2-6x+9}+\sqrt{4x^2+24x+36}\)

\(=\sqrt{\left(x-3\right)^2}+\sqrt{4\left(x+3\right)^2}\)

\(=|3-x|+|x+3|+|x+3|\ge|3-x+x+3|+0=6\)

e/ \(2E=\sqrt{x^2}+2\sqrt{x^2-2x+1}\)

\(=\sqrt{x^2}+2\sqrt{\left(x-1\right)^2}\)

\(=|x|+|1-x|+|x-1|\ge|x+1-x|+0=1\)

\(\Rightarrow E\ge\frac{1}{2}\)

a/ ĐKXĐ: \(x\ge1\)

Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm

b/ \(x\ge1\)

\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)

Đặt \(\sqrt{x-1}=a\ge0\) ta được:

\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)

- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)

- Với \(0\le a\le1\) ta được:

\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)

- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)

c/ ĐKXĐ: \(x\ge\frac{49}{14}\)

\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)

Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(7-\sqrt{14x-49}\ge0\)

\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)

Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)

d/ ĐKXĐ: \(x\ge\frac{1}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}-2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|-2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=4\)

TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=4\)

\(\Leftrightarrow\sqrt{2x-1}=5\)

\(\Leftrightarrow x=13\)

TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)

TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)

\(\sqrt{2x-1}-1-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow4=4\) (luôn đúng)

TH4: \(\frac{1}{2}\le x< 1\)

\(1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(l\right)\)

Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)

Ngại làm lần 2 quá bạn ơi 

Câu hỏi của Chuột yêu Gạo - Toán lớp 9 | Học trực tuyến

a) \(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)

\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )

b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)

\(\Leftrightarrow\left|2x-1\right|=5x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)

\(\Leftrightarrow x+3=4\)

\(\Rightarrow x=1\)