a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)

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

\(\Leftrightarrow x-3\le0\)

hay \(x\le3\)

b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)

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

\(\Leftrightarrow2x-5\le0\)

hay \(x\le\dfrac{5}{2}\)

Akai Haruma, Trần Thanh Phương, tth, lê thị hương giang

câu c đặt biến phụ xem?

a,\(Đkxđ:x\ge3\)

Ta có:

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

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

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

\(TH1:x-3=x-3\Leftrightarrow0x=0\)

\(\Rightarrow\)\(x\in R\) và \(x\ge3\)

\(TH2:x-3=3-x\Leftrightarrow2x=6\Leftrightarrow x=3\)( ko thỏa mãn điều kiện)

vậy \(\left\{x\in R/x\ge3\right\}\)

b, \(Đkxđ:x\le\dfrac{5}{2}\)

Ta có:

\(\sqrt{25-20x+4x^2}+2x=5\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=5-2x\\5-2x=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in R\\x=\dfrac{5}{2}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(\left\{x\in R/x\le\dfrac{5}{2}\right\}\)

a: \(2x^2-11x+21=3\cdot\sqrt[3]{4x-4}\)

=>\(2x^2-6x-5x+15=3\cdot\sqrt[3]{4x-4}-6\)

=>\(\left(x-3\right)\left(2x-5\right)=3\cdot\frac{4x-4-8}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}\)

=>\(\left(x-3\right)\left(2x-5\right)-3\cdot\frac{4x-12}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}=0\)

=>\(\left(x-3\right)\left\lbrack\left(2x-5\right)-3\cdot\frac{4}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}\right\rbrack=0\)

=>x-3=0

=>x=3

Bài 1:

b: ĐKXĐ: x∈R

\(x^2-x-\sqrt{x^2-x+13}=7\)

=>\(x^2-x-\sqrt{x^2-x+13}-7=0\)

=>\(x^2-x+13-\sqrt{x^2-x+13}-20=0\)

=>\(\left(\sqrt{x^2-x+13}-5\right)\left(\sqrt{x^2-x+13}+4\right)=0\)

=>\(\sqrt{x^2-x+13}-5=0\)

=>\(\sqrt{x^2-x+13}=5\)

=>\(x^2-x+13=25\)

=>\(x^2-x-12=0\)

=>(x-4)(x+3)=0

=>x=4(nhận) hoặc x=-3(nhận)

c: ĐKXĐ: \(x^2-3x+1\ge0\)

=>\(x^2-3x+\frac94-\frac54\ge0\)

=>\(\left(x-\frac32\right)^2\ge\frac54\)

=>\(\left[\begin{array}{l}x-\frac32\ge\frac{\sqrt5}{2}\\ x-\frac32\le-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{3+\sqrt5}{2}\\ x\le\frac{3-\sqrt5}{2}\end{array}\right.\)

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

=>\(x^2-3x-4+2\cdot\sqrt{x^2-3x+1}=0\)

=>\(x^2-3x+1+2\cdot\sqrt{x^2-3x+1}-5=0\)

=>\(\left(\sqrt{x^2-3x+1}+1\right)^2=6\)

=>\(\sqrt{x^2-3x+1}+1=\sqrt6\)

=>\(\sqrt{x^2-3x+1}=\sqrt6-1\)

=>\(x^2-3x+1=7-2\sqrt6\)

=>\(x^2-3x-6+2\sqrt6=0\) (1)

\(\Delta=\left(-3\right)^2-4\cdot1\cdot\left(-6+2\sqrt6\right)=9+24-8\sqrt6=33-8\sqrt6\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{3-\sqrt{33-8\sqrt6}}{2\cdot1}=\frac{3-\sqrt{33-8\sqrt6}}{2}\left(nhận\right)\\ x=\frac{3+\sqrt{33-8\sqrt6}}{2}\left(nhận\right)\end{array}\right.\)

e: ĐKXĐ: x(x+2)>=0

=>x>=0 hoặc x<=-2

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

=>\(2x^2+4x+\sqrt{x^2+2x}-3=0\)

=>\(2\cdot\left(\sqrt{x^2+2x}\right)^2+\sqrt{x^2+2x}-3=0\)

=>\(\left(2\sqrt{x^2+2x}+3\right)\left(\sqrt{x^2+2x}-1\right)=0\)

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

=>\(x^2+2x=1\)

=>\(x^2+2x+1=2\)

=>\(\left(x+1\right)^2=2\)

=>\(\left[\begin{array}{l}x+1=\sqrt2\\ x+1=-\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt2-1\left(nhận\right)\\ x=-\sqrt2-1\left(nhận\right)\end{array}\right.\)