f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

a: ĐKXĐ: \(x\in R\)

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

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

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

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)

\(\Leftrightarrow25x-4x=-8-75\)

\(\Leftrightarrow21x=-83\)

hay \(x=-\dfrac{83}{21}\)

b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)

\(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)

d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)

\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)

\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)

\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)

\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)

\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)

\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)

vậy: Phương trình vô nghiệm

Đề yc giải pt à em?

Câu b bạn có bị lỗi dấu căn không mà sao nó kéo dài cả 2 vế pt vậy :v

\(a,\sqrt{x^2-6x+9}+x=11\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=11-x\)

\(\Leftrightarrow\left|x-3\right|=11-x\\ TH_1:x\ge3\\ x-3=11-x\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(tm\right)\)

\(TH_2:x< 3\\ -x+3=11-x\\ \Leftrightarrow-x+x=11-3\\ \Leftrightarrow0=8\left(VL\right)\)

Vậy \(S=\left\{7\right\}\)

\(c,\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\) \(\left(dk:x\ge-1\right)\)

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

Đặt \(a=\sqrt{x+1}\left(a\ge0\right)\)

Pt trở thành : \(4a-3a=4\Leftrightarrow a=4\left(tmdk\right)\)

\(\Rightarrow\sqrt{x+1}=4\\ \Rightarrow\left(\sqrt{x+1}\right)^2=16\\ \Rightarrow\left|x+1\right|=16\)

\(TH_1:x\ge-1\\ x+1=16\Leftrightarrow x=15\left(tm\right)\\ TH_2:x< -1\\ -x-1=16\Leftrightarrow x=-17\left(tm\right)\)

Nhưng loại TH2 vì dk ban đầu là \(x\ge-1\)

Vậy \(S=\left\{15\right\}\)

\(d,\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\left(dk:x\ge-1\right)\\ \Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{x+1}=0\)

Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)

Tới đây bạn làm tương tự câu c nha.

bài 1:

a)\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\)

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

\(=\left(3-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)\(do2>\sqrt{3}\)

\(=6+3\sqrt{3}-2\sqrt{2}-\sqrt{6}\)

b) \(\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)

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

\(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)do\sqrt{5}>\sqrt{2}\)

\(=\sqrt{15}-\sqrt{6}+5-\sqrt{10}\)

c)\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\)

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

\(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)do\sqrt{5}>2\)

\(=5-4\)

\(=1\left(hđt.3\right)\)

d)\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{8-2\sqrt{15}}\)

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

\(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)do\sqrt{5}>\sqrt{3}\)

\(=5-3\)

\(=2\)

e)\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\)

\(=2\left(2-4+9\right)\)

\(=2.7=14\)

f)\(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)

\(=2-\sqrt{6-2\sqrt{5}}\)

\(=2-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=2-\left(\sqrt{5}-1\right)\)

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

\(=3-\sqrt{5}\)

g)\(\sqrt{3}-\sqrt{2}\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

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

\(=\sqrt{3}-\sqrt{6}-2\)

h) \(\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right)\sqrt{2}+2\sqrt{5}\)

\(=\left(2-\sqrt{6+2\sqrt{5}}\right)+2\sqrt{5}\)

\(=\left(2-\sqrt{\left(\sqrt{5}+1\right)^2}\right)+2\sqrt{5}\)

\(=2-\left(\sqrt{5}+1\right)+2\sqrt{5}\left(do\sqrt{5}>1\right)\)

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

\(=1-\sqrt{5}\)

bài 2)

a) \(\sqrt{4x^2-4x+1}=5\)

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

\(\Leftrightarrow2x-1=5\)hoặc \(\Leftrightarrow2x-1=-5\)

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

Vậy x = 3 hoặc x = -2

1: =>|2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>2x=6 hoặc 2x=-4

=>x=3 hoặc x=-2

2: \(\Leftrightarrow2\sqrt{x-3}+\dfrac{1}{3}\cdot3\sqrt{x-3}-\sqrt{x-3}=4\)

\(\Leftrightarrow\sqrt{x-3}=2\)

=>x-3=4

hay x=7

5: \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=2 hoặc x=-1