ĐKXĐ: \(x\ge2\)

Đặt \(u=\sqrt{x+3};v=\sqrt{x-2}\) Phương trình trở thành :

\(\left(u-v\right)\left(1+uv\right)=5\) Mặt khác ta thấy \(u^2-v^2=5\)

\(\Rightarrow\left(u-v\right)\left(1+uv\right)=\left(u-v\right)\left(u+v\right)\) (*)

vì \(u-v>0\) nên chia cả hai vế (*) cho \(u-v\)

Ta được: \(1+uv=u+v\) \(\Leftrightarrow uv-u-\left(v-1\right)=0\Leftrightarrow\left(v-1\right)\left(u-1\right)=0\)

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

Vậy phương trình có nghiệm duy nhất \(x=3\)

b,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}-16\sqrt{x+1}=0\) (dk \(x\ge-1\)

\(\Leftrightarrow\sqrt{x+1}\left(4-3+2-16\right)=0\)

\(\Leftrightarrow\sqrt{x+1}.-13=0\)

\(\Leftrightarrow x=-1\)

đặt x-2016=a

y-2017=b

z-2018=c

ta có\(\frac{1}{\sqrt{a}}-\frac{1}{a}+\frac{1}{\sqrt{b}}-\frac{1}{b}+\frac{1}{\sqrt{c}}-\frac{1}{c}=\frac{3}{4}\)

=>\(\left(\frac{1}{\sqrt{a}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{2}\right)^2=0\)

=>\(a=b=c=4\)

còn lại tự lm nốt

oke cao van duc

thank nhiều nha

hok tốt

Đặt \(\hept{\begin{cases}a=\sqrt{x-2009}\\b=\sqrt{y-2010}\\c=\sqrt{z-2011}\end{cases}}\)(với a,b,c>0). Khi đó phương trình đã cho trở thành

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2\)

\(\Leftrightarrow a=b=c=2\)\(\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)

Bài 1 bạn tìm quanh quanh đây, mình thấy có bài y hệt rồi nên ko làm nữa

Bài 2 như sau:

ĐKXĐ: \(x\ge\dfrac{-1}{16}\)

\(x^2-x-20-2\left(\sqrt{16x+1}-9\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-2\dfrac{\left(\sqrt{16x+1}-9\right)\left(\sqrt{16x+1}+9\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-\dfrac{32\left(x-5\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4-\dfrac{32}{\sqrt{16x+1}+9}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\Rightarrow x=5\\x+4-\dfrac{32}{\sqrt{16x+1}+9}=0\left(1\right)\end{matrix}\right.\)

Xét phương trình (1): ta có \(x+4\ge-\dfrac{1}{16}+4=\dfrac{63}{16}\) \(\forall x\ge-\dfrac{1}{16}\)

\(\sqrt{16x+1}+9\ge9\Rightarrow\dfrac{32}{\sqrt{16x+1}+9}\le\dfrac{32}{9}\) \(\forall x\ge-\dfrac{1}{16}\)

Mà \(\dfrac{63}{16}-\dfrac{32}{9}=\dfrac{55}{144}>0\) \(\Rightarrow x+4-\dfrac{32}{\sqrt{16x+1}+9}>0\) \(\forall x\ge-\dfrac{1}{16}\)

\(\Rightarrow\) pt (1) vô nghiệm

Vậy pt đã cho có nghiệm duy nhất \(x=5\)

a) \(\sqrt{x-2}+\sqrt{16x-32}=10\)

\(\Rightarrow\sqrt{x-2}+4\sqrt{x-2}=10\)

\(\Rightarrow5\sqrt{x-2}=10\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

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

ĐK \(x\ge\dfrac{1}{2}\)

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

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

\(\left|x+\sqrt{2x-1}\right|=50\)

\(\sqrt{2x-1}=50-x\)

\(\left(\sqrt{2x-1}\right)^2=\left(50-x\right)^2\)

\(\left|2x-1\right|=x^2-100x+2500\)

\(2x-1=x^2-100x+2500\)

\(x=41\)

a) ĐK x\(\ge\)2

\(\sqrt{x-2}+\sqrt{16x-32}=10\)

\(\sqrt{x-2}+\sqrt{16\left(x-2\right)}=10\)

\(\sqrt{x-2}+4\sqrt{x-2}=10\)

\(\left(1+4\right)\sqrt{x-2}=10\)

\(5\sqrt{x-2}=10\)

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

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

\(\left|x-2\right|=4\)(vì \(x\ge2\))

\(x-2=4\)

\(x=6\)

Lời giải:

a) ĐK: \(x\geq 0\)

\(4\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)

\(\Leftrightarrow 4\sqrt{x}-2\sqrt{9}.\sqrt{x}+\sqrt{16}.\sqrt{x}=5\)

\(\Leftrightarrow 4\sqrt{x}-6\sqrt{x}+4\sqrt{x}=5\)

\(\Leftrightarrow 2\sqrt{x}=5\Rightarrow \sqrt{x}=\frac{5}{2}\Rightarrow x=\frac{25}{4}\) (thỏa man)

b) ĐK: \(x\geq -5\)

PT \(\Leftrightarrow \sqrt{4}.\sqrt{x+5}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9}.\sqrt{x+5}=6\)

\(\Leftrightarrow 2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow 3\sqrt{x+5}=6\Rightarrow \sqrt{x+5}=2\)

\(\Rightarrow x+5=2^2=4\Rightarrow x=-1\) (thỏa mãn)