mị mới lớp 5 ahihi

ĐK: \(12\le x\le14\)

Sau khi nhân liên hợp chúng ta có được:

\(PT\Leftrightarrow\left(x-13\right)^2\left[1+\frac{\frac{2}{1+\sqrt{\left(x-12\right)\left(14-x\right)}}}{2+\sqrt{x-12}+\sqrt{14-x}}\right]=0\)

\(\Leftrightarrow x=13\)

Khủng khiếp tí nhưng chắc không sao:v

Akai HarumaBăng Băng 2k6HISINOMA KINIMADO

Vũ Minh TuấnNguyễn Thanh Hằng

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{26x+5}=a\ge0\\\sqrt{x^2+30}=b>0\end{matrix}\right.\)

\(\Rightarrow\frac{a^2}{b}+2a=3b\)

\(\Leftrightarrow a^2+2ab-3b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+3b\right)=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow\sqrt{26x+5}=\sqrt{x^2+30}\)

\(\Leftrightarrow x^2-26x+25=0\Rightarrow\left[{}\begin{matrix}x=1\\x=25\end{matrix}\right.\)

\(\sqrt{29-x}+\sqrt{x+3}=x^2-26x+177\left(1\right)\)

ĐK -3 =<x =<29

Với mọi a,b >=0 ta có:

\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge2ab\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Thay \(a=\sqrt{29-x};b=\sqrt{x+3}\)ta có:

\(\sqrt{29-x}+\sqrt{x+3}\le\sqrt{2\left(29-x+x+3\right)}=8\)

\(x^2-26x+177=\left(x-13\right)^2+8\ge8\)

\(\Rightarrow\sqrt{29-x}+\sqrt{x+3}\le x^2-26x+177\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{29-x}=\sqrt{x+3}\\x-13=0\end{cases}\Leftrightarrow x=13}\)

Do đó (1) <=> x=13 (tm)

b:

ĐKXĐ: \(x^2-12>=0\)

=>\(x^2\ge12\)

=>\(\left[\begin{array}{l}x\ge2\sqrt3\\ x\le-2\sqrt3\end{array}\right.\)

\(\sqrt{x^2+33}+3=2x+\sqrt{x^2-12}\)

=>\(\sqrt{x^2+33}-7=2x-8+\sqrt{x^2-12}-2\)

=>\(\frac{x^2+33-49}{\sqrt{x^2+33}+7}=2\left(x-4\right)+\frac{x^2-12-4}{\sqrt{x^2-12}+2}\)

=>\(\frac{\left(x-4\right)\left(x+4\right)}{\sqrt{x^2+33}+7}-2\left(x-4\right)-\frac{\left(x-4\right)\left(x+4\right)}{\sqrt{x^2-12}+2}=0\)

=>(x-4)\(\left(\frac{x+4}{\sqrt{x^2+33}+7}-2-\frac{x+4}{\sqrt{x^2-12}+2}\right)=0\)

=>x-4=0

=>x=4(nhận)

c:

ĐKXĐ: x>=3/2

\(3x-8\sqrt{x+14}=2\sqrt{2x-3}-28\)

=>\(3x-6-8\sqrt{x+14}+32=2\sqrt{2x-3}-2\)

=>\(3\left(x-2\right)-8\cdot\frac{x+14-16}{\sqrt{x+14}+4}=2\cdot\frac{2x-3-1}{\sqrt{2x-3}+1}\)

=>\(\left(x-2\right)\left(3-\frac{8}{\sqrt{x+14}+4}-\frac{4}{\sqrt{2x-3}+1}\right)=0\)

=>x-2=0

=>x=2(nhận)