a: ĐKXĐ: \(\begin{cases}5x^2+14x+9\ge0\\ x^2-x-20\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}\left(x+1\right)\left(5x+9\right)\ge0\\ \left(x-5\right)\left(x+4\right)\ge0\\ x\ge-1\end{cases}\)

=>x>=5

TA có: \(\sqrt{5x^2+14x+9} \le 5\sqrt{x+1} + \sqrt{x^2-x-20}\)

=>\(5x^2+14x+9 \le 25(x+1) + x^2-x-20 + 10\sqrt{(x+1)(x^2-x-20)}\)

=>\(5x^2+14x+9 \le x^2 + 24x + 5 + 10\sqrt{(x+1)^2(x-5)}\)

=>\(4x^2 - 10x + 4 \le 10(x+1)\sqrt{x-5}\)

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

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

Đặt \(t=\sqrt{x-5}\ge0\implies x=t^2+5\)

(1) sẽ trở thành: \(2(t^2+5)^2 - 5(t^2+5) + 2 \le 5(t^2+6)t\)

=>\(2(t^4 + 10t^2 + 25) - 5t^2 - 25 + 2 \le 5t^3 + 30t\)

=>\(2t^4 + 20t^2 + 50 - 5t^2 - 23 \le 5t^3 + 30t\)

=>\(2t^4 - 5t^3 + 15t^2 - 30t + 27 \le 0\)

=>\((t-1)(2t-3)(t^2 + 6) \le 0\)

=>(t-1)(2t-3)<=0

=>1<=t<=3/2

=>\(1\le\sqrt{x-5}\le\frac{3}{2}\)

=>\(1\le x-5\le\frac{9}{4}\)

\(\iff6\le x\le\frac{29}{4}\)

ĐKXĐ: \(x\in R\)

\(3x^2-5x+6=2x\cdot\sqrt{x^2-x+2}\)

=>\(3x^2-6x+x-2+8=2\cdot\sqrt{x^4-x^3+2x^2}\)

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

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

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

=>\(\left(x-2\right)\left(3x+1\right)=\dfrac{2\left(x-2\right)\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left[\left(3x+1\right)-\dfrac{2\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\right]=0\)

=>x-2=0

=>x=2(nhận)

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

\(\Leftrightarrow\left[x^2-2x\sqrt{x^2-x+2}+\left(x^2-x+2\right)\right]+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x^2-x+2}\right)^2+\left(x-2\right)^2=0\)

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

Thử lại ta thấy nghiệm \(x=2\) thỏa phương trình ban đầu.

ĐKXĐ: \(x\ge-\dfrac{4}{5}\)

Đặt \(\sqrt{5x+4}=t\ge0\Rightarrow x=\dfrac{t^2-4}{5}\)

Pt trở thành:

\(\dfrac{t^2-4}{5}-t=2\)

\(\Leftrightarrow t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{5x+4}=7\)

\(\Rightarrow5x+4=49\)

\(\Rightarrow x=9\)

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

\(PT\Leftrightarrow6\left(x+\sqrt{6x^2+6}\right)=-5x^2-2\sqrt{5}x-1\)

\(\Leftrightarrow6\left(x+\sqrt{6x^2+6}\right)=-\left(\sqrt{5}x+1\right)^2\)

\(\Rightarrow x+\sqrt{6x^2+6}\le0\)

rồi sao nữa

a: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>=0\end{matrix}\right.\)

=>3<=x<=5

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

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

=>\(\dfrac{x-3-1}{\sqrt{x-3}+1}+\dfrac{5-x-1}{\sqrt{5-x}+1}=0\)

=>\(\left(x-4\right)\left(\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}\right)=0\)

=>x-4=0

=>x=4

ĐKXĐ: \(0\le x\le5\).

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\).

PT đã cho tương đương với: \(\left(8-ab\right)\left(a-b\right)=2\left(a-b\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=6\end{matrix}\right.\).

+) \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=2,5\left(TMĐK\right)\).

+) \(ab=6\Leftrightarrow\sqrt{x\left(5-x\right)}=6\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=3\left(TMĐK\right)\end{matrix}\right.\).

Vậy...

ĐK: \(0\le x\le5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(8-ab\right)\left(a-b\right)=2\left(a^2-b^2\right)\)

\(\Leftrightarrow\left(a-b\right)\left(8-ab-2a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab+2a+2b=8\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH2: \(ab+2a+2b=8\)

\(\Leftrightarrow\sqrt{5x-x^2}+2\sqrt{5-x}+2\sqrt{x}=8\)

\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x}-3\right)\left(\sqrt{5-x}+\sqrt{x}+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5-x}+\sqrt{x}=-7\left(l\right)\\\sqrt{5-x}+\sqrt{x}=3\end{matrix}\right.\)

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

\(\Leftrightarrow5+2\sqrt{5x-x^2}=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

Vậy ...