a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

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

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)

Thấy : \(x^2-4x+16=\left(x-2\right)^2+12>0\forall x\)

P/t \(\Leftrightarrow2\left(x^2-4x+16\right)-36+\sqrt{x^2-4x+16}=0\)

Đặt \(t=\sqrt{x^2-4x+16}>0\) ; khi đó : 

\(2t^2+t-36=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-\dfrac{9}{2}\left(L\right)\end{matrix}\right.\)

Với t = 4  hay \(\sqrt{x^2-4x+16}=4\Leftrightarrow x^2-4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy ... 

Câu 1 bạn trên giải rồi mik k giải nx nha

2/ \(3\left(x^2+2\right)=10\sqrt{x^3+1}\)

\(3\left(x^2-x+1\right)+3\left(x+1\right)=10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b\ge0\end{matrix}\right.\)

pt⇔ \(3a^2+3b^2-10ab=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3b=b\\a=3b\end{matrix}\right.\)

Đến đây bạn tự giải tiếp nha

3/ \(\sqrt{3-3x}-\sqrt{3+x}=2\)

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

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

+) \(x=-2\left(TM\right)\)

+) \(x\ne-2\Rightarrow\dfrac{-3}{\sqrt{3-3x}+3}-\dfrac{1}{\sqrt{3+x}+1}=0\)

Vì VT<0 => ptvn

2 ) ĐK : \(x\ge-1\)

P/t \(\Leftrightarrow9\left(x^2+2\right)^2=100\left(x^3+1\right)\)

\(\Leftrightarrow9x^4+36x^2+36=100x^3+100\)

\(\Leftrightarrow9x^4-100x^3+36x^2-64=0\)

\(\Leftrightarrow\left(x^2-10x-8\right)\left(9x^2-10x+8\right)=0\)

\(\Leftrightarrow x^2-10x-8=0\) ( 9x^2 - 10x + 8 > 0 )

\(\Leftrightarrow x=5\pm\sqrt{33}\) ( t/m ) 

Vậy ... 

Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$

PT $\Leftrightarrow 5x-2=7^2=49$

$\Leftrightarrow 5x=51$

$\Leftrightarrow x=\frac{51}{5}=10,2$

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$

$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$

$\Leftrightarrow 8\sqrt{x-3}=24$

$\Leftrightarrow \sqrt{x-3}=3$

$\Leftrightarrow x-3=9$

$\Leftrightarrow x=12$ (tm)

Bài 1:

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$

$\Leftrightarrow (x-2)(x-3)-2.\frac{x-3}{\sqrt{x-2}+1}=0$

$\Leftrightarrow (x-3)[(x-2)-\frac{2}{\sqrt{x-2}+1}]=0$

$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$

Nếu $x-3=0$

$\Leftrightarrow x=3$ (tm) 

Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$

$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)

$\Leftrightarrow a^3+a^2-2=0$

$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$

$\Leftrightarrow (a-1)(a^2+2a+2)=0$

Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$

$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)

Vậy pt có nghiệm duy nhất $x=3$.

Bài 2:

ĐKXĐ: $x\geq 0; x\neq 4$

\(A=\frac{\sqrt{x}(\sqrt{x}-2)-\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)\sqrt{x}-2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-2\sqrt{x}}{\sqrt{x}-2}=\frac{2\sqrt{x}}{2-\sqrt{x}}\)

\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)

a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)

b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)

\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

1. Ta có \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=0\)

\(\Rightarrow\)\(\left[\left(x+2\right)\left(x+8\right)\right].\left[\left(x+4\right)\left(x+6\right)\right]+16=0\)

\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)

Đặt \(x^2+10x=t\)

Pt \(\Leftrightarrow\left(t+16\right)\left(t+24\right)+16=0\Leftrightarrow t^2+40t+400=0\Leftrightarrow t=-20\)

\(\Rightarrow x^2+10x+20=0\Rightarrow\orbr{\begin{cases}x=-5+\sqrt{5}\\x=-5-\sqrt{5}\end{cases}}\)

2. Ta có \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Rightarrow\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24=0\)\(\Rightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

Đặt \(x^2+7x=t\Rightarrow\left(t+10\right)\left(t+12\right)-24=0\Rightarrow t^2+22t+96=0\)\(\Rightarrow\orbr{\begin{cases}t=-6\\t=-16\end{cases}}\)

Với \(t=-6\Rightarrow x^2+7x+6=0\Rightarrow\orbr{\begin{cases}x=-6\\x=-1\end{cases}}\)

Với \(t=-16\Rightarrow x^2+7x+16=0\left(l\right)\)

Vậy pt có 2 nghiệm là \(\orbr{\begin{cases}x=-6\\x=-1\end{cases}}\)

Quản lí Hoàng Thị Lan Hương giúp em giải bài toán vừa đăng lên đc ko ạ.??? ^^

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1