a/ ĐKXĐ: \(x\ge-\frac{5}{2}\)

\(\sqrt{2x+5}=5\Rightarrow2x+5=25\Rightarrow x=10\)

b/ \(\sqrt{x-7}+3=0\)

Do \(\sqrt{x-7}\ge0\Rightarrow\sqrt{x-7}+3>0\Rightarrow ptvn\)

c/ ĐKXĐ: \(x\ge0\)

\(\sqrt{3x}=\sqrt{10}-1\Rightarrow3x=11-2\sqrt{10}\Rightarrow x=\frac{11-2\sqrt{10}}{3}\)

d/ \(4-7x=11\Rightarrow-7x=7\Rightarrow x=-1\)

d)

ĐK: $x\leq \frac{16}{7}$

PT $\Rightarrow 16-7x=11^2=121$

$\Rightarrow 7x=16-121=-105$

$\Leftrightarrow x=-15$ (thỏa mãn)

e) ĐK: $x\geq 3$

PT $\Rightarrow 10(x-3)=30$ (bình phương 2 vế)

$\Leftrightarrow x-3=3\Leftrightarrow x=6$

(thỏa mãn)

f)

ĐK: $x\geq 2$

PT $\Leftrightarrow \sqrt{x-2}=6$

$\Rightarrow x-2=6^2=36\Leftrightarrow x=38$ (thỏa mãn)

a)

ĐK: $x\geq \frac{-5}{2}$

PT $\Rightarrow 2x+5=25$ (bình phương 2 vế)

$\Leftrightarrow 2x=10\Leftrightarrow x=5$ (thỏa mãn)

b) ĐK: $x\geq \frac{-1}{3}$

PT $\Rightarrow 3x+1=10$ (bình phương 2 vế)

$\Leftrightarrow 3x=9\Leftrightarrow x=3$ (thỏa mãn)

c)

ĐK: $x\geq 7$

PT $\Leftrightarrow \sqrt{x-7}=3+0=3$

$\Rightarrow x-7=3^2$

$\Leftrightarrow x=16$ (thỏa mãn)

a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)

\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)

=> ptvn

d) ĐK : \(x^2+7x+7\ge0\)

Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)

\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)

\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)

\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )

\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )

f) ĐK : \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :

\(a+b-ab-1=0\)

\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)

\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)

\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)

đặt\(x^2+x+1=t\left(t>0\right)\)

\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)

bình phương 2 vế pt trở thành:

\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)

\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m

vậy pt vô nghiệm

a/ ĐKXĐ: ...

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)

\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)

\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))

\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)

\(\Leftrightarrow11a^2+6a-25=0\)

Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó

b/

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)

\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)

\(\Leftrightarrow\sqrt{a^2+3a}=2\)

\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+1=1\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\)

\(\Rightarrow3x^2-5x+7=3x^2-7x+11-6\sqrt{3x^2-7x+2}\)

\(\Leftrightarrow3\sqrt{3x^2-7x+2}=2-x\) (\(x\le2\))

\(\Leftrightarrow9\left(3x^2-7x+2\right)=x^2-4x+4\)

\(\Leftrightarrow26x^2-59x+14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)

Do biến đổi ko tương đương nên cần thay lại nghiệm vào pt ban đầu kiểm tra

d/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{x^2+3x+2}+\sqrt{x^2+6x+5}=\sqrt{2x^2+9x+7}\)

\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)

Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen

help me, pleaseee

Cần gấp lắm ạ!

a/ ĐKXĐ: \(x^2+3x+2\ge0\)

\(\Leftrightarrow3-2\sqrt{x^2+3x+2}=1-2\sqrt{x^2-x+1}\)

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

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

\(\Leftrightarrow2x=\sqrt{x^2-x+1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=x^2-x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3x^2+x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{-1+\sqrt{13}}{6}\\x=\frac{-1-\sqrt{13}}{6}\left(l\right)\end{matrix}\right.\)

b/ ĐKXĐ: \(3x^2-7x+2\ge0\)

\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\) (1)

\(\Rightarrow3x^2-5x+7=9+3x^2-7x+2-6\sqrt{3x^2-7x+2}\)

\(\Rightarrow2-x=3\sqrt{3x^2-7x+2}\) (\(x\le2\))

\(\Rightarrow\left(2-x\right)^2=9\left(3x^2-7x+2\right)\)

\(\Rightarrow x^2-4x+4=27x^2-63x+18\)

\(\Rightarrow26x^2-59x+14=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)

Do bước biến đổi thứ 2 ko phải phép tương đương nên cần thay 2 nghiệm vào (1) để kiểm tra lại, bạn tự thay nhé

c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1\\x\le-5\end{matrix}\right.\)

\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)

d/ ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\1\le x\le5\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x^2-1}+\sqrt{5-x}=\sqrt{x^2+3}\)

\(\Leftrightarrow x^2-x+4+2\sqrt{\left(x^2-1\right)\left(5-x\right)}=x^2+3\)

\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(5-x\right)}=x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\4\left(x^2-1\right)\left(5-x\right)=\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)\left[4\left(x+1\right)\left(5-x\right)-x+1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-4x^2+15x+21=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{15+\sqrt{561}}{8}\\x=\frac{15-\sqrt{561}}{8}\left(l\right)\end{matrix}\right.\)

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

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

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

a/ Giải rồi

b/ ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)

Pt trở thành:

\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)

\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)

\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)

\(\Leftrightarrow...\)

c/ Vế phải là \(181-4x\) hay \(18-14x\)?

d/ ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t>0\)

\(\Rightarrow t^2=2x+2\sqrt{x^2-16}\)

Pt trở thành:

\(\frac{t}{2}=\frac{t^2}{2}-6\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x+2\sqrt{x^2-16}=16\)

\(\Leftrightarrow\sqrt{x^2-16}=8-x\left(x\le8\right)\)

\(\Leftrightarrow x^2-16=64-16x+x^2\)

\(\Rightarrow x=...\)

e/ ĐKXD: \(x>0\)

\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)

\(\Rightarrow t^2=x+\frac{1}{4x}+1\)

Pt trở thành:

\(5t=2\left(t^2-1\right)+4\)

\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)

\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)