Bình phương đi bạn

TL:

1đk:x<1

.\(1+3x-1=9x^2\) 

     \(3x=9x^2\) 

   x=3x\(^2\) 

 =>x=0(ktm)  hoặc  x= \(\frac{1}{3}\left(tm\right)\) 

vậy x=\(\frac{1}{3}\) 

hc tốt:)

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: ...

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

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

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

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

a,\(1+\sqrt{3x+1}=3x\)(ĐK:\(x>-\frac{1}{3}\))

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

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\)

\(\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(tm\right)\end{cases}}\)

b,\(\sqrt{2+\sqrt{3x-5}}=\sqrt{x+1}\)(ĐK:\(x>-\frac{5}{3}\))

\(\Leftrightarrow2+\sqrt{3x-5}=x+1\)

\(\Leftrightarrow2+3x-5+2.2\sqrt{3x-5}=x+1\)

\(\Leftrightarrow3x-3-x-1=4\sqrt{3x-5}\)

\(\Leftrightarrow2x-4=4\sqrt{3x-5}\)

\(\Leftrightarrow4x^2-16x+16=48x-80\)

\(\Leftrightarrow4x^2-64x-64=0\)

\(\Delta=64^2-4.\left(-64\right)=4352\)

\(\orbr{\begin{cases}x_1=\frac{64-\sqrt{4352}}{8}=8-2\sqrt{17}\left(tm\right)\\x_2=\frac{64+\sqrt{4352}}{8}=8+2\sqrt{17}\left(tm\right)\end{cases}}\)

c,Cho biểu thức trong căn nhận giá trị 16 mà giải

CẢm ơn bạn nhé !

a,ĐK: x≥4

Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}=4\)

      \(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)

b, ĐK: x≥2

Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)

      \(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)

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

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

help me, pleaseee

Cần gấp lắm ạ!