ĐK: \(\left\{{}\begin{matrix}x\ge-4\\2x^2+8x+6\ge0\end{matrix}\right.\). Nhân hai vế với 4, pt tương đương với:

\(8x^2+32x+24=\sqrt{8x+32}\). Bình phương hai vế và biến đổi,pt tương đương với:

\(8\left(2x^2+7x+4\right)\left(4x^2+18x+17\right)=0\)

Và giải pt bậc 2:)

Em ko chắc đâu nha:)

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

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

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

e/ ĐKXĐ: ...

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

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

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

Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)

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

\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)

f/ ĐKXĐ: ...

\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)

Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

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

\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)

Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ

Mn giúp em vs ạ! Thanks trước!

\(c,\left(x^3-3x+2\right)\sqrt{3x-2}-2x^3+6x^2-4x=0\)

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

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

\(\Rightarrow x=1\)

Hoặc là: \(\Rightarrow\left(x+2\right)\left(x-1\right)\sqrt{3x-2}-2x\left(x-2\right)=0\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Còn cần nữa không, hôm bữa chị giải ra câu a mà quên béng mất, mấy hôm lại bận làm thuyết trình Tiếng Anh nên bỏ dở.

Giờ mà cần chị cũng chỉ làm được câu a thôi '-'

Nhìn không đủ chán rồi không dám động vào

Viết đề kiểu gì v @@

À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)

a/ ĐKXĐ: ...

\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)

Đặt \(\sqrt{x^2-5x-6}=a\ge0\)

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

\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)

b/ ĐKXĐ: ...

\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)

Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)

\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)

c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)

Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)

\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)

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

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)

e/ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)

Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)

f/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)

\(\frac{1}{a}+1+a=3a^2\)

\(\Leftrightarrow3a^3-a^2-a-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)

\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)

a: ĐKXĐ: \(x^2-5x-6>=0\)

=>(x-6)(x+1)>=0

=>\(\left[{}\begin{matrix}x>=6\\x< =-1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x-2>=0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=6\\-5x-6=-4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=6\\-x=10\end{matrix}\right.\)

=>\(x\in\varnothing\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-8x+16}=4-x\)

=>\(\sqrt{\left(x-4\right)^2}=4-x\)

=>|x-4|=4-x

=>x-4<=0

=>x<=4

c: ĐKXĐ: \(x^2-2x>=0\)

=>x(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =0\end{matrix}\right.\)

\(\sqrt{x^2-2x}=2-x\)

=>\(\left\{{}\begin{matrix}x^2-2x=\left(2-x\right)^2\\x< =2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-2x=x^2-4x+4\\x< =2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x=4\\x< =2\end{matrix}\right.\Leftrightarrow x=2\left(nhận\right)\)

d: ĐKXĐ: x>=-27/2

\(\sqrt{2x+27}-6=x\)

=>\(\sqrt{2x+27}=x+6\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+6\right)^2=2x+27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\x^2+12x+36-2x-27=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\x^2+10x+9=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+9\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-6\\x\in\left\{-9;-1\right\}\end{matrix}\right.\)

=>x=-1

Kết hợp ĐKXĐ, ta được: x=-1

a.

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x=-10\left(ktm\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

b.

\(\sqrt{x^2-8x+16}=4-x\)

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

\(\Leftrightarrow\left|x-4\right|=-\left(x-4\right)\)

\(\Leftrightarrow x-4\le0\)

\(\Rightarrow x\le4\)

c.

\(\sqrt{x^2-2x}=2-x\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x^2-2x=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\2x=4\end{matrix}\right.\)

\(\Rightarrow x=2\)

d.

\(\Leftrightarrow\sqrt{2x+27}=x+6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\x+27=\left(x+6\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x+27=x^2+12x+36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+11x+9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{85}}{2}\\x=\dfrac{-11-\sqrt{85}}{2}\left(loại\right)\end{matrix}\right.\)

a,

\(\Leftrightarrow\sqrt{1-x}=\frac{x-1}{\sqrt{6-x}+\sqrt{-5-2x}}\)

\(\Leftrightarrow-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\\-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\end{cases}}\)

b,tự nàm

c,

\(\Leftrightarrow64x^2-64x-64=64\sqrt{8x+1}\)

\(\Leftrightarrow\left(8x+1\right)^2=10\left(8x+1\right)+64\sqrt{8x+1}+55\)

đặt \(\sqrt{8x+1}=a\)

=>a⁴=10a²+64a+55

nhận thấy phương trình có dạng x⁴=ax²+bx+c

tìm số m sao cho b²-4(2m+a)(m²+c)=0

sau đó đưa về (x²+m)²=k² với k là 1 số bất kì,sau đó giải ra

b)đk \(x\ge1\)

 \(\sqrt{1+x^2+\frac{x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}=\sqrt{\frac{\left(x+1\right)^2+x^2.\left(x+1\right)^2+x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)

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

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

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

\(\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=2013\)

\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2013\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2013\)

\(\Leftrightarrow x+\left|x-2\right|=2014\)

giai 2 pt 

pt1 x+x-2=2014

x=1008

pt2 x+2-x=2014(vô lý)

ý c cái a đặt có 1 nghiệm là a=5 đó