1.

\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)

\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)

\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

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

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

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

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

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

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

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

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

\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)

\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)

\(\Leftrightarrow x^2-5x+2=0\)

b) Đặt \(u=\sqrt{1-x}\); \(v=\sqrt{1+x}\)

phương trình trở thành

\(2u-v+3uv=u^2+2\)\(\Rightarrow u^2-2u+v-3uv+2=0\)

lại có \(u^2+v^2=2\)

\(\Rightarrow u^2-2u-3uv+v+u^2+v^2=0\)

\(\Rightarrow\left(u-v-1\right)\left(2u-v\right)=0\)

đến đây thì easy rồi

a)

Đặt \(\sqrt{2x+1}=t\) ;\(\sqrt{x}=k\)

Phương trình trở thành

\(\left(3k^2+t^2\right)t-\left(3t^2+k^2\right)k-1=0\)

\(\Leftrightarrow3k^2t+t^3-3t^2k-k^3-1=0\)

\(\Leftrightarrow\left(t-k\right)\left(t^2+kt+k^2\right)-3tk\left(t-k\right)-1=0\)

\(\Leftrightarrow\left(t-k\right)^3-1=0\)

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

do t > k => t - k > 0

\(\Rightarrow\left(t-k\right)^2+t-k+1>0\)

\(\Rightarrow t-k-1=0\)

\(\Leftrightarrow t=1+k\)\(\Leftrightarrow\sqrt{2x+1}=1+\sqrt{x}\)

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

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

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

END

a)\(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)

ĐK:\(x\ge 0\)

\(\Leftrightarrow\left(5x+1\right)\sqrt{2x+1}-\left(\dfrac{31}{2}x+1\right)-\left(\left(7x+3\right)\sqrt{x}-\dfrac{31}{2}x\right)=0\)

\(\Leftrightarrow\dfrac{\left(5x+1\right)^2\left(2x+1\right)-\left(\dfrac{31}{2}x+1\right)^2}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{x\left(7x+3\right)^2-\left(\dfrac{31}{2}x\right)^2}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)

\(\Leftrightarrow\dfrac{\dfrac{1}{4}x\left(200x+19\right)\left(x-4\right)}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{\dfrac{1}{4}x\left(x-4\right)\left(196x-9\right)}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)

\(\Leftrightarrow\dfrac{1}{4}x\left(x-4\right)\left(\dfrac{200x+19}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{196x-9}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}\right)=0\)

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

Nghe t đi phần nào khó cho qua :)) b tương tự

ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

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

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

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

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

\(\Leftrightarrow a^2-3a+2+ab-b=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

\(ĐK:-5\le x\le3\)

Đặt \(\sqrt{x+5}+\sqrt{3-x}=t\ge0\Leftrightarrow t^2-8=2\sqrt{15-2x-x^2}\), PTTT:

\(t-t^2+8-2=0\\ \Leftrightarrow t^2-t-6=0\\ \Leftrightarrow t=3\left(t\ge0\right)\\ \Leftrightarrow2\sqrt{15-2x-x^2}=3^2-8=1\\ \Leftrightarrow60-8x-4x^2=1\\ \Leftrightarrow4x^2+8x-59=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+3\sqrt{7}}{2}\left(tm\right)\\x=\dfrac{-2-3\sqrt{7}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm pt là ...

ĐKXĐ: \(x\in R\)

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

=>\(\left(1-2x\right)\cdot\sqrt{x^2+1}+2x-1-2x^2-9x=0\)

=>\(\left(1-2x\right)\left(\sqrt{x^2+1}-1\right)-x\left(2x+9\right)=0\)

=>\(\left(1-2x\right)\cdot\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}-x\left(2x+9\right)=0\)

=>\(\left(1-2x\right)\cdot\dfrac{x^2}{\sqrt{x^2+1}+1}-x\left(2x+9\right)=0\)

=>\(x\left[\dfrac{x\cdot\left(1-2x\right)}{\sqrt{x^2+1}+1}-2x-9\right]=0\)

=>x=0(nhận)

Gpt là j

giải phương trình

ráng làm nốt rồi đi ngủ thoyy

1.

a) ĐK: \(x\ge2\)

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

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

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)

Vậy...

b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)

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

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

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

\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)

Vậy...

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

Nhân cả 2 vế với \(\sqrt{2}\) ta được :

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

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

\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)

Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)

\(=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)

2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)

\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)

Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)

Tương tự 2 trường hợp còn lại ta đều được \(B=0\)

Vậy \(B=0\)