sorry b, phải là cái này nha

T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(

\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)

\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)

\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)

\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)

\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

x = - 0 , 3212201247

nhân ra sau đó giải bình thừơng

\(\left(x^2+3x+2\right)\left(x^2+9x+18\right)=168x^2\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=168x^2\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=168x^2\)

\(\Leftrightarrow\left(x^2+7x+6\right)\left(x^2+5x+6\right)=168x^2\)(1)

Đặt \(x^2+5x+6=t\)

Khi đó (1) trở thành: \(\left(t+2x\right)t=168x^2\Leftrightarrow t^2+2xt-168x^2=0\)

\(\Leftrightarrow\left(t-12x\right)\left(t+14x\right)=0\Leftrightarrow\orbr{\begin{cases}t=12x\\t=-14x\end{cases}}\)

TH1: \(t=12x\Rightarrow x^2+5x+6=12x\)

\(\Leftrightarrow x^2-7x+6=0\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)

TH2: \(t=-14x\Rightarrow x^2+5x+6=-14x\Rightarrow x^2+19x+6=0\)

\(\Leftrightarrow x^2+2.x.\frac{19}{2}+\left(\frac{19}{2}\right)^2-\frac{337}{4}=0\)

\(\Leftrightarrow\left(x+\frac{19}{2}\right)^2=\frac{337}{4}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{337}-19}{2}\\x=\frac{-\sqrt{337}-19}{2}\end{cases}}\)

a/ ĐKXĐ: \(x^2+2x-6\ge0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-6}=0\left(1\right)\\\sqrt{x^2+2x-6}=2-x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2+2x-6=0\Rightarrow x=-1\pm\sqrt{7}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\6x=10\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{3}\)

Câu b nhìn ko ra hướng, ko biết đề có nhầm đâu ko :(

c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)

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

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

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

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2+x+2=\left(3-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\7x=7\end{matrix}\right.\) \(\Rightarrow x=1\)

d/

Ta có \(\sqrt{x^2+3x+4}=\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{7}{4}}>1\)

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

Nhân 2 vế của pt với \(\sqrt{x^2+3x+4}-1\)

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

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

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

\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x^2+3x+4=\left(3x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow8x^2+3x-3=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{105}}{6}\\x=\frac{-3-\sqrt{105}}{6}\left(l\right)\end{matrix}\right.\)

e/ ĐKXĐ: \(3x^2-9x+1\ge0\)

\(\Leftrightarrow3x^2-9x+1-x^2=2\left(\sqrt{3x^2-9x+1}+x\right)\)

\(\Leftrightarrow\left(\sqrt{3x^2-9x+1}+x\right)\left(\sqrt{3x^2-9x+1}+x\right)=2\left(\sqrt{3x^2-9x+1}+x\right)\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{3x^2-9x+1}+x=0\left(1\right)\\\sqrt{3x^2-9x+1}-x=2\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{3x^2-9x+1}=-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\3x^2-9x+1=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le0\\2x^2-9x+1=0\end{matrix}\right.\) \(\Rightarrow x=\frac{9\pm\sqrt{73}}{4}\left(l\right)\)

\(\left(2\right)\Leftrightarrow\sqrt{3x^2-9x+1}=x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\3x^2-9x+1=\left(x+2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\2x^2-13x-3=0\end{matrix}\right.\)

\(\Rightarrow x=\frac{13\pm\sqrt{193}}{4}\)

=33 nha mn

Help me
Cần gấp trong hôm nay

a,\(\left(2x-3\right)^2=\left(x+1\right)^2\)

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

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

\(\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Vậy...

b,\(\left(x+2\right)\left(5-3x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-3x\right)-\left(x+2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-4x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy...