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a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
\(b,x^2+3x-2=0\\ \Delta=3^2-4.1.\left(-2\right)=17\\ =>\left[{}\begin{matrix}x_1=\dfrac{-3+\sqrt{17}}{2}\\x_2=\dfrac{-3-\sqrt{17}}{2}\end{matrix}\right.\)
Mấy câu còn lại mình giải rồi
a, \(2+\sqrt{3x+4}=x\)(ĐKXĐ: \(x>\frac{3}{4}\))
\(\Leftrightarrow\sqrt{3x+4}=x-2\)
\(\Leftrightarrow\left(\sqrt{3x+4}\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow3x+4=x^2-4x+4\)
\(\Leftrightarrow x^2-4x+4-3x-4=0\)
\(\Leftrightarrow x^2-7x=0\)
\(\Leftrightarrow x\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(L\right)\\x=7\left(TM\right)\end{cases}}}\)
Vậy PT có nghiệm là \(x=7\)
b, \(\sqrt{4x^2-4x+1}-\sqrt{9x^2}=0\)
\(\Leftrightarrow\sqrt{4x^2-4x+1}=\sqrt{9x^2}\)
\(\Leftrightarrow\left(\sqrt{4x^2-4x+1}\right)^2=\left(\sqrt{9x^2}\right)^2\)
\(\Leftrightarrow4x^2-4x+1=9x^2\)
\(\Leftrightarrow9x^2-4x^2+4x-1=0\)
\(\Leftrightarrow5x^2+4x-1=0\)
\(\Leftrightarrow\left(x-\frac{1}{5}\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{5}=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)
Vậy PT có nghiệm là \(x\in\left\{-1;\frac{1}{5}\right\}\)
a) \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
Vậy...
a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)
b. \(\Leftrightarrow x^3+x+3x^2+3=0\)
\(\Leftrightarrow x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+1=0\left(vn\right)\end{matrix}\right.\)
c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)
d.
\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)-3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x-2=0\end{matrix}\right.\)
e.
\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)
\(\Leftrightarrow x\left(x^2+2x+1\right)+3\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)
T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(
\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)
\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)
\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)
\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)
\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
a. Dat \(x^2=t\left(t\ge0\right)\)
Suy ra PT:\(\orbr{\begin{cases}t^2=-4t+1\left(1\right)\left(x< 0\right)\\t^2=4t+1\left(2\right)\left(x\ge0\right)\end{cases}}\)
(1)\(\Leftrightarrow t^2+4t-1=0\)
\(\Leftrightarrow\left(t+2\right)^2-5=0\)
\(\Leftrightarrow\left(t+2+\sqrt{5}\right)\left(t+2-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=-2-\sqrt{5}\left(l\right)\\t=\sqrt{5}-2\left(n\right)\end{cases}}\)
Nghiem cua PT(1) la \(t=\sqrt{5}-2\)
(2)\(\Leftrightarrow t^2-4t-1=0\)
\(\Leftrightarrow\left(t-2\right)^2-5=0\)
\(\Leftrightarrow\left(t-2+\sqrt{5}\right)\left(t-2-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{5}\left(l\right)\\t=2+\sqrt{5}\left(n\right)\end{cases}}\)
Nghiem cua PT(2) la \(t=2+\sqrt{5}\)
Suy ra:\(\orbr{\begin{cases}x=\sqrt{\sqrt{5}-2}\\x=\sqrt{\sqrt{5}+2}\end{cases}}\)
b.\(x^3-3x^2+9x-9=0\)
\(\Leftrightarrow\left(x-3\right)^3=-18\)
\(\Leftrightarrow x-3=-\sqrt[3]{18}\)
\(\Leftrightarrow x=3-\sqrt[3]{18}\)
\(b,x^3-3x^2+9x-9=0\)
\(\Rightarrow x^2\left(x-3\right)+9\left(x-3\right)+18=0\)
\(\Rightarrow\left(x^2+9\right)\left(x-3\right)=-18\)
từ đây bạn xét các TH nhá !
Chú ý : Vì \(x^2+9\ge9\forall\) để xét ít Th hơn