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a) \(\left|3x+1\right|=\left|x+1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=\sqrt{x^2}\)
\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=x\\3x-2=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(2x-3\right)^2}\)
\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\left|x^2-1\right|+\left|x+1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)
\(\Leftrightarrow\left|x-4\right|+\left|x+2\right|=0\)
⇒ vô nghiệm
a: =>(x^2+4x-5)(x^2+4x-21)=297
=>(x^2+4x)^2-26(x^2+4x)+105-297=0
=>x^2+4x=32 hoặc x^2+4x=-6(loại)
=>x^2+4x-32=0
=>(x+8)(x-4)=0
=>x=4 hoặc x=-8
b: =>(x^2-x-3)(x^2+x-4)=0
hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)
c: =>(x-1)(x+2)(x^2-6x-2)=0
hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)
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...
a/ ĐKXĐ: \(x\ge\frac{3}{4}\)
\(\Leftrightarrow6x+1+2\sqrt{5x^2+5x}=6x+1+2\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)
\(\Leftrightarrow3x^2+5x-12=0\Rightarrow\left[{}\begin{matrix}x=-3< \frac{3}{4}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
b/ \(\Leftrightarrow x^2+x+1+2\sqrt{x^2+x+1}-3=0\)
Đặt \(\sqrt{x^2+x+1}=t>0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+1}=1\)
\(\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a) \(3x^3+6x^2-4x=0\) \(\Leftrightarrow\) \(x\left(3x^2+6x-4\right)=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{3}\\x=\dfrac{-3-\sqrt{21}}{3}\end{matrix}\right.\end{matrix}\right.\)
vậy phương trình có 2 nghiệm \(x=0;x=\dfrac{-3+\sqrt{21}}{3};x=\dfrac{-3-\sqrt{21}}{3}\)
\(a,3x^3+6x^2-4x=0\)
\(\Leftrightarrow x\left(3x^2+6x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x^2+6x-4=0\left(1\right)\end{cases}}\)
\(\Delta_{\left(1\right)}=36+4\cdot3\cdot4=84>0\)
\(\text{\Rightarrow pt có 2 nghiệm phân biệt}\)
\(x_1=\frac{-3+\sqrt{21}}{3};x_2=\frac{-3-\sqrt{21}}{3}\)
\(\text{Vậy phương trình đã cho bằng 0 khi x=0 hoặc x= }\frac{-3\pm\sqrt{21}}{3}\)
6x2-4x2-1=0
2x2-1=0
2x2=1
X=1/4
\(a,3\left(x^2+x^2\right)-2\left(x^2+x\right)-1=0\)
\(\Leftrightarrow4x^2-2x-1=0\)
\(\Delta^'=1+4=5\)
vì \(\Delta^'>0=>\)phường trình có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{1+\sqrt{5}}{4}\\x_2=\dfrac{1-\sqrt{5}}{4}\end{matrix}\right.\)
b, \(\left(x^2-4x+2\right)^2+x^2-4x-4=0\)
\(\Leftrightarrow x^4-8x^3+20x^2-16x+4+x^2-4x-4=0\)
\(\Leftrightarrow x^4-8x^3+21x^2-20x=0\)