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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) \(3x^3+6x^2-4x=0\) \(\Leftrightarrow\) \(x\left(3x^2+6x-4\right)=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{3}\\x=\dfrac{-3-\sqrt{21}}{3}\end{matrix}\right.\end{matrix}\right.\)
vậy phương trình có 2 nghiệm \(x=0;x=\dfrac{-3+\sqrt{21}}{3};x=\dfrac{-3-\sqrt{21}}{3}\)
1. \(x^3-6x^2+10x-4=0\)
<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
<=> \(\left(x-2\right)\left(x^2-4x+2\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)
Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)
=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)
\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)
1) Ta có: \(x^3-6x^2+10x-4=0\)
\(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)
+ \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
+ \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=2\)
\(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,5858;2;3,4142\right\}\)
làm tạm câu này vậy
a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)
\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)
\(\Leftrightarrow\left\{\left(x^2-x+1\right)^2+2x^2\right\}=\left(3x^2\right)^2\)
\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)
\(\Leftrightarrow\left(x^2-x+1\right)=x^2\)
\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)
Vậy...
\(\left(x-2\right)\left(x^2+6x-11\right)^2=\left(5x^2-10x+1\right)^2\) \(\Rightarrow x>2\)
\(\Rightarrow x^2+6x-11>0\)
\(pt\Leftrightarrow x-2=\left(\dfrac{5x^2-10x+1}{x^2+6x-11}\right)^2\Leftrightarrow\sqrt{x-2}=\dfrac{5x^2-10x+1}{x^2+6x-11}\)
\(\Leftrightarrow\sqrt{x-2}-1=\dfrac{5x^2-10x+1}{x^2+6x-11}-1=\dfrac{4x^2-16x+12}{x^2+6x+12}\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{x-2}+1}=\dfrac{4\left(x-1\right)\left(x-3\right)}{x^2+6x-11}\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\\dfrac{1}{\sqrt{x-2}+1}=\dfrac{4\left(x-1\right)}{x^2+6x-11}\left(1\right)\end{matrix}\right.\)
Xét (1):
\(x^2+6x-11=4\left(x-1\right)+4\left(x-1\right)\sqrt{x-2}\)
\(\Leftrightarrow x^2+2x-7-4\left(x-1\right)\sqrt{x-2}=0\)
\(\Leftrightarrow x^2-2x+1-2\left(x-1\right)\sqrt{4x-8}+4x-8=0\)
\(\Leftrightarrow\left(x-1\right)^2-2\left(x-1\right)\sqrt{4x-8}+\left(\sqrt{4x-8}\right)^2=0\)
\(\Leftrightarrow\left(x-1-\sqrt{4x-8}\right)^2=0\)
\(\Leftrightarrow x-1=\sqrt{4x-8}\)
\(\Leftrightarrow x^2-2x+1=4x-8\)
\(\Leftrightarrow\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy pt đã cho có nghiệm duy nhất \(x=3\)
Đặt \(y=x-2\), phương trình đã cho trở thành:
\( y{\left[ {{{\left( {y + 2} \right)}^2} + 6\left( {y + 2} \right) - 11} \right]^2} = {\left[ {5{{\left( {y + 2} \right)}^2} - 10\left( {y + 2} \right) + 1} \right]^2}\\ \Leftrightarrow y{\left( {{y^2} + 10y + 5} \right)^2} = {\left( {5{y^2} + 10y + 1} \right)^2}\\ \Leftrightarrow {y^5} - 5{y^4} + 10{y^3} - 10{y^2} + 5y - 1 = 0 \Leftrightarrow {\left( {y - 1} \right)^5} = 0 \Leftrightarrow y = 1 \)
Với \(y=1\) ta có \(x-2=1\) \(\Rightarrow x=3\)
Vậy \(x = 3 \)