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Giải các phương trình:
\(a,\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(b,x^4-30x^2+31x-30=0\)
a, Đặt \(x^2-5x=a\)
\(\Rightarrow\)\(a^2+10a+24=0\)
\(\Rightarrow a^2+4a+6a+24=0\)
\(\Rightarrow\left(a+4\right)\left(a+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+4=0\\a+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2-5x+4=0\left(1\right)\\x^2-5x+6=0\left(2\right)\end{cases}}}\)
Giải pt (1) ta có : \(x^2-5x+4=0\)
\(\Rightarrow x^2-4x-x+4=0\)
\(\Rightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Giải pt (2) ta có : \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy \(S=\left\{1;2;3;4\right\}\)
\(x^4-30x^2+31x-30=0\)
\(\Rightarrow x^4-30x^2+x+30x-30=0\)
\(\Rightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
Mà \(x^2-x+1>0\)với \(\forall\)\(x\)
\(\Rightarrow x^2+x-30=0\)
\(\Rightarrow x^2-5x+6x-30=0\)
\(\Rightarrow x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Vậy \(S=\left\{5;-6\right\}\)
Bài 4 :
24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ
Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0
Suy ra quãng đường AB là 36x(km)
Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)
Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)
Ta có phương trình:
\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)
Vậy quãng đường AB dài 36.2 = 72(km)
a) ta có : \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left(x\left(x-2\right)-3\left(x+2\right)\right)\left(x\left(x-1\right)-4\left(x-1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\) vậy \(x=1;x=2;x=3;x=4\)
b) ta có : \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)=0\)
\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x+1-3\right)=0\)
ta có : \(x^2+x+5>0\forall x\)
\(\Rightarrow pt\Leftrightarrow x^2+x-2=0\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\) vậy \(x=1;x=-2\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
\(\left(x^2+7x+12\right).\left(4x-16\right)-\left(x+3\right)\left(x^2-5x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^2+3x+4x+12\right).4.\left(x-4\right)-\left(x+3\right)\left(x^2-x-4x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow4\left(x+4\right)\left(x+3\right)\left(x-4\right)-\left(x+3\right)\left(x-4\right)\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(4-x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(8-x\right)=0\)
\(\Leftrightarrow\frac{\orbr{\begin{cases}x+4=0\\x-4=0\end{cases}}}{\orbr{\begin{cases}x+3=0\\8-x=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-4\\x=4\end{cases}}}{\orbr{\begin{cases}x=-3\\x=8\end{cases}}}\)
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