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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 \(S=\left\{5;-6\right\}\)
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
a) đặt \(\left(x^2+x\right)\)là \(y\)
ta có: \(3y^2-7y+4\)\(=0\)
<=>\(\left(3y-4\right)\left(y-1\right)=0\)
còn lại bạn tự xử nhé
n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)
\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{5}\)
m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)
\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)
\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=9\)
e) \(\left|5x\right|=3x-2\)
\(\Rightarrow5\cdot\left|x\right|=3x-2\)
\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
g) \(\left|-2,5x\right|=x-12\)
\(\Rightarrow2,5\cdot\left|x\right|=x-12\)
\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
a/ Đặt (x^2 - 5x) = a thì ta có
a^2 + 10a + 24 = 0
<=> (a + 4)(a + 6) = 0
Làm nốt
b/ (x - 4)(x - 5)(x - 6)(x - 7) = 1680
<=> (x - 4)(x - 7)(x - 5)(x - 6) = 1680
<=> (x^2 - 11x + 28)(x^2 - 11x + 30) = 1680
Đặt x^2 - 11x + 28 = a thì ta có
a(a + 2) = 1680
<=> (a - 40)(a + 42) = 0
Làm nốt
\(\left(x-5\right)\left(x-1\right)=2x\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-5-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy............
\(5\left(x+3\right)\left(x-2\right)-3\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow5\left(x^2+x-6\right)-3\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow2x^2-16x-60=0\)
\(\Leftrightarrow x^2-8x-30=0\)
làm tiếp nhé!!!!!
a ) \(\left(5x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=7\end{matrix}\right.\)
b ) \(\left(x^2-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-3\end{matrix}\right.\)
c )\(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d ) \(x^2+x-12=0\)
\(\Leftrightarrow x^2-4x+3x-12\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e ) \(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
g ) \(\left(x^2+1\right)\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-2\end{matrix}\right.\)
i ) \(x^4+2x^3-2x^2+2x-3=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^3-x^2+x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\\x=-3\end{matrix}\right.\)
h) \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+3x+2x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
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 \(x=1;x=-2\)