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1.
Đặt \(x^2-5x=a\Rightarrow a^2=\left(x^2-5x\right)^2\)
Thay vào pt:
\(\Rightarrow a^2+10a+24=0\)
\(\Leftrightarrow a^2+6a+4a+24=0\)
\(\Leftrightarrow a\left(a+6\right)+4\left(a+6\right)=0\)
\(\Leftrightarrow\left(a+6\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-3x-2x+6\right)\left(x^2-4x-x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-3\right)-2\left(x-3\right)\right]\left[x\left(x-4\right)-\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow x-3=0,x-2=0,x-4=0,x-1=0\)
\(\Rightarrow x=3,x=2,x=4,x=1\)
T I C K mình sẽ giải típ cho cảm ơn
Câu 1:
\((x+2)(x^2-3x+5)=(x+2)x^2\)
\(\Leftrightarrow (x+2)(x^2-3x+5)-(x+2)x^2=0\)
\(\Leftrightarrow (x+2)(x^2-3x+5-x^2)=0\)
\(\Leftrightarrow (x+2)(-3x+5)=0\Rightarrow \left[\begin{matrix} x+2=0\\ -3x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{5}{3}\end{matrix}\right.\)
Câu 2:
\(2x^2-x=3-6x\)
\(\Leftrightarrow x(2x-1)=3(1-2x)=-3(2x-1)\)
\(\Leftrightarrow x(2x-1)+3(2x-1)=0\)
\(\Leftrightarrow (2x-1)(x+3)=0\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=-3\end{matrix}\right.\)
Câu 3:
\(x^3+2x^2+x+2=0\)
\(\Leftrightarrow (x^3+2x^2)+(x+2)=0\Leftrightarrow x^2(x+2)+(x+2)=0\)
\(\Leftrightarrow (x+2)(x^2+1)=0\Rightarrow \left[\begin{matrix} x+2=0\\ x^2+1=0(\text{vô lý})\end{matrix}\right.\Rightarrow x=-2\)
Câu 5:
\(3x^2+7x-20=0\)
\(\Leftrightarrow 3x^2+12x-5x-20=0\)
\(\Leftrightarrow 3x(x+4)-5(x+4)=0\)
\(\Leftrightarrow (3x-5)(x+4)=0 \Rightarrow \left[\begin{matrix} x=\frac{5}{3}\\ x=-4\end{matrix}\right.\)
Bài 1:
1,\(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right).x^2\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-3x+5\right)-\left(x+2\right).x^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-3x+5-x^2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\dfrac{5}{3};-2\right\}\)
2,\(2x^2-x=3-6x\)
\(\Leftrightarrow2x^2-x-3+6x=0\)
\(\Leftrightarrow\left(2x^2+6x\right)-\left(x+3\right)=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\dfrac{1}{2};-3\right\}\)
3,\(x^3+2x^2+x+2=0\)
\(\Leftrightarrow x^2\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-1;-2\right\}\)
4.\(x^3+2x^2-x-2=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{1;-2\right\}\)
Nản quá không làm nữa đâu.Sorry
1: \(\Leftrightarrow\left(x+2\right)\left(x^2-3x+5-x^2\right)=0\)
=>(x+2)(-3x+5)=0
=>x=-2 hoặc x=5/3
2: \(\Leftrightarrow2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+6x-x-3=0\)
=>(x+3)(2x-1)=0
=>x=1/2 hoặc x=-3
3: \(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
=>(x+2)(x+1)(x-1)=0
hay \(x\in\left\{-2;-1;1\right\}\)
5: \(3x^2+7x-20=0\)
\(\Leftrightarrow3x^2+12x-5x-20=0\)
=>(x+4)(3x-5)=0
=>x=5/3 hoặc x=-4
a)<=>(x^2+x-3)(x^2+x-2)-12=(x-2)(x+3)(x^2+x+1)
TH1:=>x-2=0
=>x=2
TH2:x+3=0
=>x=-3
dựa vô bệt thức ta thấy
D<0=> phương trình ko có nghiệm thực
=>x=-3 hoặc 2
nhớ tick nhé
\(x\left(x-3\right)+x-3=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
KL:......................
\(x^3-5x=0\)
\(x\left(x^2-5\right)=0\)
Làm tương tự như câu a
@_@ n...h..i......ề....u q...u.....................á!
câu a bạn sai đề nha
b)
\(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)
\(x^4+x^2+1+2x^3+2x^2+2x=3x^4+3x^2+3\)
\(2\left(x^3+x^2+x\right)=2\left(x^4+x^2+1\right)\)
\(x^4-x^3+1-x=0\)
\(x^3\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^3-1\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\x^3-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
c,
<=> \(\left[\begin{matrix}x-1=0\\x^2+5x+2=0\\x^3-1=0\end{matrix}\right.\)
+/ x - 1 = 0 <=> x = 1
+/x2 + 5x + 2 =0 <=> (x + \(\frac{5}{2}\))2 - \(\frac{17}{4}\)= 0 <=> (x + \(\frac{5}{2}\))2 = \(\frac{17}{4}\)<=> x + \(\frac{5}{2}\)= \(\pm\)\(\sqrt{\frac{17}{4}}\)
<=> x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)
+/ x3 - 1 = 0 <=.> ( x - 1 )(x2 + x + 1 ) = 0
<=> x = 1
Vậy phương trình có Nghiệm là x = 1 và x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)
d,
x2 + (x + 3)(10 -2x ) = 9
<=> x2 + 10x - 2x2 + 30 - 6x -9 = 0
<=> x2 + 4x + 21 = 0
<=> 7x - x2 + 21 -3x = 0
<=> (x +3)(7-x) =0
<=> \(\left[\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.\) <=> \(\left[\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy pt có nghiệm là x = -3 và x = 7
a: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)\)
b: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3 hoặc x=2