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\(a.x^3-2x^2-x+2=0\\ \Leftrightarrow x^3-x-2x^2+2=0\\ \Leftrightarrow x\left(x^2-1\right)-2\left(x^2-1\right)=0\\\Leftrightarrow \left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-2=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-1;1;2\right\}\)
\(b.x^4+x^2+6x-8=0\\ \Leftrightarrow x^4+2x^2-x^2+6x+1-9=0\\ \Leftrightarrow x^4+2x^2+1-x^2+6x-9=0\\\Leftrightarrow \left(x^2+1\right)^2-\left(x-3\right)^2=0\\\Leftrightarrow \left(x^2+1-x+3\right)\left(x^2+1+x-3\right)=0\\\Leftrightarrow \left(x^2-x+4\right)\left(x^2+x-2\right)=0\\\Leftrightarrow \left(x^2-x+\frac{1}{4}+\frac{15}{4}\right)\left(x^2-x+2x-2\right)=0\\\Leftrightarrow \left[\left(x-\frac{1}{2}\right)^2+\frac{15}{4}\right]\left(x+2\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-1\right)=0\left(Vi\left(x-\frac{1}{2}\right)^2+\frac{15}{4}\ne0\right)\\\Leftrightarrow \left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2;1\right\}\)
1) \(x^4-2x^2-144x+1295=0\)
\(\Rightarrow\)Cậu xem lại đề thử xem nhé !
2) \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-1\right)-24=0\)
\(\Leftrightarrow x^4+2x^3-x^2-2x-24=0\)
\(\Leftrightarrow x^4+x^3+4x^2+x^3+x^2+4x-6x^2-6x-24=0\)
\(\Leftrightarrow x^2\left(x^2+x+4\right)+x\left(x^2+x+4\right)-6\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\)\(x+3=0\)
hoặc \(x-2=0\)
hoặc \(x^2+x+4=0\)
\(\Leftrightarrow\)\(x=-3\left(tm\right)\)
hoặc \(x=2\left(tm\right)\)
hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)
3) \(x^4-2x^3+4x^2-3x-10=0\)
\(\Leftrightarrow x^4+x^3-3x^3-3x^2+7x^2+7x-10x-10=0\)
\(\Leftrightarrow x^3\left(x+1\right)-3x^2\left(x+1\right)+7x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-3x^2+7x-10\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-2x^2-x^2+2x+5x-10\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x-2\right)-x\left(x-2\right)+5\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)=0\)
\(\Leftrightarrow\)\(x+1=0\)
hoặc \(x-2=0\)
hoặc \(x^2-x+5=0\)
\(\Leftrightarrow x=-1\left(tm\right)\)
hoặc \(x=2\left(tm\right)\)
hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là :\(S=\left\{-1;2\right\}\)
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
Vậy \(x=5\)hoặc \(x=1\)
\(b,x^2-2x=24\)
\(\Leftrightarrow x^2-2x+1-1=24\)
\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)
\(\Leftrightarrow x-1=\pm5\)
\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
Vậy \(x=6\) hoặc \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
Bài 1:
a) (5x-4)(4x+6)=0
\(\Leftrightarrow\orbr{\begin{cases}5x-4=0\\4x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=4\\4x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{5}\\y=\frac{-3}{2}\end{cases}}}\)
b) (x-5)(3-2x)(3x+4)=0
<=> x-5=0 hoặc 3-2x=0 hoặc 3x+4=0
<=> x=5 hoặc x=\(\frac{3}{2}\)hoặc x=\(\frac{-4}{3}\)
c) (2x+1)(x2+2)=0
=> 2x+1=0 (vì x2+2>0)
=> x=\(\frac{-1}{2}\)
bài 1:
a) (5x - 4)(4x + 6) = 0
<=> 5x - 4 = 0 hoặc 4x + 6 = 0
<=> 5x = 0 + 4 hoặc 4x = 0 - 6
<=> 5x = 4 hoặc 4x = -6
<=> x = 4/5 hoặc x = -6/4 = -3/2
b) (x - 5)(3 - 2x)(3x + 4) = 0
<=> x - 5 = 0 hoặc 3 - 2x = 0 hoặc 3x + 4 = 0
<=> x = 0 + 5 hoặc -2x = 0 - 3 hoặc 3x = 0 - 4
<=> x = 5 hoặc -2x = -3 hoặc 3x = -4
<=> x = 5 hoặc x = 3/2 hoặc x = 4/3
c) (2x + 1)(x^2 + 2) = 0
vì x^2 + 2 > 0 nên:
<=> 2x + 1 = 0
<=> 2x = 0 - 1
<=> 2x = -1
<=> x = -1/2
bài 2:
a) (2x + 7)^2 = 9(x + 2)^2
<=> 4x^2 + 28x + 49 = 9x^2 + 36x + 36
<=> 4x^2 + 28x + 49 - 9x^2 - 36x - 36 = 0
<=> -5x^2 - 8x + 13 = 0
<=> (-5x - 13)(x - 1) = 0
<=> 5x + 13 = 0 hoặc x - 1 = 0
<=> 5x = 0 - 13 hoặc x = 0 + 1
<=> 5x = -13 hoặc x = 1
<=> x = -13/5 hoặc x = 1
b) (x^2 - 1)(x + 2)(x - 3) = (x - 1)(x^2 - 4)(x + 5)
<=> x^4 - x^3 - 7x^2 + x + 6 = x^4 + 4x^3 - 9x^2 - 16x + 20
<=> x^4 - x^3 - 7x^2 + x + 6 - x^4 - 4x^3 + 9x^2 + 16x - 20 = 0
<=> -5x^3 - 2x^2 + 17x - 14 = 0
<=> (-x + 1)(x + 2)(5x - 7) = 0
<=> x - 1 = 0 hoặc x + 2 = 0 hoặc 5x - 7 = 0
<=> x = 0 + 1 hoặc x = 0 - 2 hoặc 5x = 0 + 7
<=> x = 1 hoặc x = -2 hoặc 5x = 7
<=> x = 1 hoặc x = -2 hoặc x = 7/5
\(a,\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
\(9x^2-3x-6x+2=9x^2+6x+1\)
\(-9x+2-6x-1=0\)
\(-15x+1=0\)
\(-15x=-1\)
\(x=\frac{1}{15}\)
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:
\(3x^2+7x-20=0\)
\(\Leftrightarrow 3x^2+12x-5x-20=0\)
\(\Leftrightarrow 3x(x+4)-5(x+4)=0\)
\(\Leftrightarrow (3x-5)(x+4)=0 \Rightarrow \left[\begin{matrix} x=\frac{5}{3}\\ x=-4\end{matrix}\right.\)
a) Đặt x2 + 3x + 2 = a
<=> a(a + 1) - 2 = 0
<=> a2 + a - 2 = 0
<=> a2 + 2a - a - 2 = 0
<=> (a - 1)(a + 2) = 0
<=> \(\left[{}\begin{matrix}a-1=0\\a-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x^2+3x+2-1=0\\x^2+3x+2-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left(x^2+3x+\frac{9}{4}\right)=\frac{5}{4}\\x\left(x+3\right)=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left(x+\frac{3}{2}\right)^2=\frac{5}{4}\\\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x+\frac{3}{2}=\sqrt{\frac{5}{4}}\\x+\frac{3}{2}=-\sqrt{\frac{5}{4}}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{\sqrt{5}-3}{2}\\x=\frac{-\sqrt{5}-3}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy S = {\(\frac{\sqrt{5}-3}{2}\); \(\frac{-\sqrt{5}-3}{2}\); 0; 3}
b) Đặt x2 + x = b
<=> (b - 2)(b - 3) = 12
<=> n2 - 3b - 2b + 6 - 12 = 0
<=> b2 - 5b - 6 = 0
<=> b2 - 6b + b - 6 = 0
<=> (b - 6)(b + 1) = 0
<=> \(\left[{}\begin{matrix}b-6=0\\b+1=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+1=0\left(vn\right)\end{matrix}\right.\)
<=> x2 + 3x - 2x - 6 = 0
<=> (x + 3)(x - 2) = 0
<=> x = -3 hoặc x = 2
Vậy S = {-3; 2}
c) x(x + 1)(x - 1)(x + 2) = 24
<=> (x2 + x)(x2 + x - 2) - 24 = 0
Đặt x2 + x = t
<=> t(t - 2) - 24 = 0
<=> t2 - 2t - 24 = 0
<=> t2 - 6t + 4t - 24 = 0
<=> (t + 4)(t - 6) = 0
<=> \(\left[{}\begin{matrix}t+4=0\\t-6=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x^2+x+4=0\left(vn\right)\\x^2+x-6=0\end{matrix}\right.\)(* vn là vô nghiệm)
<=> x2 + 3x - 2x - 6 = 0
<=> (x + 3)(x - 2) = 0
<=> x = -3 hoặc x = 2
Vậy S = {-3; 2}
d) (x + 1)(x + 2)(x + 3)(x + 4) - 24 = 0
<=> (x2 + 5x +4)(x2 + 5x + 6) - 24 = 0
Đặt x2 + 5x = y
<=> (y + 4)(y + 6) - 24 = 0
<=> y2 + 10y + 24 - 24 = 0
<=> y(y + 10) = 0
<=> \(\left[{}\begin{matrix}y=0\\y+10=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\left(vn\right)\end{matrix}\right.\)
<=> x(x + 5) = 0
<=> x= 0 hoặc x = -5
Vậy S = {0; -5}