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Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)
\(4x^2-40x+100=18x+9\)
\(4x^2-58x+91=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)
\(c,x^3+3x^2-6x-8=0\)
\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)
\(Th1:x+4=0\Leftrightarrow x=-4\)
\(Th2:x-2=0\Leftrightarrow x=2\)
\(Th3:x+1=0\Leftrightarrow x=-1\)
\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)
\(< =>4x^2-40x+100=18x+9\)
\(< =>4x^2+58x+91=0\)
\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
a, (3x+1)(7x+3)=(5x-7)(3x+1)
<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0
<=> (3x+1)(7x+3-5x+7)=0
<=> (3x+1)(2x+10)=0
<=> 2(3x+1)(x+5)=0
=> 3x+1=0 hoặc x+5=0
=> x= -1/3 hoặc x=-5
Vậy...
a) (3x - 2)(4x + 5) = 0
⇔ 3x - 2 = 0 hoặc 4x + 5 = 0
1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3
2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4
Vậy phương trình có tập nghiệm S = {2/3;−5/4}
b) (2,3x - 6,9)(0,1x + 2) = 0
⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3
2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.
Vậy phương trình có tập hợp nghiệm S = {3;-20}
c) (4x + 2)(x2 + 1) = 0 ⇔ 4x + 2 = 0 hoặc x2 + 1 = 0
1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2
2) x2 + 1 = 0 ⇔ x2 = -1 (vô lí vì x2 ≥ 0)
Vậy phương trình có tập hợp nghiệm S = {−1/2}
d) (2x + 7)(x - 5)(5x + 1) = 0
⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2
2) x - 5 = 0 ⇔ x = 5
3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5
Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
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)
voi x=0 ta thay 0 o phai la no pt
voi x<>0 chia ca 2 ve cho x^2 ta dc
x^2-3x+6-3/x+1/x^2=0
(x^2+1/x^2)-3(x+1/x)+6=0 dat a=x+1/x ta co (x+1/x)^2=a^2=>x^2+1/x^2=a^2-2
=>a^2-3a+4=0=>pt vo no :(
